2009 Gravitation Lecture Notes

2009 Tampines Junior College

TAMPINES JUNIOR COLLEGE 2009 JC1 H2 PHYSICS

GRAVITATIONAL FIELD

OBJECTIVES Candidates should be able to:

(a) show an understanding of the concept of a gravitational field as an example of field of force and define gravitational field strength as force per unit mass.

(b) recall and use Newton’s law of gravitation in the form 2

21

r

mGmF .

(c) derive, from Newton’s law of gravitation and the definition of gravitational field

strength, the equation 2r

GMg for the gravitational field strength of a point

mass.

(d) recall and apply the equation 2r

GMg for the gravitational field strength of a

point mass to new situations or to solve related problems.

(e) show an appreciation that on the surface of the Earth g is approximately constant and is equal to the acceleration of free fall.

(f) define potential at a point as the work done in bringing unit mass from infinity to the point.

(g) solve problem using the equation r

GM for the potential in the field of a

point mass.

(h) recognise the analogy between certain qualitative and quantitative aspects of gravitational and electric fields.

(i) analyse circular orbits in inverse square law fields by relating the gravitational force to the centripetal acceleration it causes.

(j) show an understanding of geostationary orbits and their application.

REFERENCES

1. Physics, Robert Hutchings

2. Advanced Level Physics, Nelkon & Parker

3. Fundamentals of Physics, Halliday and Resnick

Tampines Junior College Gravitational Field

Mrs Lim YH (2009) 2

1 GRAVITATIONAL FORCE 1.1 What is gravitational force?

Every particle of matter in the universe attracts every other particle. This attractive force is called the gravitational force. The more massive and closer the particles are, the greater is the gravitational force between them. Gravitational force is one of the fundamental forces that exist in nature.

1.2 Newton’s law of gravitation

The Newton’s law of gravitation states that every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

Consider two particles of masses m1 and m2, and distance r apart.

The gravitational force between the two particles of mass m1 and m2 is given by

221

r

mGmF

where G is the gravitational constant which has a value of 6.67 10–11 N m2 kg–2 and r is the distance between the two particles.

Note: 1. The two particles exert equal and opposite attractive force on each other.

The force is directed along the line joining the two particles.

2. For a sphere of uniform density, the force that it exerts on other objects can be obtained by treating the sphere as a point mass at the centre of the sphere.

Question: Can Newton’s law of gravitation be applied to non-point masses or

non-spherical masses? Answer: If the objects involved are neither point masses nor spherical masses,

the formula is still applicable if the objects are placed sufficiently far apart such that their sizes become negligible compared to the distance separating them.

m1 m2

F F

r

Important Concepts/ insights.


Mrs Lim YH (2009) 3

Quick Check 1 Earth and the Moon are gravitationally attracted to each other. Does the more massive Earth attract the moon with, (a) a greater force (b) the same force (c) or less force, than the moon attracts Earth? Quick Check 2

Compared to the weight of that object at sea level, the weight of an object on peak of Himalayas (about 8848 m above sea level) will be (a) the same. (b) slightly smaller. (c) much smaller. Quick Check 3 On the ground the gravitational force on an object is W. What is the gravitational force at a height R, where R is the radius of the earth? A) 0.25W B) 0.5W C) W D) 2 W

* See worked example 1 in tutorial.

2 GRAVITATIONAL FIELD Gravitational field is a region in which a mass experiences a gravitational force.

2.1 Field of a mass

The region around a mass is a gravitational field. The following diagrams show the field lines around a point mass and around a uniform sphere (such as the Earth). The lines are always radial and directed towards the centre of mass. Note that the field lines are imaginary lines.

The field around the mass is not uniform. The field is stronger at points near the mass (lines are closer) and becomes weaker further away (lines are further apart). This can be deduced from the density of the field lines. The quantity that indicates the strength of the field at a point is known as the gravitational field strength.



Mrs Lim YH (2009) 4

2.2 Gravitational field strength

The strength of the gravitational field at several positions can be compared by measuring the gravitational force acting on a standardised mass placed at these positions.

The gravitational field strength at a point is defined as the gravitational force per unit mass acting at that point.

The usual symbol for field strength is g and its units are N kg–1.

2.3 Deriving the field strength equation 2r

MGg

Consider a point mass M. The point mass sets up a gravitational field around it. Let us derive an expression for the field strength at point P at a distance r from the point mass.

If we place a small mass ms at P, the force experienced by it is 2

s

r

mMGF

Force per unit mass acting at P is then sm

F =

s

2s

mr

mMG

= 2r

MG

Field strength at P (which is force per unit mass) is therefore given by

2r

MGg

2.4 Representation of the Field Strength

The diagram shows the arrow representing the field strength g at point P due to a point mass M. Field strength is a vector, its direction is towards the point mass producing the field.

Note: Field strength g is inversely proportional to the square of the distance

(g 1/r 2). Such a field is known as an inverse square law field. A minus sign actually appears on the right hand side of the formulae for the

gravitational force and gravitational field strength, so that they look like

PM

r

PM

g

r



Mrs Lim YH (2009) 5

221

r

mGmF

and

2r

MGg

The minus sign is used to indicate the attractive nature of the force (rather than repulsive). It NEED NOT be included when we are calculating a value for the force or the field strength, except during graph sketching.

Quick Check 4 The gravitational field generated by (or due to) the more massive Earth will be (a) greater (b) the same (c) or less than the gravitational field generated by the moon? Quick Check 5 The gravitational field (due to the Earth) experienced by an object on the peak of Himalayas (about 8848 m above sea level) is (a) the same as (b) slightly smaller than (c) much smaller than the gravitational field (due to the Earth) experienced by that object at sea level. Quick Check 6 Karen’s mass is 50 kg and Kenny’s mass is 75 kg. They are both sitting in the TPJC Auditorium (considering only their interaction with Earth only) They will experience the same a) Gravitational force b) Gravitational field strength * See worked example 2 in tutorial.

Quick Check 7

If a third object sits between two massive objects like (say when the moon orbits round the earth, it will experience the Sun’s gravitational field as well). What will be the total gravitational field? How should they be added?

(A) as a scalar

(B) as a vector

*Ssee worked example 3 in tutorial



Mrs Lim YH (2009) 6

2.5 Graphical representation of Earth’s field strength

The graph shows how the field strength g varies with distance r from the centre of the Earth. The field strength follows the inverse square law outside the Earth. The negative sign for field strength indicates that the direction is towards the Earth’s centre.

2r

MGg

2

ER

MG

** (see appendix 6)

2.6 Acceleration of free fall

Consider a mass m at the surface of the Earth. Let F be the gravitational force exerted on the mass. By definition, field strength at the Earth surface would be

g = F / m

If the mass is allowed to fall freely, its acceleration would be

a = F / m = g We thus see that the field strength g at the Earth surface is numerically equal to the acceleration of free fall.

The field strength g at the Earth surface is 9.81N kg–1 (or 9.81 m s-2) and is approximately constant near the Earth surface because of the large size of the Earth.

3 GRAVITATIONAL POTENTIAL ENERGY

The potential energy of a body is the energy it possesses due to its position in a field of force. Gravitational potential energy is therefore the energy a mass possesses when it is placed in a gravitational field.

R field strength g distance r

Earth M



Mrs Lim YH (2009) 7

3.1 Gravitational Potential Energy Near the Surface of the Earth

Consider the diagram on the right. We have already learnt that when lifting a mass m through a vertical height h (where h << RE), the potential energy gained by the mass is Increase in gravitational p.e. = mgh

or

Strictly speaking, the quantity mgh represents the change in potential energy when the mass is lifted through a vertical height h rather than the absolute amount it possesses at a height h. We conveniently choose the lower level (usually the ground level) as the zero potential energy level so that mgh is the extra energy a mass has when it is at height h compared to when it is at the zero potential energy level.

At this point we would also like to recall from the earlier topic on Work, Energy and Power that increase in potential energy is actually work done against gravity. It is based on this understanding that the formula U = mgh is derived.

3.2 Formal Definition of Gravitational Potential Energy

The gravitational potential energy of a mass at a point is defined as the work done by an external force in bringing the mass from infinity to that point.

Here, the work is done by an external force acting in opposite direction to the gravitational attraction. Alternatively, the gravitational potential energy of a mass at a point can also be understood as the work done against gravity in bringing the mass from infinity to that point.

To illustrate the meaning of the gravitational potential energy, consider the following diagram of a point mass M. The region around the point mass M is the gravitational field set up by the mass. The small mass m will be accelerated to P automatically by the gravitational attraction acting on it. The function of the external force is therefore to prevent it from gaining speed (or kinetic energy) as it moves from infinity to P.

The symbol used is U and the unit is the Joule (J).

m

h

m U = mgh

external force

gravitational force P

M

m

at infinity



Mrs Lim YH (2009) 8

Mathematically, the definition can be written as

Hence

So will the two equations give the same answer if we do some calculations? Lets find out …. A 3-kg mass is projected to a height h above the Earth’s surface. Taking the radius of the

Earth is 6400 km and mass of Earth = 6 x 1024 kg. Use U = mgh and r

GMmU to

find the change in gravitational potential energy if (i) h = 1000 m (ii) h = 1000 km (i) Using U = mgh, U = 3 x 9.8 x 1000 = 2.94 x 104 J

Using r

GMmU ,

(ii) Using U = mgh U = 3 x 9.8 x 1000 000 = 2.94 x 107 J

Using r

GMmU ,

r

GMmU

r

GMm

r

GMm

drr

GMm

drFU

r

r

r

G

2

J102.93 4

33

2411

106400

1

106401

1310610672.6

1

1000

1

EE RRGMmU

J102.54 6

33

2411

3

106400

1

107400

1310610672.6

1

101000

1

EE RRGMmU

M and m are point or spherical masses



Mrs Lim YH (2009) 9

Conclusion: Both equations gave close answers in (i) but gave quite different answers in (ii). Why? Note:

Strictly speaking, the term r

GMmU gives the total potential energy of the whole

system of two masses M and m. When two masses M and m are placed at a distance r apart, it is not right to say that each mass possesses this amount of potential energy. Rather, we should regard this as the energy shared by the two masses.

Consistency between PE = mgh and PE = r

GMm

Now, using r

GMmPE

PEi- = ER

GMm

PEf = h

ER

GMm

hRR

hGMm

hRR

RhRGMm

hR

1

R

1GMm

R

GMm

hR

GMmU

EE

EE

EE

EE

EE

Now if h is small compared to than RE, then 2

1

h

1

EEE RRR

Therefore

mgh

hR

GMm

R

hGMmΔU

2E

2E

Hence U = mgh is a special case of the more general r

GMmPE .

It is valid only if the change in height h is small compared to the radius of the Earth, when g remains constant.

h

RE

mass m raised from surface of Earth through height h

PEf

PEi



Mrs Lim YH (2009) 10

3.3 Projection of a Body from the Surface of a Planet

When a body of mass m is projected from the surface of the Earth with an amount of kinetic energy EK, the furthest distance away from the centre of the Earth, r, that it will reach is obtained by using conservation of energy, i.e. assuming no energy loss. The total energy of the body is conserved, i.e. sum of kinetic energy and potential energy remains constant throughout. In such a case, the kinetic energy of the body is converted to potential energy as it moves away from the Earth: M = mass of Earth, RE = radius of Earth

KE + PE at surface of Earth = KE + PE at furthest distance away

EK + (ER

GMm ) = 0 + (

r

GMm )

Note: Cannot use EK = mgh as the value of h may not be small compared to

radius of Earth. This problem may also be represented graphically by the energy-distance graph:

The total energy ET is obtained by adding EK to ER

GMm . The furthest distance r is

obtained from the intersection of the line representing ET and the potential energy curve because at this point, all the energy is in the form of potential energy.

ET = Ep + Ek

PotentialEnergy

RE

EK

Ep R

GMmE

EK

r dist. from centre of Earth 0




3.3.1 Condition For Escape to Infinity (Escape Speed) By law of conservation of energy, since total energy (KE + PE) at infinity is greater or equal to zero, therefore the condition for escape is

KE + PE 0

i.e. ½ mv2 + (-GMm/RE) 0

v ER

MG2

So the escape speed is

v =ER

MG2

4 GRAVITATIONAL POTENTIAL

The quantity gravitational potential energy depends on both the size of the mass as well as the position of the mass. We would like to define a energy quantity that is only dependent on the position (just like gravitational field strength). This quantity is the gravitational potential.

The gravitational potential at a point is defined as the work done by an external agent in bringing a unit mass from infinity to that point.

The usual symbol for gravitational potential is and its SI unit is J kg-1.

Suppose we want to find the gravitational potential at a particular point P. Theoretically, to determine the potential at P, we would bring in a unit mass (1 kg) all the way from infinity to P. The work done in bringing the unit mass from infinity to P is the gravitational potential at P.

For example, if -2 J of work is done in bringing a unit mass from infinity to P, the gravitational potential at P would be -2 J kg–1. Here, the work is done by an external force and not the gravitational attraction. The value can be negative if the external force is opposite to the displacement vector.

v

to infinity

rocket

Earth

KE 0 PE = 0




4.1 Formula for Gravitational Potential From the definition of gravitational potential, we see that its relation with gravitational potential energy is given by

and hence

We see that the gravitational potential at a point only depends on its distance r from the source of gravitation (the mass M). If a mass m is placed at a point with a gravitational potential , its gravitational potential energy is given by U = m.

Note: Both the gravitational potential and the gravitational potential energy are scalar

quantities. The negative sign present in their formulae is part of their numerical values and cannot be left out.

* see worked example 4 and 5 in tutorial.

5 RELATIONSHIP BETWEEN FIELD STRENGTH AND POTENTIAL

The relationship between gravitational field strength and gravitational potential is

dr

dg

You can easily verify that this relation for the point mass case. The gravitational field strength at a point is equal to the negative of the potential gradient at that point. The minus sign indicates that the potential falls when moving in the direction of the field. The field strength is numerically equal to the potential gradient.

m

U

r

GM




Consider the following graph that shows how the gravitational potential varies with the distance r from the centre of a planet.

The gradient at any point on the curve represents the field strength at that point.

Similarly, by multiplying mass on both sides of the above equation, we obtain the relation between the gravitational force F and the gravitational potential energy U:

dr

dUF

* See appendix 7 for more details and work example 6 in tutorial for calculations.

6 ORBITS

What keeps a satellite orbiting around the Earth? It is the Earth’s gravitational attraction that holds a satellite in its orbit. Without gravitational attraction, the satellite would move in a straight line rather than in a circular path. Once the satellite is in orbit, it does not need any rocket motor to keep it in orbit.

There are many examples of naturally occurring orbital motion in space. For instance, the Moon orbits around the Earth; all the nine planets including the Earth orbit around the Sun.

6.1 Kinematics of circular orbits

Satellites and planets may move in orbits that are circular or elliptical. In this syllabus, we shall only deal with problems that involve circular orbits. In circular orbits, the satellites and planets move with constant speed.

gradient gives field strength

gravitational potential

distance r




Consider a satellite of mass m moving around the Earth of mass M in a circular orbit of radius r as shown:

There is only one force acting on the satellite, the gravitational force exerted by the Earth. The gravitational force on the satellite is used as the centripetal force that keeps the satellite in a circular orbit. As a result, the satellite does not fall towards the Earth.

gravitational force = centripetal force

2r

mMG =

r

vm 2

= 2rm

How is T related to r ?


2r

mMG = 2rm =

22

T

rm

r3 = 224

TGM

6.2 Energies Associated with Satellites

Imagine a satellite of mass m moves in a circular orbit about the Earth of mass M. The radius of the orbit is r. Derive an expression for

(a) the kinetic energy T of the satellite,

Gravitational force = centripetal force

2r

mMG =

r

vm 2

kinetic energy, T = 221 vm =

r

mMG

2

F r




(b) the gravitational potential energy V of the satellite,

Gravitational potential energy, V = r

mMG

(c) the total energy E of the satellite. Total energy, E = T + V

= r

mMG

2 +

r

mMG

= r

mMG

2

Note: A negative total energy indicates that the satellite does not have sufficient energy to escape from Earth’s gravitational field to infinity.

6.3 Geostationary (or Geosynchronous) orbit

If the period of the satellite’s orbit is 24 hours, and if the satellite orbits over the equator and in the same direction as the Earth’s rotation, then the satellite will always appear above the same point on the Earth. Such a satellite is called a geosynchronous or geostationary satellite, and its orbit a geosynchronous or geostationary orbit. To an observer on the ground, the geostationary satellite seems to be stationary relative to the observer. Note that a geostationary orbit can (i) only occur above the equator and (ii) travels from west to east.

Since the period of orbit of a satellite depends on the radius of the orbit (T2 r3), a geostationary satellite is placed at a particular orbital radius. The following calculations show how this orbital radius is found.

To find the geostationary orbit: gravitational force = centripetal force

2r

mMG = 2rm =

22

T

rm

r = 31

224

T

MG

For geostationary orbit, we substitute T = 24 hours = 86400 s, M = 6.0 1024 kg, and obtain

r =

31

22

2411

)86400(4

)100.6()1067.6(

= 4.23 107 m

The radius of a geostationary orbit is 4.23 107 m.




Summary: (i) General Equations:

Note: The formulas inside the boxes are applicable only to point masses and uniform spheres while the other formulas (inside ellipses) are valid in all situations.

(ii) Orbital Motion


2r

mMG =

r

vm 2

= 2rm



Appendices Appendix 1 - SIR ISSAC NEWTON AND GRAVITATION

There is a popular story that Newton was sitting under an apple tree, an apple fell on his head, and he suddenly thought of the Universal Law of Gravitation. As in all such legends, this is almost certainly not true in its details, but the story contains elements of what actually happened.

What Really Happened with the Apple?

Probably the more correct version of the story is that Newton, upon observing an apple fall from a tree, began to think along the following lines: The apple accelerated since it started with zero velocity. Thus there must be a force that acts on the apple to cause this acceleration. Let's call this force "gravity", and the associated acceleration the "acceleration due to gravity". Then imagine the apple tree is twice as high. Again, we expect the apple to be accelerated toward the ground, so this suggests that gravity reaches to the top of the tallest apple tree.

Sir Isaac's Most Excellent Idea

Now came Newton's truly brilliant insight: if the force of gravity reaches to the top of the highest tree, might it not reach all the way to the orbit of the Moon! Then, the orbit of the Moon about the Earth could be a consequence of the gravitational force, because the acceleration due to gravity could change the velocity of the Moon in just such a way that it followed an orbit around the earth.

This can be illustrated with the thought experiment shown in Fig. 2. Suppose we fire a cannon horizontally from a high mountain; the projectile will eventually fall to earth, as indicated by the shortest trajectory in the figure, because of the gravitational force directed toward the center of the Earth and the associated acceleration. (Remember that an acceleration is a change in velocity and that velocity is a vector, so it has both a magnitude and a direction. Thus, an acceleration occurs if either or both the magnitude and the direction of the velocity change.)

But as we increase the muzzle velocity for our imaginary cannon, the projectile will travel further and further before returning to earth. Finally, Newton reasoned that if the cannon projected the cannon ball with exactly the right velocity, the projectile would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. That is, the cannon ball would have been put into orbit around the Earth. Newton concluded that the orbit of the Moon was of exactly the same nature: the Moon continuously "fell" in its path around the Earth because of the acceleration due to gravity, thus producing its orbit.

Fig. 1

Fig. 2



By such reasoning, Newton came to the conclusion that any two objects in the Universe exert gravitational attraction on each other, with the force having a universal form:

The constant of proportionality G is known as the universal gravitational constant. It is termed a "universal constant" because it is thought to be the same at all places and all times, and thus universally characterizes the intrinsic strength of the gravitational force.

Appendix 2 - Experiment to measure g using falling body

The diagram shows the experimental setup used to measure the acceleration of free fall using a falling body.

A steel ball is released from P and an electronic timer is triggered to start by the release. After falling a distance s, it passes a light gate, which causes the timer to stop. By measuring the distance s and the time t of fall as recorded by the timer, the free-fall acceleration g can be determined from the equation

s = ½ g t 2

g = 2

2

t

s

Appendix 3 - True weight and apparent weight

The true weight of a body is equal to the gravitational force on the body and is determined only by the position of the body. The apparent weight is the force that the body exerts on its support. It will not be equal to the true weight if the body is undergoing acceleration. The apparent weight is zero i.e. the body is weightless for the following cases:

Law of Universal Gravitation Every object in the universe attracts every other object with a force directed along the line of centres for the two objects that is proportional to the product of their masses and inversely proportional to the square of the separation between the two objects.

221

Gr

mmGF

s

P

light beam

light gate

to electronic

timer

ball



(i) a body falling freely under gravity

(ii) a space vehicle orbiting the Earth In the second case, the gravitational force on the man is used entirely to provide the centripetal force required to keep him in circular motion in his orbit. The astronaut does not exert any force on the vehicle. He experiences weightlessness. Note: A body's true weight is zero only at the point where there is no gravitational field. This can happen in

outer space (or deep space) where the gravity effects of the Earth and other planets are zero. Another example of such a point is to be found between the Earth and the Moon where the two gravitational fields cancel.

Appendix 4 - Factors affecting g on Earth The factors affecting g are: (a) altitude and latitude (b) density (c) rotation of the earth

It is assumed in calculations that the Earth is spherical, has uniform density (homogeneous) and that it does not rotate. Thus in finding g, we only consider the distance from the centre of the Earth. However, the earth is actually ellipsoidal i.e. it is flattened at the poles and bulges at the Equator (points at the poles are closer than points on the equator). It has non-uniform density due to unequal deposits of mineral in different parts of the Earth (variations in density enable oil prospecting). The Earth rotates about a polar axis with an angular speed ω= 2/T = 2/(24 x 60 x 60) = 7.27 x 10-5 rad s-1. All objects on the surface of the Earth are undergoing circular motion with same value of ω except at the poles where radius of circle is zero. Part of the gravitational force (and thus go) is used to provide the centripetal force for the objects. The amount of force depends on the latitude since the radius of path is different.

Appendix 5 - Effect of the rotation of the Earth on g The Earth rotates from the West to the East about the North South axis with a period of 24 hours. Any object on the surface at the equator is rotating with the Earth with an angular velocity ω. It thus experiences a centripetal acceleration. F = gravitational force of attraction acting on object. N = normal force acting on object. By Newton's second law,

net force = F - N = mRω2 where R is radius of Earth

But = mg (true weight)

where g = acceleration of free fall at Earth's surface = 9.81 m s-2.

Thus, N = mg - mRω2

= m(g - Rω2)

R

GmMF

2

N

F



By Newton's third law, N = force exerted by object on Earth's surface. � apparent weight, mg' = N

= m(g - Rω2) i.e. apparent acceleration g' = g - Rω2

Appendix 6 – Variation of g The Earth may be taken to be uniform sphere of radius r and density ρ. For r less than the radius of the earth, the Mass is NOT CONSTANT but depends on the radius.

Hence 2r

GMg =

2

3

3

4

r

Gr =

3

4 Gr a linear relationship with r

Appendix 7 – Derivation of relations

By definition,

r

drFU ||

Hence, dr

dUF

Since = U/m, we have

drr

gr

drmF

mU

||||

Thus, dr

dg

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2009 Gravitation Lecture Notes

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