CHAPTER 11

THE GRAVITATIONAL FIELD

• Newton’s Law of Gravitation • Inertial mass and gravitational mass • Gravitational potential energy

• Satellites

• Kepler’s 3rd Law

• Escape speed

Newton’s Law of Gravitation

A force of attraction

occurs between two

masses given by

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. He found that

the density of the Earth is 5.48 times the density of water,

Cavendish’s measurements allowed the first determination

of the mass of the Earth; it is within 1% of today’s

accepted value.

r m1 m2

i.e., ρEarth = 5.48×1000 kg ⋅m−3 = 5480 kg ⋅m−3.

∴mE = volume × density = 43πRE

3 ρE

= 43π(6.40 ×106m)3 × (5480 kg ⋅m−3) = 6.02 ×1024 kg.

F = G

m1m2

r2.

mE mA

r = h + RE

The force on the apple (i.e.,

its weight) is

When released the acceleration of the apple is

Very close to the Earth’s surface, and

where is the Earth’s radius. We know

and

From above, the acceleration due to gravity a distance r

from a mass m is:

a =

FEAmA

= GmE

(h + RE )2.

RE >> h a ! g.

∴G =

RE2

mEg,

RE g = 9.81 m/s2,

mE = 6.02 ×1024 kg RE = 6.4 ×106 m,

∴G = 6.67 ×10−11 N ⋅m2 /kg2.

FEA = mAg(r) = GmEmA

r2

= GmEmA

(h + RE )2.

g(r) = G m

r2.

This expression provides us a way of determining the

acceleration of gravity at the surface of a massive object

like a planet. If the mass is and the radius is then

Examples:

• Mars:

• Moon:

In deriving these expressions we have assumed that the

Earth and planets are spherical. In fact, the Earth is really

a “flattened” sphere … look …

gp = Gmp

R2p

.

mp

Rp

mass = 6.42 ×1023 kgradius = 3.39 ×103 m

gMars = 3.7 m/s2

mass = 7.35×1022 kgradius = 1.74 ×103 m

gMoon = 1.6 m/s2

Req

RP

For the Earth

In fact, Since ones

weight is mg, one weighs less

• up a mountain,

• in an airplane,

than at sea-level!

At 30,000 ft:

Req > Rp.

g = G

mE

RE2

∴geq < gP.

geq ~ 9.83 m/s2 and gP ~ 9.78 m/s2.

Δww

~ −0.3%.

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 = minert =

Fa

,

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 = mgrav =

r2FMmGM

.

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., mgrav = miner. This is known as the

principle of equivalence.

Examples of gravitational and inertial mass:

Gravitational mass:

• related to the time it takes an object to fall from a

certain height.

• the weight of an object on a spring scale.

Inertial mass:

• the acceleration given to an object by a

compressed spring.

In fact, g is a vector and is often referred to as the

gravitational field, i.e., the region over which one mass

influences another. It is analogous to a magnetic field and

an electric field. Since is a vector, the resultant

gravitational field of a collection of mass is the vector sum

of the individual contributions.

(!g)

!g

Question 11.1: Three 5.0 kg masses are located at the

positions shown in the x-y plane. What is the resultant

force on the mass at caused by the other two masses? A

A

F1x = −Gm1mA

r12

= −(6.67 ×10−11N ⋅m2 /kg2)(5.0 kg)(5.0 kg)

(0.40 m)2

= −1.04 ×10−8N.

F2 = Gm2mA

r22

= (6.67 ×10−11N ⋅m2 /kg2) (5.0 kg)(5.0 kg)

(0.50 m)2

= 6.67 ×10−9 N.

∴F2x = −(6.67 ×10−9 N)cosθ = −0.80(6.67 ×10−9 N)= −(5.34 ×10−9 N),

and F2y = (6.67 ×10−9 N)sinθ = 0.60(6.67 ×10−9 N)= (4.00 ×10−9 N).

A

∴FRx = F1x + F2x

= (−1.04 ×10−8N) + (−5.34 ×10−9 N)

= −1.57 ×10−8N,

and FRy = F2y = 0.40 ×10−8N.

∴!FR = (−1.57 ×10−8N)i + (0.40 ×10−8N) j,

and !FR = 1.62 ×10−8N.

Question 11.2: Five objects, each of mass are

equally spaced on the arc of a semicircle of radius

An object of mass is located at

the center of curvature of the arc.

(a) What is the gravitational force acting on the

mass?

(b) What is the gravitational field at the center of

curvature of the arc?

M = 3.00 kg

R = 10.0 cm. m = 2.00 kg

2.00 kg

(a)  The total force components acting on m are

and

Fx = GMm

R2+G

Mm

R2cos45! −G

Mm

R2cos45! −G

Mm

R2

= 0,

Fy = G Mm

R2sin45! +G Mm

R2+G Mm

R2sin45!

= G Mm

R21+ 2sin45!( ),

= (6.67 ×10−11 N ⋅m2 /kg2)(3.00 kg)(2.00 kg)

(0.10 m)2(2.414)

= 9.66 ×10−8 N.

∴!F = (9.66 ×10−8 N) j.

(b) The gravitational force acting on m is where

is the gravitational field. !F = m!g,

∴ !g =!Fm

= (9.66 ×10−8 N) j2.00 kg

= (4.83×10−8 m/s2) j.

!g Question 11.3: The Earth rotates about its N-S axis.

Assuming the Earth is a sphere, is the free-fall acceleration

measured at the equator the same as, greater than, or less

than the gravitational acceleration for a stationary Earth,

i.e., g = G

mE

RE2

?

View of the Earth looking at the

South pole and the FBD of

forces acting on an object

(O) at the equator; is the

normal force (equal to the

weight as measured on a scale) and is the gravitational

force. The net force is the centripetal force, i.e.,

where is the measured weight of the object. Then the

effective value of free-fall acceleration is

Since the Earth rotates though in 24 h,

Therefore, the effect is negligible.

!FN

!Fg = m!g

!a

FN

Fg

Fg − FN = mg − m ′g = mω2RE,

m ′g

′g = g −ω2RE, i.e., ′g < g.

2π

ω = 2π(24 × 60 × 60 s)

= 7.27 ×10−5 rad/s.

∴ω2RE = (7.27 ×10−5 rad/s)2(6.37 ×106 m) = 0.034 m/s2,

i.e., Δgg

= −0.0035.

O Gravitational potential energy

Above, we saw that the

acceleration of gravity at a

distance h above the Earth’s

surface was

g = G

mE

r2= G

mE

(h + RE )2.

Near the Earth’ surface, we showed that the

gravitational potential energy of an object of mass m at

a height h above the surface is

However, far from the Earth’s surface we must take into

account that In chapter 6, we defined a change

in potential energy as

where is a (conservative) force acting on an object

and is a general displacement.

In this case, for the gravitational force

Ug = mgh = mg(r − RE ).

g ∝ r−2.

dU = −!Fid!s,

!F

d!s

dU = −

!Fg id!r = Fg dr = G

mEm

r2dr.

∴U(r) = G

mEmr2∫ dr = GmEm

drr2∫ = −G

mEmr

+ U!,

where U! is an integration constant. We choose U! = 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., ΔU, which is independent of our choice of U!.

r

U(r)

0

U(RE )

RE U(r) = −G

mEmr

←−G

mEmRE

= −mgRE

∴U(r) = G

mEm

r2dr∫ = GmEm dr

r2=∫ −G

mEmr

+ U!,

We have

U(r) = −GmEm

r= −G

mEmRE + h( ) = −G

mEm

RE 1 + hRE( ).

For small values of h, h

RE<< 1, then

U(r) = −G

mEmRE

1 +h

RE

⎛

⎝ ⎜

⎞

⎠ ⎟ −1

= −GmEmRE

1−h

RE

⎛

⎝ ⎜

⎞

⎠ ⎟

= −G

mEmRE

+ GmEmRE

2 h.

But G

mEmRE

= U(RE ) is constant and G

mERE

2 = g.

∴U(r) = U(RE ) + mgh ,

i.e., in moving an object a height h above the Earth’s surface increases the gravitational potential energy by

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

Note that the weight of an object is ! F g = m! g , i.e.,

! g is a

vector and is called the gravitational field.

Question 11.4: A object is released from rest when it is

above the surface of a planet whose mass is

and radius Neglecting

air resistance,

(a) what is the speed of the object just before it strikes

the surface?

(b) Compare that speed with the speed it would have

achieved if the gravitational field were constant, i.e.,

the value at the surface.

4.0 ×106m

M = 4.0 ×1024kg R = 5.0 ×106m.

h

Ki = 0 : U(R + h)

Kf = 1

2mvf

2 : U(R)

(a)  Using the conservation of

mechanical energy, i.e.,

we obtain Kf + Uf = Ki + Ui,

12

mvf2 −G Mm

R= 0 −G Mm

(R + h).

∴vf2 = 2(6.67 ×10−11N ⋅m2 /kg2)(4.0 ×1024kg)

× 1

5.0 ×106m− 1

9.0 ×106m

⎛⎝⎜

⎞⎠⎟

= 4.743×107(m/s)2.∴vf = 6.89 ×103 m/s.

(b) Assuming g is constant, g at the surface is

We have

g = G M

R2= (6.67 ×10−11N ⋅m2 /kg2) (4.0 ×1024kg)

(5.0 ×106m)2

= 10.7 m/s2.

ΔK = −ΔU. ∴ 1

2m ′v 2 = mgh,

i.e., ′v = 2gh = 2(10.7 m/s2)(4.0 ×106m)

= 9.25×103 m/s.

Question 11.5: When farthest from Earth, i.e., at apogee,

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

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

357,600 km. If the mass of the Earth is

what are the orbital speeds of the Moon at apogee and

perigee?

5.89 ×1024kg,

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

! L = ! r × m! v .

But at perigee and apogee ! r ⊥! v ,

so L = mvr .

∴mva ra = mvprp, i.e., va = vp

rpra

.

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

12

mva2 − G

MEmra

=12

mv p2 − G

MEmrp

.

∴va

2 − 2GMEra

= vp2 − 2G

MErp

.

Hence vp

2 rpra

⎛

⎝ ⎜

⎞

⎠ ⎟

2− 2G

MEra

= vp2 − 2G

MErp

,

i.e., vp

2 1−rp2

ra2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = 2GME

1rp

−1ra

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ .

ra rp vp

va

Substituting for the given values

From earlier

vp2 0.2257( ) = 2.638×105 (m/s)2.

∴vp = 2.638×105 (m/s)2

0.2257= 1080 m/s.

va = vprpra

= (1080 m/s) (3.576 ×108m)

(4.064 ×108m)= 950 m/s.

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.

∴G

mEmr2 =

mv2

r.

So, in a stable orbit

v2 =

GmEr

.

The time for one orbit T =

2πrv

i.e., v2 =

4π2r2

T2 = GmE

r= ′ g r,

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

v

r m

ME

constant

∴T2 =

4π2

GmEr3 i.e., T2 ∝ r3.

~ Kepler’s 3rd Law of planetary motion ~ Note also

v = G

mEr

.

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 .. ?

Treating the Moon as a satellite around the Earth …

The speed of the Moon is

Then, from above

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

its distance from the Earth and its orbital period.

v = circumference of orbittime of orbit

= 2πrT

= 2π(3.84 ×108 m)(27.3× 24 × 60 × 60 s)

= 1.023 km/s.

v = G

mEr

mE = v2 rG

= (1.023×103 m/s)2(3.84 ×108 m)

6.67 ×10−11 N ⋅m2 /kg2

= 6.02 ×1024 kg.

Using a similar approach we can calculate the mass of the

Sun by considering the Earth as a satellite …

v = 2π(1.5×1011 m)(365× 24 × 60 × 60 s)

= 2.99 ×104 m/s.

∴mS = v2 rG

= (2.99 ×104 m/s)2 1.5×1011 m

6.67 ×10−11 N ⋅m2 /kg2

= 2.01×1030 kg.

Question 11.6: A 200 kg satellite circles the planet Roton

every 2.8 h in an orbit of radius If the radius

of Roton is what is

(a) the magnitude of the free-fall acceleration at the

surface of Roton, and

(b) the kinetic energy of the satellite?

1.20 ×107 m.

5.0 ×106m,

(a) The free-fall acceleration at the surface of a planet is

where M is its mass and R is its radius. However, we need

to find M. Using Kepler’s 3rd Law we have

where is the radius of the orbit and T is the orbital time.

Now

Hence

g = G M

R2,

T2 = 4π2

GMRs,

Rs

T = (2.8 h)(3600 s/h) = 1.008×104s,

∴M = 4π2

(6.67 ×10−11N ⋅m2 /kg2)

(1.20 ×107 m)3

(1.008×104s)2

= 1.01×1025kg.

g = (6.67 ×10−11N ⋅m2 /kg2) 1.01×1025kg

(5.0 ×106m)2

= 26.9 m/s2.

i.e., M = 4π2

GRs

3

T2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

,

(b) For a satellite in orbit, the centripetal force

mv2

Rs= G Mm

Rs2

i.e., K = 12

mv2 = G Mm2Rs

= (6.67 ×10−11N ⋅m2 /kg2) (1.01×1025kg)(200 kg)

2(1.20 ×107 m)

= 5.61×109 J.

Question 11.7: Europa orbits Jupiter with a period of 3.55

days at an average distance of from Jupiter’s

center.

(a) Assuming the orbit is circular, what is the mass of

Jupiter?

(b) Another moon, Callisto, orbits Jupiter every 16.7

days. What is its average distance from Jupiter?

6.71×108 m

(a)  Kepler’s 3rd Law states that

where and are the orbital period and orbital radius

of Europa, respectively, and is the mass of Jupiter.

(b) Also where subscript ca refers to

Callisto.

Teu

2 = 4π2

GMJReu

3 ,

Teu Reu

MJ

∴MJ =4π2Reu

3

GTeu2

= 4π2(6.71×108 m)3

(6.67 ×10−11 N ⋅m2 /kg2)(3.55× 24 × 60 × 60 s)

= 1.90 ×1027 kg.

Tca

2 = 4π2

GMJRca

3 ,

∴TcaTeu

⎛

⎝⎜⎞

⎠⎟

2

=RcaReu

⎛

⎝⎜⎞

⎠⎟

3

, i.e., Rca = ReuTcaTeu

⎛

⎝⎜⎞

⎠⎟

23

= (6.71×108 m) 16.7 d3.55 d

⎛⎝⎜

⎞⎠⎟

23

= 1.88×109 m.

Escape speed

The escape speed for a planet is defined as the minimum

initial speed of a projectile that enables it to escape the

influence of the planet’s gravitational field. Then

For the Earth

For an object launched from the Earth’s surface

• if the projectile will be “bound” and

will orbit the Earth in a circle or ellipse.

• if the projectile will be “unbound” and

will follow a hyperbolic path and leave the Earth and never

return.

Kf = 0 : U(∞) = 0

Ki =

12

mve2 : U(R)

Ki + Ui = Kf + Uf

i.e., 12

mve2 + U(R) = 0.

∴ 12

mve2 = G Mm

R,

i.e., ve = 2GMR

= 2gR.

ve = 2(9.81 m/s2)(6.37 ×106 m) = 11.2 km/s.

vi <11.2 km/s,

vi >11.2 km/s,

Question 11.8: Does it take more fuel, less fuel or the same

amount of fuel for a rocket to travel from the Earth to the

Moon as from the Moon to the Earth?

The work-kinetic energy theorem tells us that the

minimum work done to escape the Earth or Moon is the

same as the change in kinetic energy to achieve escape

speed.

But

The escape speed is Since and

then In fact,

Since the Earth’s escape speed is greater than the Moon’s

escape speed, it requires more work, i.e., more fuel, for a

spacecraft to travel from the Earth to the Moon than vice

versa.

∴W = ΔK = Ki − Kf .

Kf = 0,

∴W = 1

2mve

2.

ve = 2gR. RE > RM

gE > gM , ve(Earth) > ve(Moon).

ve(Earth) = 11.2 km/sve(Moon) = 2.36 km/s.

Question 11.9: Each planet in the solar system orbits at

its own particular speed around the Sun. If the orbital

speeds of the planets were doubled, which of the

following is true? Each planet would

A: assume a closer orbit around the Sun,

B: assume a more distant orbit around the Sun, or

C: escape the solar system.

D: Any or all of the above depending on the planet.

The escape speed from the Earth’s surface is

where are the mass and radius of the Earth,

respectively. So, the escape speed of a planet at a distance

R from the center of the Sun is

Earlier, we found that the orbital speed of a planet around the Sun is

So, if the orbital speed is doubled,

which is greater than the escape speed. Therefore, each

planet will escape the solar system (answer C). Note that

the orbital speed need only to be increased by a factor of

for a planet to escape the solar system.

ve =

2GMSR

.

ve =

2GMERE

,

ME and RE

v =

GMSR

.

′v ⇒ 2

GMSR

,

2

Question 11.10: A simple model of a black hole would be

to consider it a spherical object whose mass 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,

(a) Show that

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.

Take

RS.

RS = 2GM

c2,

c = 3×108 m/s.

(a)  From earlier, we showed that the escape speed is

For a black hole,

(b)

ve = 2GM

R.

ve = c and R = RS,

i.e., c = 2GM

RS⇒ RS = 2GM

c2.

M = 10MSun = 10(2.0 ×1030 kg) = 2.0 ×1031 kg.

∴RS = 2(6.67 ×10−11 N ⋅m2 /kg2)(2.0 ×1031 kg)

(3×108 m/s)2

= 2.96 ×104 m (29.6 km).