Chapter 4 Gravitation Physics Beyond 2000 Gravity Newton ry/newtongrav.html.

Chapter 4 Gravitation

Physics Beyond 2000

Gravity

• Newton

• http://csep10.phys.utk.edu/astr161/lect/history/newtongrav.html

• http://www.britannica.com/bcom/eb/article/9/0,5716,109169+2+106265,00.html

• http://www.nelsonitp.com/physics/guide/pages/gravity/g1.html

Gravity

• The moon is performing circular motion round the earth.

• The centripetal force comes from the gravity.

Fc

earthmoon

v

Gravity

• Newton found that the gravity on the moon is the same force making an apple fall.

W

Ground

Newton’s Law of Gravitation

• Objects attract each other with gravitational force.• In the diagram,

m1 and m2 are the masses of the objects and r is the distance between them.

m1 m2

F F

r

Newton’s Law of Gravitation

• Every particle of matter attracts every other particle with a force whose magnitude is

m1 m2

F F

r

221.

r

mmGF G is a universal constant

G = 6.67 10-11 m3kg-1s-2

Note that the law applies to particles only.

Example 1

• Find how small the gravitation is.

Shell Theorem

• Extends the formula

to spherical objects like a ball, the earth, the sun and all planets.

221.

r

mmGF

Theorem 1a. Outside a uniform spherical shell.• The shell attracts the external particle as if a

ll the shell’s mass were concentrated at its centre.

221.

r

mmGF

F F

r

m1m2O

Theorem 1b. Outside a uniform sphere.

• The sphere attracts the external particle as if all the sphere’s mass were concentrated at its centre.

221.

r

mmGF

F F

r

m1m2O

Example 2 Outside a uniform sphere.

• The earth is almost a uniform sphere.

221.

r

mmGF

F

r

m1 m2O

earth

F

Theorem 2a. Inside a uniform spherical shell.

• The net gravitational force is zero on an object inside a uniform shell.

m2

m1

The two forceson m2 cancel.

Theorem 2b. Inside a uniform sphere.

221.

r

mmGF

m2m1

r

F

where m1 is the mass of the corewith r the distance from the centre to the mass m2

Example 3

• Inside a uniform sphere.

m2m1

r

F

Gravitational Field

• A gravitational field is a region in which any mass will experience a gravitational force.

• A uniform gravitational field is a field in which the gravitational force in independent of the position.

• http://saturn.vcu.edu/~rgowdy/mod/g33/s.htm

Field strength, g

• The gravitational field strength, g, is the gravitational force per unit mass on a test mass.

test massF

m

m

Fg

F is the gravitational force

m is the mass of the test mass

g is a vector, in the same direction of F.SI unit of g is Nkg-1.

Field strength, g

• The gravitational field strength, g, is the gravitational force per unit mass on a test mass.

test massF

m

m

Fg

F is the gravitational force

m is the mass of the test mass

SI unit of g isNkg-1.

Field strength, g, outside an isolated sphere of mass M

• The gravitational field strength, g, outside an isolated sphere of mass M is

2r

GMg

Prove it by placing a test mass m at a point X with distance r from the centre of the isolated sphere M.

M r

field strengthat X

XO

Example 4

• The field strength of the earth at the position of the moon.

Field strength, g

• Unit of g is Nkg-1.

• g is also a measure of the acceleration of the test mass.

• g is also the acceleration due to gravity, unit is ms-2.

Field strength, g

• Field strength, g.

• Unit Nkg-1.

• A measure of the strength of the gravitational field.

• Acceleration due to gravity, g.

• Unit ms-2.• A description of

the motion of a test mass in free fall.

Field lines

• We can represent the field strength by drawing field lines.

• The field lines for a planet are radially inward.

Radial fieldplanet

Field lines

• We can represent the field strength by drawing field lines.

• The field lines for a uniform field are parallel.

Uniform field

earth’s surface

Field lines

• The density of the field lines indicates the relative field strength.

g1= 10 Nkg-1 g2= 5 Nkg-1

Field lines

• The arrow and the tangent to the field lines indicates the direction of the force acting on the test mass.

test mass

direction of the force

The earth’s gravitational field

• Mass of the earth Me 5.98 1024 kg

• Radius of the earth Re 6.37 106 m

Re

O

Gravity on the earth’s surface, go

• The gravitational field go near the earth’s surface is uniform and

2e

eo R

GMg

The value of go 9.8 Nkg-1

Example 5

• The gravity on the earth’s surface, go.

Apparent Weight

• Use a spring-balance to measure the weight of a body.

• Depending on the case, the measured weight R (the apparent weight) is not equal to the gravitational force mgo.

R

mgo

Apparent Weight

• The reading on the spring-balance is affected by the following factors:

1. The density of the earth crust is not uniform.

2. The earth is not a perfect sphere.

3. The earth is rotating.

Apparent Weight

1.The density of the earth crust is not uniform.

• Places have different density underneath. Thus the gravitational force is not uniform.

Apparent Weight

2. The earth is not a perfect sphere.

Points at the poles are closer to the centre than points on the equators.

rpole < requator

gpole > gequator N-pole

S-pole

Equator

Apparent Weight

3. The earth is rotating. Except at the pole, all

points on earth are performing circular motion with the same angular velocity . However the radii of the circles may be different.

X

Y

Apparent Weight

3. The earth is rotating. Consider a mass m is at p

oint X with latitude .

The radius of the circle is r = Re.cos .

X

rRe

m

O

Y

Apparent Weight

3. The earth is rotating. The net force on the mass

m must be equal to the centripetal force.

X

rRe

m

2mrFF cnet cos... 2

eRm

O

YFc

Note that Fnet points to Y.

Apparent Weight R

3. The earth is rotating. The net force on the mass m

must be equal to the centripetal force.

So the apparent weight (normal reaction) R does not cancel the gravitational force mgo.

X

r m

O

YFc

R

mgo

co FgmR

Apparent Weight R

3. The earth is rotating. The apparent weight R is

not equal to the gravitational force mgo in magnitude.

X

r m

O

YFc

R

mgo

co FgmR

Apparent weight R on the equator

2eo mRRmg 2eo mRmgR

emgThe apparent field strengthon the equator is

2eoe Rgg

mgo R

Apparent weight R at the poles

0 Rmgo

omgR pmg

The apparent field strengthat the poles is

op gg

mgo

R

Example 6

• Compare the apparent weights.

Apparent weight at latitude

X

r m

O

YFc

R

mgo

2mrFF cnet cos... 2

eRm

co FgmR

Note that the apparent weight Ris not exactly along the line throughthe centre of the earth.

Variation of g with height and depth

• Outside the earth at height h.

h = height of the mass m from the earth’s surface

r m

hRe

Meg

O



r m

hRe

Meg

O

2

22

2

).(

).(

r

Rg

r

R

R

GMr

GMg

eo

e

e

e

e

where go is the field strength on the earth’s surface.

2

1

rg



r m

hRe

Meg

O

where go is the field strength on the earth’ssurface.

2

2

)1.(

).(

eo

e

eo

R

hg

hR

Rgg


• Outside the earth at height h close to the earth’s surface. h<<Re.

r m

hRe

Meg

O

where go is the field strength on the earth’ssurface.

)2

1.(e

o R

hgg

ee R

h

R

h 21)1( 2


• Below the earth’s surface.

Re

rO d

Me

g

Only the core withcolour gives thegravitational force.

r = Re-d



Re

rO d

Me

g

Find the mass Mr of

3)(e

er R

rMM

r = Re-d



Re

rO d

Me

g 32

2

)(e

e

r

R

r

r

GMr

GMg

r = Re-d



Re

rO d

Me

g )1(

)(2

eo

ee

e

R

dg

R

r

R

GMg

r = Re-dg r


earth

ggo

0r distance from the centreof the earthRe

1. r < Re , g r.

2. r > Re , 2

1

rg

Gravitational potential energy Up

• Object inside a gravitational field has gravitational potential energy.• When object falls towards the earth, it gains kinetic energy and

loses gravitational potential energy.

This objectpossesses Up

earth

Zero potential energy

• By convention, the gravitational potential energy of the object is zero when its separation x from the centre of the earth is .

Up = 0earth

x O

Negative potential energy

• For separation less than r, the gravitational potential energy of the object is less than zero. So it is negative.

Up < 0earth

O r


• Definition 1

• It is the negative of the work done by the gravitational force FG as the object moves from infinity to that point.

earth

O r

FG

dx


• Definition 1

earth

O r

FG

dx

WU p


• Definition 2

• It is the negative of the work done by the external force F to bring the object from that point to infinity.

earth

O r

F

dx

Me

m


• Definition 2

earth

O r

F

dx

r

P dxFU .

Me

m


2x

mGMFFrom e

r

mGMU eP

earth

O r

Me

m

Example 7

• Conservation of kinetic and gravitational potential energy.

Example 8

• - Work done

= gravitational potential energy

Example 9

• Two particles are each in the other’s gravitational field.

• Thus each particle possesses gravitational energy.

System of three particles• Each particle is in another two particles’

gravitational fields.

• Each particle possesses gravitational potential energy due to the other two particles.

m

M1 M2

r1r2

Up of

2

2

1

1

r

GmM

r

GmM

System of three particlesUp of

Up of

m

M1 M2

r1r2

r3

2

2

3

12

r

GmM

r

MGM

1

1

3

12

r

GmM

r

MGM

Example 10

• Up of the moon due to the earth’s gravitational field.

r

earth

moon

What is the Up of the earth due to the moon’s gravitational field?

Escape speed ve

• Escape speed ve is the minimum projection speed required for any object to escape from the surface of a planet without return.

ve

Escape speed ve

• Escape speed ve is the minimum projection speed required for any object to escape from the surface of a planet without return.

ve

Escape speed ve

• On the surface of the planet, the body possesses both kinetic energy Uk and gravitational potential energy Up.

veR

mM

Uk

2

2

1emv

UP

R

GMm

Escape speed ve

• If the body is able to escape away, it means the body still possesses kinetic energy at infinity.

• Note that the gravitational energy of the body at infinity is zero.

veR

mM

Escape speed ve

veR

mM

• If there is not any loss of energy on projection,

the total energy of the body at lift-off

= the total energy of the body at infinity

Escape speed ve

veR

mM

)(2

1 2

r

GMmmve = kinetic energy at infinity

≧0

Escape speed ve

veR

mM

RgR

GMv o2

2where go is the gravitational acceleration on the surfaceof the earth.

Escape speed ve

veR

mM

RgR

GMv oe 2

2

So the escape speed from earth is

Escape speed ve

veR

mM

Example: Find ve

Gravitational potential V

• Definition: The gravitational potential at a point is the

gravitational potential energy per unit test mass.

m

UV

where U is the gravitational potential energy of a mass m at the point


• Definition: The gravitational potential at a point is the

gravitational potential energy per unit test mass.

m

UV unit of V is J kg-1


• Example 12 – to find the change in gravitational potential energy.

• ΔU = U – Uo

• If ΔU >0, there is a gain in U.

• If ΔU <0, there is a loss in U.

Equipotentials

• Equipotentials are lines or surfaces on which all points have the same potential.

• The equipotentials are always perpendicular to the field lines.

Equipotentials• The equipotentials around the earth are ima

ginary spherical shells centered at the earth’s centre.

Equipotentials• The field is radial.

Equipotentials• The equipotentials near the earth’s surface a

re parallel and evenly spaced surface.

• The field is uniform.

surface

Equipotentials

• Example 13 – Earth’s equipotential.

Potential V and field strength g

dr

dVg

rr

MGV

2r

GMg

Potential V and field strength g

rr

MGV

2r

GMg

If we consider the magnitude of g only,

dr

dVg

Earth-moon system

• http://tycho.usno.navy.mil/vphase.html• The potential is the sum of the potentials due to the earth and the moon.

earth

moonP

D

r

D-r

Me

Mm

http://tycho.usno.navy.mil/vphase.html





Earth-moon system

earth

moonP

D

r

D-r

Me

Mm

rD

GM

r

GMV meP

Earth-moon system

earth

moon

r

V

0

Earth-moon system

earth

moon

r

V

0

dr

dVg

Earth-moon system

earth

moon

r

V

0

dr

dVg

g

Earth-moon system

earth

moon

r

V

0g=0

g = 0 at a point X between the earthand the moon. X is a neutral point.

X

Earth-moon system

earth

moon

r

V

0

g>0

g points to the centre of the earth if it is positive.

X

Earth-moon system

earth

moon

r

V

0g<0

g points to the centre of the moon if it is negative.

X

Earth-moon systemGiven: Me = 5.98 × 1024 kg Mm = 7.35 × 1022 kg D = 3.84 × 108 m G = 6.67 × 10-11 Nm2kg-2

Find: the position X at which g = 0.

earth

moon

X

x

Hint: 0dr

dVg

Earth-moon systemGiven: Me = 5.98 × 1024 kg Mm = 7.35 × 1022 kg D = 3.84 × 108 m G = 6.67 × 10-11 Nm2kg-2

Find: the position X at which g = 0.

earth

moon

X

x

Answer: x = 3.46 × 108 m

Earth-moon system

• Example 14 – potential difference near the earth’s surface.

Orbital motion• The description of the motion of a planet

round the sun.

sun

Orbital motion• Kepler’s law:

1. The law of orbits.

All planets move in elliptical orbits, with the sun at one focus.

sun


2. The law of areas.

The area swept out in a given time by the line joining any planet to the sun is always the same.

sun


3. The law of periods.

The square of the period T of any planet about the sun is proportional to the cube of their mean distance r from the sun.

sun 32 rT

Orbital motion

• Basically, we only study the simple case of circular orbit.

r

Orbital motionA satellite of mass m performs circular motion round the earth with speed vc .The radius of the orbit is r.

r

satellite

earth

vc

Orbital motion

The centripetal force is provided by the gravitational force.

r

satellite

earth

vc

Fc

Orbital motion

Show that

r

satellite

earth

vc

Fc

r

GMv ec where Me is the

mass of the earth

Orbital motion

• Example 15 – find the speed of a satellite.

r

satellite

earth

vc

Proof of Kepler’s 3rd law in a circular orbit

32 rT

r1

satellite 1

earth

vc1

r2satellite 2

vc2

Proof of Kepler’s 3rd law in a circular orbit

r1

satellite 1

earth

vc1

r2satellite 2

vc2

Note that the proof is true for satellites round the same planet.

Kepler’s 3rd law

• Example 16 – apply Kepler’s 3rd law.

Satellites

• Natural satellites – e.g. moon.

• Artificial satellites –

e.g. communication satellites,

weather satellites.

http://weather.yahoo.com/graphics/satellite/US.html

http://www.smgaels.org/physics/97/MGRAHLFS.HTM

http://weather.yahoo.com/graphics/satellite/US.html




Geosynchronous satellites

• A geosynchronous satellite is above the earth’s equator.

• It rotates about the earth with the same angular speed as the earth and in the same direction.

• It seems stationary by observers on earth.


ω

equator

axis

satellite

h

Re

Geosynchronous satellitesFind the radius of the orbit of a geosynchornoussatellite.

ω

equator

axis

satelliteh Re

rs

h + Re = rs


rs = 4.23×107 m

ω

equator

axis

satelliteh Re

r

h + Re = rs


h = 3.59×107 m

ω

equator

axis

satelliteh Re

r

h + Re = rs

Parking OrbitNote that there is only one such orbit.It is called a parking orbit.

ω

equator

axis

satelliteh Re

r

h + Re = rs

Satellites Near the Earth’s surface

• Assume that the orbit is circular with radius r Re , the radius of the earth.

• The gravitational field strength go is almost a constant (9.8 N kg-1).

• The gravitational force provides the required centripetal force.

Satellites Near the Earth’s surface

r Re satellite

earth

vr

Find vr

Energy and Satellite Motion

• Find v and the kinetic energy Uk of the satellite.

r

satellite

earth Me

v

m

Energy and Satellite Motion• The satellite in the orbit possesses both

kinetic energy and gravitational energy.

r

satellite

earth Me

v

m


r

satellite

earth Me

v

m

r

GMv e2

r

mGMU ek 2

Note that Uk > 0


• Find Up the gravitational potential of the satellite.

r

satellite

earth Me

v

m


r

satellite

earth Me

v

mr

mGMU ep

Note that Up < 0


r

satellite

earth Me

v

m

Find U, the total energy of the satellite.


r

satellite

earth Me

v

m

r

mGMUUU epk 2

Note that U < 0


r

satellite

earth Me

v

m

U : Up : Uk = -1 : -2 : 1

Falling to the earth

r

satellite

earth Me

v

m

The satellite may lose energy due to airresistance. The total energy becomes more negative and r becomes less.

r

mGMU e

2


r

satellite

earth Me

v

m

The satellite follows a spiral path towardsthe earth.

r

mGMU e

2


r

satellite

earth Me

v

m

As r decreases, the kinetic energy of the satellite increases and the satellite moves faster.

r

mGMU ek 2


Example 17 – Loss of energy

Weightlessness in spacecraft

mg

vv

The astronaut is weightless.


• We fell our weight because there is normal reaction on us.

mg

Normal reaction

ground


• If there is not any normal reaction on us, we feel weightless. e.g. free falling

mg


v

mg The gravitational forcemg on the astronaut isthe required centripetalforce. He does not requireany normal reaction toact on him.


v

mg

The astronaut isweightless.

http://www.nasm.edu/galleries/gal109/NEWHTF/HTF611A.HTM




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