Motion in a Circle Formulae Need to recall (i) = 2 T : angular velocity T: period (ii) v = r v : linear velocity r : radius of circle : angular velocity (iii) a= 2 v r 2 a : centripetal acceleration ora= r (iv) F= 2 m v r 2 F: centripetal force acting on a body in circular motion orF= mr m: mass of bo dy in circu lar motion Definitions 1. The radian is the angle subtended by an arc length equal to the radius of the circl e. 2. Angular displacement is the angle turned about the centre of the circle. 3. Angular velocity ω is the rate of change of angula r displacement. (units : rad s −1 ) 4. Centripetal forceis the resultant force acting on an object in uniform circular motion and is directed towards the centre of the circle. Sample Explanation Questions 1. What is meant by uniform circular motion? Uniform circular motion is the motion of an object moving in circular path at constant speed. 2. Why is there no work done on an object moving in uniform circular motion? There is no work done on the object, as there is no displacement in the direction of the resultant force (resultant force is always perpendicular to the velocity). 3. Explain why there is no change in kinetic energy of the object moving in uniform circular motion even though there is a resultant force acting on it. In uniform circular motion, the resultant force is always perpendicular to the velocity. This changes the direction of motion, but the speed is constant. Hence no change in the kinetic energy of the object. 1 a isfm a so o @ g m a il. co m Ce llNo . 03467030039
F centripetal force acting on a body in circular motion
or F = mr m mass of body in circular motion
Definitions
1 The radian is the angle subtended by an arc length equal to the radius of the circle
2 Angular displacement is the angle turned about the centre of the circle
3 Angular velocity ω is the rate of change of angular displacement (units rad sminus1)
4 Centripetal force is the resultant force acting on an object in uniform circular motion and isdirected towards the centre of the circle
Sample Explanation Questions
1 What is meant by uniform circular motion
Uniform circular motion is the motion of an object moving in circular path at constant speed
2 Why is there no work done on an object moving in uniform circular motion
There is no work done on the object as there is no displacement in the direction of theresultant force (resultant force is always perpendicular to the velocity)
3 Explain why there is no change in kinetic energy of the object moving in uniformcircular motion even though there is a resultant force acting on it
In uniform circular motion the resultant force is always perpendicular to the velocity Thischanges the direction of motion but the speed is constant Hence no change in the kineticenergy of the object
1 A gravitational field is a region of space where a mass will experience a gravitational forcewhen placed in that field
2 Gravitational field strength at a point is defined as the force per unit mass acting on asmall mass placed at that point
3 Newtonrsquos law of gravitation states that the gravitational force of attraction between twopoint masses is proportional to the product of the masses and inversely proportional to thesquare of their separation
4 The gravitational potential at a point is defined as the work done per unit mass (by anexternal agent) in bringing a small mass from infinity to that point
5 Gravitational potential energy of an object at a point is defined as the work done (by anexternal agent) in bringing the object from infinity to that point
6 Geostationary orbits are orbits of satellites orbiting around the Earth such that these
satellites would appear stationary when observed from the Earth
FormulaeNeed to recall (negative sign in the equations signify attractive nature)
(i) g =m
F g gravitational field strength at a point
F = mg F gravitational forcem mass of the body
1 2
2
Gmm
r(ii) F = G gravitational constantm1 m2 mass of 2 bodiesr separation of the 2 bodies
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times106 m Its mass isassumed to be concentrated at its centre
Given that the gravitational field strength at the Earthrsquos surface is 981N kgndash1 show thatthe mass of the Earth is 599times1024 kg
[2]
(b) A satellite is placed in geostationary orbit around the Earth
(i) Calculate the angular speed of the satellite in its orbit
angular speed = rad sndash1 [3]
(ii) Using the data in (a) determine the radius of the orbit
radius = m [3]
4 If an object is projected vertically upwards from the surface of a planet at a fast enoughspeed it can escape the planetrsquos gravitational field This means that the object can arrive atinfinity where it has zero kinetic energy The speed that is just enough for this to happen isknown as the escape speed
(a) (i) By equating the kinetic energy of the object at the planetrsquos surface to its total gainof potential energy in going to infinity show that the escape speed v is given by
v 2 =
where R is the radius of the planet and M is its mass
where g is the acceleration of free fall at the planetrsquos surface
[3]
1 (a) Define gravitational potential
[2]
(b) Explain why values of gravitational potential near to an isolated mass are all negative
[3]
(c) The Earth may be assumed to be an isolated sphere of radius 64 times103 km with its massof 60times1024 kg concentrated at its centre An object is projected vertically from thesurface of the Earth so that it reaches an altitude of 13 times104 km
Calculate for this object
(i) the change in gravitational potential
change in potential = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip Jkgndash1
3 A binary star consists of two stars that orbit about a fixed point C as shown in Fig 31
Fig 31
The star of mass M 1 has a circular orbit of radius R 1 and the star of mass M 2 has a circularorbit of radius R 2 Both stars have the same angular speed ω about C
(a) State the formula in terms of G M 1 M 2 R 1 R 2 and ω for
(i) the gravitational force between the two stars
(ii) the centripetal force on the star of mass M 1
2 An α-particle (42He) is moving directly towards a stationary gold nucleus (19779Au)
The α-particle and the gold nucleus may be considered to be solid spheres with the chargeand mass concentrated at the centre of each sphere
When the two spheres are just touching the separation of their centres is 96 x 10ndash15m
(a) The α-particle and the gold nucleus may be assumed to be an isolated systemCalculate for the α-particle just in contact with the gold nucleus
(i) its gravitational potential energy
gravitational potential energy = J [3]
(ii) its electric potential energy
electric potential energy = J [3]
(b) Using your answers in (a) suggest why when making calculations based on anα-particle scattering experiment gravitational effects are not considered
[1]
(c) In the α-particle scattering experiment conducted in 1913 the maximum kinetic energyof the available α-particles was about 6 MeV Suggest why in this experiment the radiusof the target nucleus could not be determined
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 A gravitational field is a region of space where a mass will experience a gravitational forcewhen placed in that field
2 Gravitational field strength at a point is defined as the force per unit mass acting on asmall mass placed at that point
3 Newtonrsquos law of gravitation states that the gravitational force of attraction between twopoint masses is proportional to the product of the masses and inversely proportional to thesquare of their separation
4 The gravitational potential at a point is defined as the work done per unit mass (by anexternal agent) in bringing a small mass from infinity to that point
5 Gravitational potential energy of an object at a point is defined as the work done (by anexternal agent) in bringing the object from infinity to that point
6 Geostationary orbits are orbits of satellites orbiting around the Earth such that these
satellites would appear stationary when observed from the Earth
FormulaeNeed to recall (negative sign in the equations signify attractive nature)
(i) g =m
F g gravitational field strength at a point
F = mg F gravitational forcem mass of the body
1 2
2
Gmm
r(ii) F = G gravitational constantm1 m2 mass of 2 bodiesr separation of the 2 bodies
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times106 m Its mass isassumed to be concentrated at its centre
Given that the gravitational field strength at the Earthrsquos surface is 981N kgndash1 show thatthe mass of the Earth is 599times1024 kg
[2]
(b) A satellite is placed in geostationary orbit around the Earth
(i) Calculate the angular speed of the satellite in its orbit
angular speed = rad sndash1 [3]
(ii) Using the data in (a) determine the radius of the orbit
radius = m [3]
4 If an object is projected vertically upwards from the surface of a planet at a fast enoughspeed it can escape the planetrsquos gravitational field This means that the object can arrive atinfinity where it has zero kinetic energy The speed that is just enough for this to happen isknown as the escape speed
(a) (i) By equating the kinetic energy of the object at the planetrsquos surface to its total gainof potential energy in going to infinity show that the escape speed v is given by
v 2 =
where R is the radius of the planet and M is its mass
where g is the acceleration of free fall at the planetrsquos surface
[3]
1 (a) Define gravitational potential
[2]
(b) Explain why values of gravitational potential near to an isolated mass are all negative
[3]
(c) The Earth may be assumed to be an isolated sphere of radius 64 times103 km with its massof 60times1024 kg concentrated at its centre An object is projected vertically from thesurface of the Earth so that it reaches an altitude of 13 times104 km
Calculate for this object
(i) the change in gravitational potential
change in potential = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip Jkgndash1
3 A binary star consists of two stars that orbit about a fixed point C as shown in Fig 31
Fig 31
The star of mass M 1 has a circular orbit of radius R 1 and the star of mass M 2 has a circularorbit of radius R 2 Both stars have the same angular speed ω about C
(a) State the formula in terms of G M 1 M 2 R 1 R 2 and ω for
(i) the gravitational force between the two stars
(ii) the centripetal force on the star of mass M 1
2 An α-particle (42He) is moving directly towards a stationary gold nucleus (19779Au)
The α-particle and the gold nucleus may be considered to be solid spheres with the chargeand mass concentrated at the centre of each sphere
When the two spheres are just touching the separation of their centres is 96 x 10ndash15m
(a) The α-particle and the gold nucleus may be assumed to be an isolated systemCalculate for the α-particle just in contact with the gold nucleus
(i) its gravitational potential energy
gravitational potential energy = J [3]
(ii) its electric potential energy
electric potential energy = J [3]
(b) Using your answers in (a) suggest why when making calculations based on anα-particle scattering experiment gravitational effects are not considered
[1]
(c) In the α-particle scattering experiment conducted in 1913 the maximum kinetic energyof the available α-particles was about 6 MeV Suggest why in this experiment the radiusof the target nucleus could not be determined
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times106 m Its mass isassumed to be concentrated at its centre
Given that the gravitational field strength at the Earthrsquos surface is 981N kgndash1 show thatthe mass of the Earth is 599times1024 kg
[2]
(b) A satellite is placed in geostationary orbit around the Earth
(i) Calculate the angular speed of the satellite in its orbit
angular speed = rad sndash1 [3]
(ii) Using the data in (a) determine the radius of the orbit
radius = m [3]
4 If an object is projected vertically upwards from the surface of a planet at a fast enoughspeed it can escape the planetrsquos gravitational field This means that the object can arrive atinfinity where it has zero kinetic energy The speed that is just enough for this to happen isknown as the escape speed
(a) (i) By equating the kinetic energy of the object at the planetrsquos surface to its total gainof potential energy in going to infinity show that the escape speed v is given by
v 2 =
where R is the radius of the planet and M is its mass
where g is the acceleration of free fall at the planetrsquos surface
[3]
1 (a) Define gravitational potential
[2]
(b) Explain why values of gravitational potential near to an isolated mass are all negative
[3]
(c) The Earth may be assumed to be an isolated sphere of radius 64 times103 km with its massof 60times1024 kg concentrated at its centre An object is projected vertically from thesurface of the Earth so that it reaches an altitude of 13 times104 km
Calculate for this object
(i) the change in gravitational potential
change in potential = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip Jkgndash1
3 A binary star consists of two stars that orbit about a fixed point C as shown in Fig 31
Fig 31
The star of mass M 1 has a circular orbit of radius R 1 and the star of mass M 2 has a circularorbit of radius R 2 Both stars have the same angular speed ω about C
(a) State the formula in terms of G M 1 M 2 R 1 R 2 and ω for
(i) the gravitational force between the two stars
(ii) the centripetal force on the star of mass M 1
2 An α-particle (42He) is moving directly towards a stationary gold nucleus (19779Au)
The α-particle and the gold nucleus may be considered to be solid spheres with the chargeand mass concentrated at the centre of each sphere
When the two spheres are just touching the separation of their centres is 96 x 10ndash15m
(a) The α-particle and the gold nucleus may be assumed to be an isolated systemCalculate for the α-particle just in contact with the gold nucleus
(i) its gravitational potential energy
gravitational potential energy = J [3]
(ii) its electric potential energy
electric potential energy = J [3]
(b) Using your answers in (a) suggest why when making calculations based on anα-particle scattering experiment gravitational effects are not considered
[1]
(c) In the α-particle scattering experiment conducted in 1913 the maximum kinetic energyof the available α-particles was about 6 MeV Suggest why in this experiment the radiusof the target nucleus could not be determined
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
where g is the acceleration of free fall at the planetrsquos surface
[3]
1 (a) Define gravitational potential
[2]
(b) Explain why values of gravitational potential near to an isolated mass are all negative
[3]
(c) The Earth may be assumed to be an isolated sphere of radius 64 times103 km with its massof 60times1024 kg concentrated at its centre An object is projected vertically from thesurface of the Earth so that it reaches an altitude of 13 times104 km
Calculate for this object
(i) the change in gravitational potential
change in potential = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip Jkgndash1
3 A binary star consists of two stars that orbit about a fixed point C as shown in Fig 31
Fig 31
The star of mass M 1 has a circular orbit of radius R 1 and the star of mass M 2 has a circularorbit of radius R 2 Both stars have the same angular speed ω about C
(a) State the formula in terms of G M 1 M 2 R 1 R 2 and ω for
(i) the gravitational force between the two stars
(ii) the centripetal force on the star of mass M 1
2 An α-particle (42He) is moving directly towards a stationary gold nucleus (19779Au)
The α-particle and the gold nucleus may be considered to be solid spheres with the chargeand mass concentrated at the centre of each sphere
When the two spheres are just touching the separation of their centres is 96 x 10ndash15m
(a) The α-particle and the gold nucleus may be assumed to be an isolated systemCalculate for the α-particle just in contact with the gold nucleus
(i) its gravitational potential energy
gravitational potential energy = J [3]
(ii) its electric potential energy
electric potential energy = J [3]
(b) Using your answers in (a) suggest why when making calculations based on anα-particle scattering experiment gravitational effects are not considered
[1]
(c) In the α-particle scattering experiment conducted in 1913 the maximum kinetic energyof the available α-particles was about 6 MeV Suggest why in this experiment the radiusof the target nucleus could not be determined
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
3 A binary star consists of two stars that orbit about a fixed point C as shown in Fig 31
Fig 31
The star of mass M 1 has a circular orbit of radius R 1 and the star of mass M 2 has a circularorbit of radius R 2 Both stars have the same angular speed ω about C
(a) State the formula in terms of G M 1 M 2 R 1 R 2 and ω for
(i) the gravitational force between the two stars
(ii) the centripetal force on the star of mass M 1
2 An α-particle (42He) is moving directly towards a stationary gold nucleus (19779Au)
The α-particle and the gold nucleus may be considered to be solid spheres with the chargeand mass concentrated at the centre of each sphere
When the two spheres are just touching the separation of their centres is 96 x 10ndash15m
(a) The α-particle and the gold nucleus may be assumed to be an isolated systemCalculate for the α-particle just in contact with the gold nucleus
(i) its gravitational potential energy
gravitational potential energy = J [3]
(ii) its electric potential energy
electric potential energy = J [3]
(b) Using your answers in (a) suggest why when making calculations based on anα-particle scattering experiment gravitational effects are not considered
[1]
(c) In the α-particle scattering experiment conducted in 1913 the maximum kinetic energyof the available α-particles was about 6 MeV Suggest why in this experiment the radiusof the target nucleus could not be determined
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
3 A binary star consists of two stars that orbit about a fixed point C as shown in Fig 31
Fig 31
The star of mass M 1 has a circular orbit of radius R 1 and the star of mass M 2 has a circularorbit of radius R 2 Both stars have the same angular speed ω about C
(a) State the formula in terms of G M 1 M 2 R 1 R 2 and ω for
(i) the gravitational force between the two stars
(ii) the centripetal force on the star of mass M 1
2 An α-particle (42He) is moving directly towards a stationary gold nucleus (19779Au)
The α-particle and the gold nucleus may be considered to be solid spheres with the chargeand mass concentrated at the centre of each sphere
When the two spheres are just touching the separation of their centres is 96 x 10ndash15m
(a) The α-particle and the gold nucleus may be assumed to be an isolated systemCalculate for the α-particle just in contact with the gold nucleus
(i) its gravitational potential energy
gravitational potential energy = J [3]
(ii) its electric potential energy
electric potential energy = J [3]
(b) Using your answers in (a) suggest why when making calculations based on anα-particle scattering experiment gravitational effects are not considered
[1]
(c) In the α-particle scattering experiment conducted in 1913 the maximum kinetic energyof the available α-particles was about 6 MeV Suggest why in this experiment the radiusof the target nucleus could not be determined
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
3 A binary star consists of two stars that orbit about a fixed point C as shown in Fig 31
Fig 31
The star of mass M 1 has a circular orbit of radius R 1 and the star of mass M 2 has a circularorbit of radius R 2 Both stars have the same angular speed ω about C
(a) State the formula in terms of G M 1 M 2 R 1 R 2 and ω for
(i) the gravitational force between the two stars
(ii) the centripetal force on the star of mass M 1
2 An α-particle (42He) is moving directly towards a stationary gold nucleus (19779Au)
The α-particle and the gold nucleus may be considered to be solid spheres with the chargeand mass concentrated at the centre of each sphere
When the two spheres are just touching the separation of their centres is 96 x 10ndash15m
(a) The α-particle and the gold nucleus may be assumed to be an isolated systemCalculate for the α-particle just in contact with the gold nucleus
(i) its gravitational potential energy
gravitational potential energy = J [3]
(ii) its electric potential energy
electric potential energy = J [3]
(b) Using your answers in (a) suggest why when making calculations based on anα-particle scattering experiment gravitational effects are not considered
[1]
(c) In the α-particle scattering experiment conducted in 1913 the maximum kinetic energyof the available α-particles was about 6 MeV Suggest why in this experiment the radiusof the target nucleus could not be determined
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
2 An α-particle (42He) is moving directly towards a stationary gold nucleus (19779Au)
The α-particle and the gold nucleus may be considered to be solid spheres with the chargeand mass concentrated at the centre of each sphere
When the two spheres are just touching the separation of their centres is 96 x 10ndash15m
(a) The α-particle and the gold nucleus may be assumed to be an isolated systemCalculate for the α-particle just in contact with the gold nucleus
(i) its gravitational potential energy
gravitational potential energy = J [3]
(ii) its electric potential energy
electric potential energy = J [3]
(b) Using your answers in (a) suggest why when making calculations based on anα-particle scattering experiment gravitational effects are not considered
[1]
(c) In the α-particle scattering experiment conducted in 1913 the maximum kinetic energyof the available α-particles was about 6 MeV Suggest why in this experiment the radiusof the target nucleus could not be determined
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
2 An α-particle (42He) is moving directly towards a stationary gold nucleus (19779Au)
The α-particle and the gold nucleus may be considered to be solid spheres with the chargeand mass concentrated at the centre of each sphere
When the two spheres are just touching the separation of their centres is 96 x 10ndash15m
(a) The α-particle and the gold nucleus may be assumed to be an isolated systemCalculate for the α-particle just in contact with the gold nucleus
(i) its gravitational potential energy
gravitational potential energy = J [3]
(ii) its electric potential energy
electric potential energy = J [3]
(b) Using your answers in (a) suggest why when making calculations based on anα-particle scattering experiment gravitational effects are not considered
[1]
(c) In the α-particle scattering experiment conducted in 1913 the maximum kinetic energyof the available α-particles was about 6 MeV Suggest why in this experiment the radiusof the target nucleus could not be determined
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
2 An α-particle (42He) is moving directly towards a stationary gold nucleus (19779Au)
The α-particle and the gold nucleus may be considered to be solid spheres with the chargeand mass concentrated at the centre of each sphere
When the two spheres are just touching the separation of their centres is 96 x 10ndash15m
(a) The α-particle and the gold nucleus may be assumed to be an isolated systemCalculate for the α-particle just in contact with the gold nucleus
(i) its gravitational potential energy
gravitational potential energy = J [3]
(ii) its electric potential energy
electric potential energy = J [3]
(b) Using your answers in (a) suggest why when making calculations based on anα-particle scattering experiment gravitational effects are not considered
[1]
(c) In the α-particle scattering experiment conducted in 1913 the maximum kinetic energyof the available α-particles was about 6 MeV Suggest why in this experiment the radiusof the target nucleus could not be determined
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 The orbit of the Earth mass 60 times 1024 kg may be assumed to be a circle of radius15 times 1011m with the Sun at its centre as illustrated in Fig 11
Fig 11
The time taken for one orbit is 32times
10
7
s
(a) Calculate
(i) the magnitude of the angular velocity of the Earth about the Sun
angular velocity = rads ndash1 [2]
(ii) the magnitude of the centripetal force acting on the Earth
force = N [2]
Earthmass 60 x 1024 kg
Sun
15 x 1011 m
(b) (i) State the origin of the centripetal force calculated in (a)(ii)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 The Earth may be considered to be a sphere of radius 64 times106 m with its mass of60 times1024 kg concentrated at its centreA satellite of mass 650kg is to be launched from the Equator and put into geostationaryorbit
(a) Show that the radius of the geostationary orbit is 42 times107 m
[3]
(b) Determine the increase in gravitational potential energy of the satellite during its launchfrom the Earthrsquos surface to the geostationary orbit
energy = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip J [4]
(c) Suggest one advantage of launching satellites from the Equator in the direction ofrotation of the Earth
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
4 A rocket is launched from the surface of the Earth
Fig 41 gives data for the speed of the rocket at two heights above the Earthrsquos surface afterthe rocket engine has been switched off
Fig 41
The Earth may be assumed to be a uniform sphere of radius R = 638 times106 m with its massM concentrated at its centre The rocket after the engine has been switched off hasmass m
(a) Write down an expression in terms of
(i) G M m h 1 h 2 and R for the change in gravitational potential energy of the rocket
[1]
(ii) m v 1and v 2 for the change in kinetic energy of the rocket
[1]
(b) Using the expressions in (a) determine a value for the mass M of the Earth
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [3]
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) Explain what is meant by a gravitational field
[1]
(b) A spherical planet has mass M and radius R The planet may be considered to have allits mass concentrated at its centre A rocket is launched from the surface of the planet such that the rocket moves radially
away from the planet The rocket engines are stopped when the rocket is at a height R above the surface of the planet as shown in Fig 11
R
R 2R
planet
Fig 11
The mass of the rocket after its engines have been stopped is m
(i) Show that for the rocket to travel from a height R to a height 2R above the planetrsquossurface the change ΔE P in the magnitude of the gravitational potential energy ofthe rocket is given by the expression
ΔE P =GMm
6R
[2]
(ii) During the ascent from a height R to a height 2R the speed of the rocket changesfrom 7600 m sndash1 to 7320 m sndash1 Show that in SI units the change ΔE K in the kineticenergy of the rocket is given by the expression
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(i) Use the expressions in (b) to determine a value for the mass M of the planet
M = helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip kg [2]
(ii) State one assumption made in the determination in (i)
[1]
1 (a) Explain
(i) what is meant by a radian
[2]
(ii) why one complete revolution is equivalent to an angular displacement of 2π rad
[1]
(b) An elastic cord has an unextended length of 130 cm One end of the cord is attached to
a fixed point C A small mass of weight 50 N is hung from the free end of the cord Thecord extends to a length of 148 cm as shown in Fig 11
148cm
C
smallmass
Fig 11
The cord and mass are now made to rotate at constant angular speed ω in a verticalplane about point C When the cord is vertical and above C its length is the unextendedlength of 130 cm as shown in Fig 12
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(ii) A small mass is attached to a string The mass is rotating about a fixed point P atconstant speed as shown in Fig 11
mass rotatingat constant speed
P
Fig 11
Explain what is meant by the angular speed about point P of the mass
[2]
(b) A horizontal flat plate is free to rotate about a vertical axis through its centre as shownin Fig 12
M
d
plate
Fig 12
A small mass M is placed on the plate a distance d from the axis of rotation The speed of rotation of the plate is gradually increased from zero until the mass is
seen to slide off the plate
The maximum frictional force F between the plate and the mass is given by the
expression
F = 072W
where W is the weight of the mass M The distance d is 35 cm
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
Determine the maximum number of revolutions of the plate per minute for the mass M toremain on the plate Explain your working
number = [5]
(c) The plate in (b) is covered when stationary with mud
Suggest and explain whether mud near the edge of the plate or near the centre will firstleave the plate as the angular speed of the plate is slowly increased
[2]
1 A spherical planet has mass M and radius R The planet may be assumed to be isolated in space and to have its mass concentrated at itscentreThe planet spins on its axis with angular speed ω as illustrated in Fig 11
R
mass m
pole of
planet
equator of
planet
Fig 11
A small object of mass m rests on the equator of the planet The surface of the planet exertsa normal reaction force on the mass
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(i) the gravitational force between the planet and the object
[1]
(ii) the centripetal force required for circular motion of the small mass
[1]
(iii) the normal reaction exerted by the planet on the mass
[1]
(b) (i) Explain why the normal reaction on the mass will have different values at theequator and at the poles
[2]
(ii) The radius of the planet is 64 times 106 m It completes one revolution in 86 times 104 sCalculate the magnitude of the centripetal acceleration at
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) The Earth may be considered to be a uniform sphere of radius 638 times 103 km with itsmass concentrated at its centre
(i) Define gravitational field strength
[1]
(ii) By considering the gravitational field strength at the surface of the Earth show thatthe mass of the Earth is 599 times 1024 kg
[2]
(b) The Global Positioning System (GPS) is a navigation system that can be used anywhereon Earth It uses a number of satellites that orbit the Earth in circular orbits at a distanceof 222 times 104 km above its surface
(i) Use data from (a) to calculate the angular speed of a GPS satellite in its orbit
angular speed = rad s ndash1 [3]
(ii) Use your answer in (i) to show that the satellites are not in geostationary orbits
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(b) An isolated star has radius R The mass of the star may be considered to be a pointmass at the centre of the star
The gravitational field strength at the surface of the star is g s
On Fig 11 sketch a graph to show the variation of the gravitational field strength of thestar with distance from its centre You should consider distances in the range R to 4R
0R 2R 3R 4R
distance
02g s
04g s
06g s
08g s
gravitational
field strength
surfaceof star
10g s
Fig 11[2]
(c) The Earth and the Moon may be considered to be spheres that are isolated in spacewith their masses concentrated at their centres
The masses of the Earth and the Moon are 600 times 1024 kg and 740 times 1022 kgrespectively
The radius of the Earth is R E and the separation of the centres of the Earth and theMoon is 60R E as illustrated in Fig 12
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(i) Explain why there is a point between the Earth and the Moon at which thegravitational field strength is zero
[2]
(ii) Determine the distance in terms of R E from the centre of the Earth at which thegravitational field strength is zero
distance = R E [3]
(iii) On the axes of Fig 13 sketch a graph to show the variation of the gravitationalfield strength with position between the surface of the Earth and the surface of theMoon
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(c) Two protons are isolated in space Their centres are separated by a distance R Each proton may be considered to be a point mass with point charge Determine the magnitude of the ratio
force between protons due to electric fieldforce between protons due to gravitational field
ratio = [3]
1 (a) Newtonrsquos law of gravitation applies to point masses
(i) State Newtonrsquos law of gravitation
[2]
(ii) Explain why although the planets and the Sun are not point masses the law alsoapplies to planets orbiting the Sun
[1]
(b) Gravitational fields and electric fields show certain similarities and certain differences State one aspect of gravitational and electric fields where there is
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) A moon is in a circular orbit of radius r about a planet The angular speed of the moonin its orbit is ω The planet and its moon may be considered to be point masses that areisolated in space
Show that r and ω are related by the expression
r3ω
2 = constant
Explain your working
[3]
(b) Phobos and Deimos are moons that are in circular orbits about the planet MarsData for Phobos and Deimos are shown in Fig 11
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(ii) The period of rotation of Mars about its axis is 246 hours Deimos is in an equatorial orbit orbiting in the same direction as the spin of Mars
about its axis
Use your answer in (i) to comment on the orbit of Deimos
[1]
1 The planet Mars may be considered to be an isolated sphere of diameter 679 times 106 m withits mass of 642 times 1023 kg concentrated at its centreA rock of mass 140 kg rests on the surface of Mars
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(ii) show that its gravitational potential energy is ndash177 times 107 J
[2]
(b) Use the information in (a)(ii) to determine the speed at which the rock must leave thesurface of Mars so that it will escape the gravitational attraction of the planet
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(b) Gravitational fields and electric fields are two examples of fields of force State one similarity and one difference between these two fields of force
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(b) The gravitational potential φ at distance r from point mass M is given by the expression
φ = ndashGM
r
where G is the gravitational constant
Explain the significance of the negative sign in this expression
[2]
(c) A spherical planet may be assumed to be an isolated point mass with its massconcentrated at its centre A small mass m is moving near to and normal to the surfaceof the planet The mass moves away from the planet through a short distance h
State and explain why the change in gravitational potential energy ΔE P of the mass isgiven by the expression
ΔE P = mgh
where g is the acceleration of free fall
[4]
(d) The planet in (c) has mass M and diameter 68 times 103 km The product GM for this planetis 43 times 1013 N m2 kgndash1
A rock initially at rest a long distance from the planet accelerates towards the planetAssuming that the planet has negligible atmosphere calculate the speed of the rock asit hits the surface of the planet
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(ii) Tidal action on the Earthrsquos surface causes the radius of the orbit of the Moon toincrease by 40 cm each year
Use your answer in (i) to determine the change in one year of the gravitationalpotential energy of the Moon Explain your working
energy change = J [3]
1 (a) State Newtonrsquos law of gravitation
[2]
(b) A satellite of mass m is in a circular orbit of radius r about a planet of mass M For this planet the product GM is 400 times 1014 N m2 kgndash1 where G is the gravitational
constant
The planet may be assumed to be isolated in space
(i) By considering the gravitational force on the satellite and the centripetal forceshow that the kinetic energy E K of the satellite is given by the expression
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) State what is meant by a gravitational field
[2]
(b) In the Solar System the planets may be assumed to be in circular orbits about the Sun Data for the radii of the orbits of the Earth and Jupiter about the Sun are given in
Fig 11
radius of orbit km
EarthJupiter
150 times 108
778 times 108
Fig 11
(i) State Newtonrsquos law of gravitation
[3]
(ii) Use Newtonrsquos law to determine the ratio
gravitational field strength due to the Sun at orbit of Earth
gravitational field strength due to the Sun at orbit of Jupiter
ratio = [3]
(iii) Use your answers in (ii) to explain whether the linear speed of the satellite increasesdecreases or remains unchanged when the radius of the orbit decreases
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) Explain what is meant by a geostationary orbit
[3]
(b) A satellite of mass m is in a circular orbit about a planet The mass M of the planet may be considered to be concentrated at its centre Show that the radius R of the orbit of the satellite is given by the expression
R 3 =
GMT 2
4π2
where T is the period of the orbit of the satellite and G is the gravitational constant Explain your working
[4]
(c) The Earth has mass 60 times 1024 kg Use the expression given in (b) to determine theradius of the geostationary orbit about the Earth
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(b) The Moon may be considered to be an isolated sphere of radius 174 times 103 km with itsmass of 735 times 1022 kg concentrated at its centre
(i) A rock of mass 450 kg is situated on the surface of the Moon Show that the changein gravitational potential energy of the rock in moving it from the Moonrsquos surface toinfinity is 127 times 107 J
[1]
(ii) The escape speed of the rock is the minimum speed that the rock must be givenwhen it is on the Moonrsquos surface so that it can escape to infinity
Use the answer in (i) to determine the escape speed Explain your working
speed = m sndash1 [2]
(c) The Moon in (b) is assumed to be isolated in space The Moon does in fact orbit theEarth
State and explain whether the minimum speed for the rock to reach the Earth from thesurface of the Moon is different from the escape speed calculated in (b)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(b) A star and a planet are isolated in space The planet orbits the star in a circular orbit ofradius R as illustrated in Fig 11
starmass M
R
planet
Fig 11
The angular speed of the planet about the star is ω
By considering the circular motion of the planet about the star of mass M show that ωand R are related by the expression
R3ω
2 = GM
where G is the gravitational constant Explain your working
[3]
(c) The Earth orbits the Sun in a circular orbit of radius 15 times 108 km The mass of the Sunis 20 times 1030 kg
A distant star is found to have a planet that has a circular orbit about the star The radiusof the orbit is 60 times 108 km and the period of the orbit is 20 years
Use the expression in (b) to calculate the mass of the star
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(b) A stone of mass m has gravitational potential energy E P at a point X in a gravitational fieldThe magnitude of the gravitational potential at X is φ
State the relation between m E P and φ
[1]
(c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated atits centre The gravitational potential at the surface of the planet is minus 630 times 107 J kgminus1
A stone of mass 130 kg is travelling towards the planet such that its distance from the centreof the planet changes from 6R to 5R
Calculate the change in gravitational potential energy of the stone
change in energy = J [4]
1 The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet
(a) A stone travelling at speed v is in a circular orbit of radius r about the planet as illustrated inFig 11
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(b) A second stone initially at rest at infinity travels towards the planet as illustrated in Fig 12
x
planet
V 0
stone
Fig 12 (not to scale)
The stone does not hit the surface of the planet
(i) Determine in terms of the gravitational constant G and the mass M of the planet thespeed V 0 of the stone at a distance x from the centre of the planet Explain your working
You may assume that the gravitational attraction on the stone is due only to the planet
[3]
(ii) Use your answer in (i) and the expression in (a) to explain whether this stone could entera circular orbit about the planet
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 An isolated spherical planet has a diameter of 68 times 106 m Its mass of 64 times 1023 kg may beassumed to be a point mass at the centre of the planet
(a) Show that the gravitational field strength at the surface of the planet is 37 N kgminus1
[2]
(b) A stone of mass 24 kg is raised from the surface of the planet through a vertical height of
1800 m Use the value of field strength given in (a) to determine the change in gravitational potential
energy of the stone Explain your working
change in energy = J [3]
(c) A rock initially at rest at infinity moves towards the planet At point P its height above thesurface of the planet is 35D where D is the diameter of the planet as shown in Fig 11
path of
rockP
35D
planet
D
Fig 11
Calculate the speed of the rock at point P assuming that the change in gravitational potentialenergy is all transferred to kinetic energy
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
2 A large bowl is made from part of a hollow sphere
A small spherical ball is placed inside the bowl and is given a horizontal speed The ball follows ahorizontal circular path of constant radius as shown in Fig 21
14cm
ball
Fig 21
The forces acting on the ball are its weight W and the normal reaction force R of the bowl on theball as shown in Fig 22
ballwall of
bowl
W
R
Fig 22
The normal reaction force R is at an angle θ to the horizontal
(a) (i) By resolving the reaction force R into two perpendicular components show that the
resultant force F acting on the ball is given by the expression
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(ii) no difference OR lines closer near surface of smaller sphere B1 [3]
(b) (i) F G = GMmR 2 C1
= (667 X 10-11 x 598 x 1024)(6380 x 103)2
= 980 N A1
(ii) F C = mR ω 2 C1
ω = 2 π T C1
F C = (4π2 x 6380 x 103)864 x 104)2
= 00337 N A1
(iii) F G - F C = 977 N A1 [6]
(c) because acceleration (of free fall) is (resultant) force per unitmass B1acceleration = 977 m s-2 B1 [2]
3 (a) (i) (force) = GM 1M 2(R 1 + R 2)2 B1
(ii) (force) = M 1R 1ω 2 or M 2R 2ω
2 B1 [2]
(b) ω = 2π(126 x 108) or 2πT C1
= 499 x 10-8 rad s-1 A1 [2]
allow 2 sf 159π x 10 -8 scores 12
(c) (i) reference to either taking moments (about C) or same (centripetal)force B1M 1R 1 = M 2R 2 or M 1R 1ω
2 = M 2R 2ω 2 B1
hence M 1M 2 = R 2R 1 A0 [2](ii) R 2 = 34 x 32 x 1011 m = 24 x 1011 m A1
R 1 = (32 x 1011) ndash R 2 = 80 x 1010 m (allow vice versa) A1 [2]if values are both wrong but have ratio of four to three then allow12
(d) (i) M 2 = (R 1 + R 2)2 x R 1 x ω
2 I G (any subject for equation) C1
= (32 x 1011)2 x 80 x 1010 x (499 x 10-8)2(667 x 10-11) C1= 306 x 1029 kg A1
(ii) less massive (only award this mark if reasonable attempt at (i)) B1 [4]
(917 x 1029 kg for more massive star)Total [12]
1 (a) θ (rad) = 2π x (103360) 1
= 0180 rad (nb 3 sig fig) 1 [2]
(b) (i) tan θ = 0182 (nb 3 sig fig) 1
(ii) percentage error = (00020180) x 100 1
= 11 () 1 [3]
(allow 00020182 and allow 1 4 sig fig)
1 (a) (i) angular speed = 2πT C1
= 2π(32 times 107)
= 196 times 10-7 rad s-1 A1 [2]
(ii) force = mr ω 2 or force = mv
2 r and v = r ω C1
= 60 times 1024times 15 times 1011
times (196 times 10-7)2
= 346 times 1022 N A1 [2]
(b) (i) gravitationgravitygravitational field (strength) B1 [1]
(ii) F = GMmx2 or GM = r 3ω 2 C1
346 times 1022 = (667 times 10-11times M times 60 times 1024)(15 times 1011)2 C1
M = 195 times 1030 kg A1 [3]
1 (a) GM R 2 = R ω 2 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
ω = 2π (24 times 3600) helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
667 times 10ndash11times 60 times 1024 = R
3times ω
2
R 3 = 757 times 1022 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip M1
R = 423 times 107 m helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A0 [3]
(b) (i) ∆Φ = GM R e ndash GM R o helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= (667 times 10ndash11times 60 times 1024) ( 1 64 times 106 ndash 1 42 times 107)
= 531 times 107 J kgndash1 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
∆E P = 531 times 107times 650 helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip C1
= 345 times 1010 J helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip A1 [4]
(c) eg satellite will already have some speed in the correct direction hellip B1 [1]
1 (a) centripetal force is provided by gravitational force B1mv
2 r = GMm r 2 B1
hence v = radic(GM r ) A0 [2]
(b) (i) E K (= frac12mv 2) = GMm 2r B1 [1]
(ii) E P = - GMm r B1 [1]
(iii) E T = - GMm r + GMm 2r C1= - GMm 2r A1 [2]
(c) (i) if E T decreases then - GMm 2r becomes more negativeor GMm 2r becomes larger M1so r decreases A1 [2]
(ii) E K = GMm 2r and r decreases M1so (E K and) v increases A1 [2]
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
(iii) reaction force = GMm R 2 ndash mR ω 2 (allow ecf) B1 [1]
(b) (i) either value of R in expression R ω 2 varies
or mR ω 2 no longer parallel to GMm R 2 normal to surface B1
becomes smaller as object approaches a pole is zero at pole B1 [2]
(ii) 1 acceleration = 64 times 106
times (2π 86 times 104
)2
C1= 0034 m sndash2 A1 [2]
2 acceleration = 0 A1 [1]
(c) eg lsquoradiusrsquo of planet varies density of planet not constantplanet spinningnearby planets stars(any sensible comments 1 mark each maximum 2) B2 [2]
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) g = GM R 2 C186 times (06 times 107)2 = M times 667 times 10ndash11 C1M = 46 times 1024 kg A1 [3]
(c) (i) either potential decreases as distance from planet decreasesor potential zero at infinity and X is closer to zeroor potential α ndash1r and Y more negative M1so point Y is closer to planet A1 [2]
(ii) idea of ∆φ = frac12v 2 C1(68 ndash 53) times 107 = frac12v 2
v = 55 times 103 msndash1 A1 [2]
Section A
1 (a) F prop Mm R 2 helliphellip(words or explained symbols) M1either M and m are point massesor R gtgt diameter of masses hellip(do not allow lsquosizersquo ) A1 [2]
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) (i) weight = GMmr 2 C1= (667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106)2 C1= 520 N A1 [3]
(ii) potential energy = ndashGMmr C1= ndash(667 times 10ndash11 times 642 times 1023 times 140)(frac12 times 679 times 106) M1= ndash177 times 107 J A0 [2]
(b) either frac12mv 2 = 177 times 107 C1
v 2 = (177 times 107 times 2)140 C1
v = 503 times 103 msndash1 A1or frac12mv
2 = GMmr (C1)v 2 = (2 times 667 x 10ndash11 times 642 times 1023)(679 times 1062) (C1)
v = 502 times 103 msndash1 (A1) [3]
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1either as r decreases objectmassbody does workor work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fallgravitational field strength) B1
g = GMr 2 B1if r h g is constant B1
∆E P = force times distance moved M1= mgh A0
or ∆E P = m∆φ (C1)= GMm(1r 1 ndash 1r 2) = GMm(r 2 ndash r 1)r 1r 2 (B1)
if r 2 asymp r 1 then (r 2 ndash r 1) = h and r 1r 2 = r 2 (B1)g = GMr 2 (B1)
∆E P = mgh (A0) [4]
(d) frac12mv 2 = m∆φ
v 2 = 2 times GMr C1= (2 times 43 times 1013) (34 times 106) C1
v = 50 times 103msndash1 A1 [3](Use of diameter instead of radius to give v = 36 times 10 3 msndash1 scores 2 marks)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) force proportional to product of masses and inversely proportional tosquare of separation (do not allow square of distanceradius) M1either point masses or separation size of masses A1 [2]
(b) (i) ω = 2π (273 times 24 times 3600) or 2π (236 x 106) M1= 266 times 10ndash6 radsndash1 A0 [1]
(ii) GM = r
3ω 2
or GM = v
2
r C1M = (384 times 105 times 103)3 times (266 times 10ndash6)2 (667 times 10ndash11) M1= 60 times 1024kg A0 [2]
(special case uses g = GMr 2 with g = 981 r = 64 times 106 scores max 1 mark)
(c) (i) grav force = (60 times 1024) times (74 times 1022) times (667 times 10ndash11)(384 times 108)2 C1= 20 times 1020N (allow 1SF ) A1 [2]
(ii) either ∆E P = Fx because F constant as x radius of orbit B1
∆E P = 20 times 1020 times 40 times 10ndash2 C1= 80 times 1018J (allow 1SF ) A1 [3]
or ∆E P = GMmr 1 ndash GMmr 2 C1Correct substitution B1
80 times 1018J A1(∆E P = GMmr 1 + GMmr 2 is incorrect physics so 03)
1 (a) force is proportional to the product of the masses andinversely proportional to the square of the separation M1either point masses or separation gtgt size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1mv 2r = GMmr 2 and E K = frac12mv 2 M1hence E K = GMm2r A0 [2]
(ii) 1 ∆E K = frac12 times 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1= 926 times 107J (ignore any sign in answer ) A1 [2]
(allow 10 times 108J if evidence that E K evaluated separately for each r )
2 ∆E P = 400 times 1014 times 620 times (730 times 106ndash1 ndash 734 times 106ndash1) C1 = 185 times 108J (ignore any sign in answer ) A1 [2]
(allow 18 or 19 times 108J)
(iii) either (730 times 106)ndash1 ndash (734 times 106)ndash1 or ∆E K is positive EK increased M1speed has increased A1 [2]
1 (a) region of space area volume B1where a mass experiences a force B1 [2]
(b) (i) force proportional to product of two masses M1
force inversely proportional to the square of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [3]
(ii) field strength = GM x 2 or field strength prop 983089 983087 x 983090 C1
ratio = (778 times 108)2 (15 times 108)2 C1= 27 A1 [3]
(c) (i) either centripetal force = mR ω 2 and ω = 2π T
or centripetal force = mv 2 R and v = 2πR T B1gravitational force provides the centripetal force B1
either GMm R 2 = mR ω 2 or GMm R 2 = mv 2 R
M1
M = 4π2R 3 GT 2 A0 [3] (allow working to be given in terms of acceleration)
(ii) M = 4π2 times (15 times 1011)3 667 times 10ndash11times (316 times 107)2 C1
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
1 (a) equatorial orbit above equator B1satellite moves from west to east same direction as Earth spins B1period is 24 hours same period as spinning of Earth B1 [3](allow 1 mark for lsquoappears to be stationaryoverheadrsquo if none of above marks scored)
(b) gravitational force providesis the centripetal force B1
GMmR 2 = mR ω 2 or GMmR 2 = mv 2R M1
ω = 2π T or v = 2πR T or clear substitution M1
clear working to give R 3 = (GMT 2 4π2) A1 [4]
(c) R 3 = 667 times 10ndash11times 60 times 1024
times (24 times 3600)2 4π2 C1
= 757 times 1022 C1
R = 42 times 107 m A1 [3]
(missing out 3600 gives 18 times 10 5 m and scores 23 marks)
1 (a) work done in moving unit mass M1from infinity (to the point) A1 [2]
(b) (i) gravitational potential energy = GMm x
energy = (667 times 10ndash11times 735 times 1022
times 45) (174 times 106) M1
energy = 127 times 107 J A0 [1]
(ii) change in grav potential energy = change in kinetic energy B1
frac12 times 45 times v 2 = 127 times 107
v = 24 times 103 msndash1 A1 [2]
(c) Earth would attract the rock potential at Earth(rsquos surface) not zero lt0 at Earth potential due to Moon not zero M1escape speed would be lower A1 [2]
1 (a) force proportional to product of the two masses and inversely proportional to thesquare of their separation M1either reference to point masses or separation gtgt lsquosizersquo of masses A1 [2]
(b) gravitational force provides the centripetal force B1
GMmR 2 = mR ω
2 M1where m is the mass of the planet A1
GM = R 3ω 2 A0 [3]
(c) ω = 2π T C1
either M star M Sun = (R star R Sun)3times (T Sun T star )
2
M star = 43times (frac12)2 times 20 times 1030 C1
= 32 times 1031kg A1 [3]
or M star = (2π)2 R star 3 GT
2 (C1)
= (2π)2 times (60 times 1011)3 667 times 10ndash11times (2 times 365 times 24 times 3600)2 (C1)
