1

CHAPTER 11

THE GRAVITATIONAL FIELD(GRAVITY)

• Newton’s Law of Gravitationo gravitational and inertial mass

• Gravitational potential energyo satelliteso Kepler’s 3rd Law o escape speed

• Gravitational field inside spheres*o hollow sphere o solid sphere

* You study using notes provided.

GRAVITATIONAL FIELD:

The groundwork for Newton’s great contribution to

understanding gravity was laid by three majors players:

Copernicus (1473-1543)

Galileo (1564-1642)

Kepler (1571-1630)

• shape of planetary orbits

• speed around an orbit

• relating time around orbit to distance from

the Sun

2

A value for G, the universal gravitational constant, can

be obtained from measurements by Henry Cavendish

(1731-1810) of the specific gravity of the Earth, as we

shall show shortly. Cavendish’s value for the specific

gravity of the Earth was 5.48, i.e., the density of the

Earth is 5.48 times the density of water:

∴ = × ⋅ = ⋅− −ρEarth 5 48 1000 54803 3. kg m kg m .

∴ = × =m volume density RE E E43

3π ρ

= ×( ) × = ×43

6 40 10 5480 6 02 106 3 24π . . kg.

So, Cavendish’s experiment allowed the first

determination of the mass of the Earth; it is within 1% of

today’s accepted value.

m1 m2

r

F Gm m

r= 1 2

2 distance between cm’s

Consider the apple that allegedly fell in Newton’s

orchard.

Newton’s 3rd Law tells us that:

r rF FEA AE= − .

Using the 2nd Law we have:

m a m aA A E Er r

= .

But mA ≈ 0 1. kg and m kgE ≈ ×6 1024 .

∴ ≈ × −a aE A1 67 10 25. .

Since near the Earth’s surface, a gA = ,

aE = × −1 64 10 24. m/s2.

So, the Earth also experiences an acceleration towards

the apple, but it is insignificant!

mE

mA

rFEA

rFAE

3

The force on the apple (i.e., its weight)

F Gm m

rG

m m

R hEA

E A E A

E= =

+2 2( ),

and the acceleration of the apple is

aFm

Gm

R hEA

A

E

E= =

+( )2.

Very close to Earth’s surface, h RE<< and a g≡ , so

G

Rm

gE

E=

2

where RE is the Earth’s radius. We know g = 9 81. m/s2,

m kgE = ×6 02 1024. and R mE ≈ ×6 4 106. ,

∴ = × ⋅−G 6 67 10 11. N m /kg2 2.

r h RE= +mE

mA

Let’s take acloser look

Re-arranging the expression for the universal

gravitational constant, gives us a method for determining

the acceleration at the surface

of a massive object, is valid

universally. So, if we know the

mass ( mp) and radius ( rp) of a

planet we can determine the

acceleration due to gravity at the

surface because G is a universal

constant!

g Gm

Rpp

p= 2

Here are some examples ...

g G

m

R= 2

Earthmass = ×6 1024 kgradius = 6378km

g = 9 8. m/s2

4

In deriving the expression for g we have made an

assumption ... that the Earth and planets are spherical !

Marsmass = ×6 42 1023. kg

radius = 3393kmg gMars = =3 7 0 38. .m/s2

Moonmass = ×7 35 1022. kg

radius = 1738kmg gMoon = =1 6 0 165. .m/s2

... it’s really a “flattened” sphere with R Req p> .

∴ <g geq p.

In fact, gp ~ .9 83m/s2 and geq ~ .9 78m/s2. Since ones

weight is mg, one weighs less

• up a mountain,

• in an airplane,

than at sea-level!

At 30,000ft: ∆ww

~ . %−0 3 .

Rp

Req

g Gm

RE

E= 2

But look atthe Earth!

5

Gravitational and inertial mass

So far, without stating it explicitly we have used the

same “mass” (m) in Newton’s Law of gravitation as was

defined by Newton’s 2nd Law in chapter 4. Note that in

the definition of the inertial mass of an object,

inertial mass = =mFainert ,

there is no mention of gravity. The mass used in

Newton’s Law of gravitation is called the gravitational

mass, which can be determined by measuring the

attractive force exerted on it by another mass (M) a

distance r away, viz:

gravitational mass = =mr F

GMgravMm

2.

There is no mention of acceleration in this expression.

Although these are two different concepts of mass,

Newton asserted that, for the same object, these masses

are identical, i.e., m mgrav iner= . This is known as the

principle of equivalence.

Gravitational potential energy

The acceleration of gravity

at a distance h above the

Earth’s surface is:

g G

m

rG

m

h RE E

E= =

+2 2( ).

(Note: it is never zero so you are never completely

weightless!!)

r h RE= +

9.81

Height above Earth’s surface h ( ×103km)

g ( m/s2)

R m

m kg

E

E

= ×

= ×

6 38 10

5 98 10

6

24

.

.

8 0.

6 0.

4 0.

2 0.

0 5 10 15 20 25

6

Near the Earth’s surface the gravitational potential

energy of an object of mass m is

U mgh mg r Rg E= = −( ),

where Ug = 0 at the Earth’s surface. Far from the Earth

we must take into account the fact that g r∝ −2. In

chapter 6 we defined the change in potential energy as

dU F ds= − •r r

,

where rF is a (conservative) force acting on the object

and dsr is a general displacement.

For the gravitational force,

dU F ds F r ds F r ds Gm m

rdrg g g

E= − • = − − • = • =r r r r

( ˆ) ˆ 2 .

dr ds r ds= = •r r

cos ˆφ

mEdsr

drr

rFg m

φ

∴ = ∫ = ∫ = − +U r G

m m

rdr Gm m

dr

rG

m mr

UEE

E( ) 2 2 o,

where Uo is an integration constant. We choose Uo = 0,

i.e., we define the gravitational potential energy to be

zero at r = ∞.

We obtain a negative value for U RE( ) because we

defined U( )∞ = 0. In earlier chapters we usually defined

U RE( ) = 0. However, it does not cause difficulties as

we normally work with changes in potential energy, i.e.,

∆U, which is independent of our choice of Uo.

r

U r( )

0

U RE( )

RE U r G

m mrE( ) = −

← − = −Gm mR

mgRE

EE

7

We have

U r Gm m

rG

m mR h

Gm m

R hR

E E

E

E

E E

( ) = − = −+( )

= −+( )1

.

For small values of h, h

RE<<1, then

U r G

m mR

hR

Gm mR

hR

E

E E

E

E E( ) = − +

= − −

−

1 11

= − +Gm mR

Gm mR

hE

E

E

E2 .

But G

m mR

U RE

EE= ( ) is constant and

G

mR

gE

E2 = .

∴ = +U r U R mghE( ) ( ) ,

i.e., in moving an object a height h above the Earth’s

surface increases the gravitational potential energy by

∆U mgh= ,

providing h RE<< , a result we used in chapters 6 and 7.

Note that the weight of an object is r rF mgg = , i.e.,

rg is a

vector and is called the gravitational field.

Since rg is a vector field, the resultant gravitational field

of a system of masses is a vector sum of the individual

contributions.

Question 1: Five objects, each of mass 3.00kg are are

equally spaced on the arc of a semicircle of radius

10.0cm. An object of mass 2.00kg is located at the

center of curvature of the arc. (a) What is the

gravitational force on the 2.00kg mass? (b) What is the

gravitational field at the center of curvature of the arc?

3 00. kg

3 00. kg 3 00. kg

3 00. kg

3 00. kg2 00. kg

10 0. cm

8

(a) The force acting on m due to each of the masses M is

F GmM

RG

mM

RG

mM

RG

mM

Rx = + − −2 2 2 245 45sin sino o

= 0,

and

F GmM

RG

mM

RG

mM

Ry = + +2 2 245 45sin sino o

= +( )GmM

R2 1 2 45sin o

=× × ×

×−6 67 10 2 00 3 00

0 1002 414

11

2. . .

( . ).

= × −9 66 10 8. N (in the j direction).

MM

MM

M

Rm x

y

j

i

(b) The gravitational force acting on m is r rF mg= .

∴ = =+

=× −r

r

gFm

F i F j

mjx y

ˆ ˆ . ˆ

.9 66 10

2 00

8

= × −4 83 10 8. j m/s2.

9

Question 2: A solid sphere of radius R and density ρo

has its center at the origin. There is a hollow, spherical

cavity of radius r

R=

2 within the sphere and centered at

xR

=2

, as shown. Find the gravitational field at points on

the x-axis for x R> .

R rx

y

We will model the system

as a sphere of mass

M R=

43

3π ρo

and a sphere of radius R

2

with negative mass ( −m).

Then g x g x g xsolid cavity( ) ( ) ( )= +

= +−

−( )G

M

xG

m

x R2 22

( )

=

+

− ( )

−( )

G R

x

G R

x R

ρ π ρ πo o43

43 2

2

3

2

3

2

=

−

−( )

GR

x x R

43

1 1

8 2

3

2 2πρo .

When x R=

g R GR

( ) =2

3πρo .

R rx

y

10

Question 3: An object is fired upward from the surface

of the Earth with an initial speed of 4.0km/s. Find the

maximum height it achieves. Compare that height with

the height it would have achieved if the gravitational

field were constant.

We use conservation of energy, i.e., K U K Ui i f f+ = + .

Using the expression for

gravitational potential

energy, we get12

2mv Gm mRi

E

E−

= −+

0 Gm m

R hE

E( ).

Solving for h we find

hR

GmR v

R

gRv

E

E

E i

E

E

i

=

−

=

−

21

212 2

,

where g =

GmR

E

E2 is the gravitational field at the

Earths’s surface.

∴ =×

× × ×

×−

h6370 10

2 9 81 6370 10

4 0 101

3

3

3 2.

( . )

= ×9 35 105. m (935km).

K mvi i=12

2:U RE( )

Kf = 0:U R hE( )+

h

11

If the gravitational field (g) were constant, then

∆ ∆K U mgh= − = ′,

i.e., ′ = =××

=hv

gkmi

2 3 2

24 0 102 9 81

815 5( . )

.. .

In the first part we found the “correct” height was

hR

gRv

E

E

i

=

−

212

.

But, from above, v

ghi

2

2= ′, where ′h is the height for a

constant gravitational field.

∴ =

′−

=′− ′( )

hR

Rh

h RR h

E

E

E

E1.

Check h km km( ).

.=

×−( )

=815 5 63706370 815 5

935 .

Satellites

Consider a satellite, mass m, in a circular orbit, radius r,

around the Earth (or a planet around the Sun). The

gravitational force between the satellite and the Earth

provides the centripetal acceleration.

∴ =Gm m

r

mvr

E2

2.

So, in a stable orbit

vGm

rE2 = .

The time for one orbit Tr

v=

2π

i.e., vr

TG

mr

g rE22 2

24

= = = ′π

,

where ′g is the gravitational field acting on the satellite.

v

rm

ME

12

constant

∴ =TGm

rE

22

34π i.e., T r2 3∝ .

~ Kepler’s 3rd Law of planetary motion ~

Note also

v Gm

rE= .

i.e., the velocity of a satellite in a stable orbit does not

depend on its mass only the mass of the central object!

... well, it provides us a way of

working out the mass of a planet.

Take the Earth-Moon system for

example ...

so what .. ?

Consider the Moon as the satellite around the Earth.

The speed of the Moon is

vr

T= =

distance around orbittime of orbit

2π

=× ×× × ×

=2 3 84 10

27 3 24 60 601023

8π ..

m/s.

From above we have: m vrGE = 2

=× ×

×= ×−

1023 3 84 10

6 67 106 0 10

2 8

1124.

.. kg,

which the the same as the result we got before.

Note: we didn’t need to know the mass of the Moon,

only its distance from the Earth and its orbital period.

3840 km

v

v Gm

rE=

13

∴ =× ×× × ×

= ×v2 1 5 10

365 24 60 602 99 10

114π .

. m/s.

So m vrGS = = × ×

×

× −2 4 2

11

112 99 101 5 10

6 67 10( . )

.

.

= ×2 0 1030. kg.

Using a similar approach we candetermine the mass of the Sun bythinking of the Earth as a satellite!Look ...

1 5 108. × km vr

TG

mrS= =

2π

Question 4: If K is the kinetic energy of a satellite in a

stable Earth orbit and Ug is the potential energy of the

Earth-satellite system, what is the relationship between

K and Ug?

14

We saw that the centripetal force of an Earth-satellite

system is a result of the gravitational attraction between

the Earth and the satellite, i.e.,

F G

m m

r

mvrg

E= =2

2,

so v G

mrE2 = .

∴ = =K mv G

m mrE1

212

2 .

But, the potential energy is U Gm m

rgE= − ,

∴ = −KUg

2 .

Therefore, the total mechanical energy is

E K UU

gg= + = 2

= −

12

Gm m

rE .

Note that E < 0 so the satellite is “bound”.

Ug

K

E

Energy

r

Question 5: Europa orbits Jupiter with a period of 3.55d

at an average distance of 6 71 108. × m from Jupiter’s

center. (a) Assuming the orbit is circlar, what is the mass

of Jupiter? (b) Another moon, Callisto, orbits Jupiter

with a period of 16.7d. What is its average distance from

Jupiter?

15

(a) In algebraic form, Kepler’s 3rd Law states that

T

GMRe

Je

22

34=

π,

where Te and Re are the orbital period and the orbital

radius of Europa, respectively, and MJ is the mass of

Jupiter.

∴ = =

× ×

× × × ×−MGT

RJe

e4 4 6 71 10

6 67 10 3 55 24 3600

2

23

2 8 3

11 2π π ( . )

. ( . )

= ×1 90 1027. kg.

(b) Also, TGM

RcJ

c2

234

=π

, where subscript c refers to the

moon Callisto.

∴

=

TT

RR

c

e

c

e

2 3

, i.e., R RTTc e

c

e=

23

= × × ( ) = ×6 71 10 16 73 55 1 88 108 23 9. .. . m.

Work done to put a satellite in orbit

The work done against the gravitational force in moving

an object a distance drr is

dW F dr GM m

rdrE= • =

r r2 (

r rF r ).

(Note, if dr ( = h) is very small compared with r, e.g.,

taking r as the Earth’s radius (6400 km) and h ≈ a few

meters, so g is constant, then

dW m G

mR

h mghE

E=

=2 ,

a result we obtained before for the work done in lifting

an object a short distance.)

drr

m

ME rr

16

Hence, the work done lifting a satellite to a height h is

W GM m

rdr GM m

rdrE

R

R h

ER

R h

E

E

E

E=

∫ =

∫

+ +

2 21

= −

= −+

−

+GM m

rGM m

R h RER

R h

EE EE

E1 1 1

= −− +

+

=

+

GM mR R hR R h

mGMR

hRR hE

E E

E E

E

E

E

E

( )( ) ( )2

=+

mg

hRR h

E

E( ).

By the work-energy theorem, the launch speed required

for the satellite to reach a height h is given by

12

2mv mghR

R hE

E=

+

( ),

i.e., v ghR

R hE

E=

+

2

( ).

The escape speed (ve) is defined as the minimum speed

required for an object to “escape” the Earth’s

gravitational field. Then

K U K Ui i f f+ = + ,

i.e., 12

02mv U Re E− =( ) .

∴ =

12

2mv GM mRe

E

E,

so, vGMR

gReE

EE= =

22 .

For the Earth, we find

ve = × × × = ×2 9 81 6370 10 1 12 103 4. . m/s

= 11 2. km/s.

For an object launched from Earth,

• if ve <11 2. km/s the object will be “bound” and will

orbit the Earth in a circle or ellipse,

• if ve >11 2. km/s the object will be “unbound” and

will follow a hyperbolic path and leave the Earth and

never return.

K mvi i=

12

2:U RE( )

Kf = 0: U( )∞ = 0

17

Question 6: One way of thinking about a black hole is to

consider a spherical object whose density is so large that

the escape speed at the surface is greater than the speed

of light, c. The critical radius for the formation of a

black hole is called the Schwarzschild radius, RS. (a)

Show that RGM

cS =

22 , where M is the mass of the black

hole. (b) Calculate the Schwarzschild radius for a black

hole whose mass is equal to 10 solar masses.

(a) The escape speed is given by

v

GMRe =

2.

For a black hole, v ce = and R RS= ,

i.e., cGMR

RGM

cSS= ⇒ =

2 22 .

(b) If M M kgSun= = × × = ×10 10 1 99 10 1 99 1030 31. . ,

∴ =× × × ×

×RS

2 6 67 10 1 99 10

3 10

11 31

8 2. .

( )

= ×2 95 104. m (29.5km).

18

Gravitational field inside a hollow sphere

The gravitational field inside a hollow sphere is zero, i.e.,rg = 0 (r R< ).

We can show this result by

considering a small (point)

mass mo situated at some

point inside the sphere. If

the distance from the mass

to points on the shell diametrically opposite are r1 and r2,

then the areas A1 and A2 and the volumes V A r1 1= ∆

and V A r2 2= ∆ are proportional to r12 and r2

2,

respectively. If the density of the shell is ρ, then

m V r1 1 12= ∝ρ and m V r2 2 2

2= ∝ρ ,

i.e., mr

mr

1

12

2

22= and G

m mr

Gm m

ro o1

12

2

22= .

Thus, the net force on mo is zero. That argument can be

made for all diameterically opposite areas A1 and A2

over the sphere. So, the gravitational field inside a

hollow sphere is zero.

A1m1

∆r

r1r2

A2m2∆φ

mo

In fact, the result is true for any thickness of shell. We

can apply the previous result by dividing the shell into a

continuum of concentric shells

and summing the fields at the the

point of interest. Thus, rg = 0,

inside all hollow spheres.

Gravitational field inside a solid sphere

To determine the gravitational field inside a solid sphere,

a distance r from the center, we use the previous result.

The mass of the part of the sphere

outside r makes no contribution to

the gravitational field at or inside

r. Only the mass ′M inside radius

r contributes to the gravitational field at r. This mass

produces a field equal to that of a point mass ′M at the

center of the sphere. If M is the mass of the sphere, then

R

r

19

′ = =Mr

RM

r

RM

4343

3

3

3

3

π

π.

Then, using an earlier expression for g, we find

rg r R G

M

r

GM

Rr( )≤ =

′=2 3 ,

inside the sphere.

The magnitude of the gravitational field is zero at the

origin and increases linearly with r. But, as before, it is

directed towards the center of the sphere.

The stipulation for this result is that the sphere is of

uniform mass density.

g GM

Rrr = 3

g r( )

rR

g GM

rr = 2

Question 7: Agamemnon is a planet of radius 4850km

and density 5500kg/m3. A straight shaft is drilled

completely through the planet Agamemnon. If a stone is

dropped from the top of the shaft, what would its speed

be (a) at the center, and (b) at a distance of 2500km

above the center? (c) Would the stone reach the opening

at the other end of the shaft?

20

(a) If we let a mass m fall down the shaft, to a radius r,

the work done by the gravitational field ∆W is

r rF dr F dr mg drg

R

r

gR

r

rR

r∫ • = ∫ = ∫ ,

where gr is the gravitational

field at radius r. But

g GM

rG

M

Rrr =

′=2 3 ,

where ′M is the mass contained within the sphere of

radius r.

∴ = − ∫ = −[ ]∆W

GMm

Rrdr

GMm

RR r

R

r

3 32 2

2.

But M R= = × ×

43

43

4850 10 55003 3 3π ρ π( )

= ×2 63 1024. kg,

and from the work-energy theorem, ∆ ∆W K= .

With R km= 4850 and r = 0,

∆ ∆W KGM

Rm m= = =

× × ×

× ×

−

26 67 10 2 63 10

2 4850 10

11 24

3. .

(J).

Rr

i.e., 12

1 81 102 7mv m= ×. .

∴ = × × =v 2 1 81 10 60147. m/s.

(b) With R km= 4850 and r km= 2500 ,

∆ ∆W K m= = ×1 33 107. (J).

∴ = × × =v 2 1 33 10 51547. m/s.

(c) Yes, the stone would just reach the opposite opening.

Since the scenario is symmetrical, its speed at the

opening would be zero, having done

∆W m= ×1 81 107. Joules of work against the

gravitational field.

7500

2500

5000

v(m/s)

rR

0 11

6014m/s5154m/s

21

Question 8: When farthest from Earth, i.e., at apogee,

the Moon-Earth distance is 406,400km. When closest to

Earth, i.e., at perigee, the Moon-Earth distance is

357,600km. If the mass of the Earth is 5 89 1024. × kg,

what are the orbital speeds of the Moon at apogee and

perigee?

In the previous chapter we saw that providing there are

no external forces/torques, then the angular momentum

of a system like the Earth-Moon is conserved. The

angular momentum of the Moon is r r rL r mv= × .

But at perigee and apogee r rr v⊥ ,

so L mvr= .

∴ =mv r mv ra a p p, i.e., v vr

ra pp

a= .

Also, mechanical energy is conserved, i.e.,

12

12

2 2mv GM m

rmv G

M mra

E

ap

E

p− = − .

∴ − = −v GMr

v GMra

E

ap

E

p

2 22 2 .

Hence v

r

rG

Mr

v GMrp

p

a

E

ap

E

p

22

22 2

− = − ,

i.e., v

r

rGM

r rpp

aE

p a

22

21 21 1

−

= −

.

rarp vp

va

22

∴−

=

−

v

r r

rGM

r r

r rpa p

aE

a p

a p

22 2

2 2

so vr r

rGM

rpa p

a

E

p

2 2+

= ,

i.e., vGM r

r r rpE a

p a p=

+2( )

=× × × × × ×

× × × + ×( )−2 6 67 10 5 98 10 4 064 10

3 576 10 4 064 10 3 576 10

11 24 8

8 8 8. . .

. . .

= 1090m/s (1.090km/s).

∴ = = ×

×

×v v

r

ra pp

a1090

3 576 10

4 064 10

8

8.

.

= 959m/s (0.959km/s).