Chapter 13 – Student Solutions Manual

1. The magnitude of the force of one particle on the other is given by F = Gm1m2/r2,
where m1 and m2 are the masses, r is their separation, and G is the universal gravitational
constant. We solve for r:

r=
Gm1m2
=
( 6.67 ×10 −11
N ⋅ m 2 / kg 2 ) ( 5.2kg )( 2.4kg )
= 19 m
F 2.3 × 10−12 N

7. At the point where the forces balance GM e m / r12 = GM s m / r22 , where Me is the mass of
Earth, Ms is the mass of the Sun, m is the mass of the space probe, r1 is the distance from
the center of Earth to the probe, and r2 is the distance from the center of the Sun to the
probe. We substitute r2 = d − r1, where d is the distance from the center of Earth to the
center of the Sun, to find

Me Ms
= .
( d − r1 )
2 2
r1

Taking the positive square root of both sides, we solve for r1. A little algebra yields

r1 =
d Me
=
(150 ×10 m ) 9
5.98 ×1024 kg
= 2.60 ×108 m.
Ms + Me 1.99 ×10 kg + 5.98 ×10 kg
30 24

Values for Me, Ms, and d can be found in Appendix C.

17. The acceleration due to gravity is given by ag = GM/r2, where M is the mass of Earth
and r is the distance from Earth’s center. We substitute r = R + h, where R is the radius
of Earth and h is the altitude, to obtain ag = GM /(R + h)2. We solve for h and obtain
h = GM / ag − R . According to Appendix C, R = 6.37 × 106 m and M = 5.98 × 1024 kg,
so

h=
( 6.67 ×10 −11
m 3 / s 2 ⋅ kg )( 5.98 × 1024 kg )
− 6.37 × 106 m = 2.6 × 106 m.
( 4.9m / s )
2

29. (a) The density of a uniform sphere is given by ρ = 3M/4πR3, where M is its mass and
R is its radius. The ratio of the density of Mars to the density of Earth is

3
ρ M M M RE3 ⎛ 0.65 × 104 km ⎞
= = 0.11 ⎜ ⎟ = 0.74.
ρ E M E RM3 ⎝ 3.45 × 10 km ⎠
3
(b) The value of ag at the surface of a planet is given by ag = GM/R2, so the value for
Mars is
2
M R2 ⎛ 0.65 × 104 km ⎞
⎟ ( 9.8 m/s ) = 3.8 m/s .
2 2
ag M = M 2E a g E = 0.11⎜
⎝ 3.45 × 10 km ⎠
3
M E RM

(c) If v is the escape speed, then, for a particle of mass m

1 2 mM 2GM
mv = G ⇒ v= .
2 R R

For Mars, the escape speed is

2(6.67 × 10 −11 m3 /s 2 ⋅ kg) ( 0.11) ( 5.98 × 1024 kg )

v= = 5.0 × 103 m/s.
3.45 × 10 m
6

37. (a) We use the principle of conservation of energy. Initially the particle is at the
surface of the asteroid and has potential energy Ui = −GMm/R, where M is the mass of
the asteroid, R is its radius, and m is the mass of the particle being fired upward. The
initial kinetic energy is 1 2 mv 2 . The particle just escapes if its kinetic energy is zero when
it is infinitely far from the asteroid. The final potential and kinetic energies are both zero.
Conservation of energy yields −GMm/R + ½mv2 = 0. We replace GM/R with agR, where
ag is the acceleration due to gravity at the surface. Then, the energy equation becomes
−agR + ½v2 = 0. We solve for v:

v= 2a g R = 2(3.0 m/s 2 ) (500 × 103 m) = 1.7 × 103 m/s.

(b) Initially the particle is at the surface; the potential energy is Ui = −GMm/R and the
kinetic energy is Ki = ½mv2. Suppose the particle is a distance h above the surface when it
momentarily comes to rest. The final potential energy is Uf = −GMm/(R + h) and the final
kinetic energy is Kf = 0. Conservation of energy yields

GMm 1 2 GMm
− + mv = − .
R 2 R+h

We replace GM with agR2 and cancel m in the energy equation to obtain

1 2 a R2
− ag R + v =− g .
2 ( R + h)
The solution for h is
 2a g R 2 2(3.0 m/s 2 ) (500 × 103 m) 2
h= −R = − (500 × 103 m)
2a g R − v 2 2(3.0 m/s 2 ) (500 × 103 m) − (1000 m/s) 2
= 2.5 × 105 m.

(c) Initially the particle is a distance h above the surface and is at rest. Its potential energy
is Ui = −GMm/(R + h) and its initial kinetic energy is Ki = 0. Just before it hits the
asteroid its potential energy is Uf = −GMm/R. Write 1 2 mv 2f for the final kinetic energy.
Conservation of energy yields

GMm GMm 1 2
− =− + mv .
R+h R 2

We substitute agR2 for GM and cancel m, obtaining

ag R 2 1 2
− = − ag R + v .
R+h 2

The solution for v is

2a g R 2 2(3.0 m/s 2 )(500 × 103 m) 2

v = 2ag R − = 2(3.0 m/s 2 ) (500 × 103 m) −
R+h (500 × 103 m) + (1000 × 103 m)
= 1.4 × 103 m/s.

39. (a) The momentum of the two-star system is conserved, and since the stars have the
same mass, their speeds and kinetic energies are the same. We use the principle of
conservation of energy. The initial potential energy is Ui = −GM2/ri, where M is the mass
of either star and ri is their initial center-to-center separation. The initial kinetic energy is
zero since the stars are at rest. The final potential energy is Uf = −2GM2/ri since the final
separation is ri/2. We write Mv2 for the final kinetic energy of the system. This is the sum
of two terms, each of which is ½Mv2. Conservation of energy yields

GM 2 2GM 2
− =− + Mv 2 .
ri ri
The solution for v is

GM (6.67 × 10−11 m3 / s 2 ⋅ kg) (1030 kg)

v= = = 8.2 × 104 m/s.
ri 1010 m

(b) Now the final separation of the centers is rf = 2R = 2 × 105 m, where R is the radius of
either of the stars. The final potential energy is given by Uf = −GM2/rf and the energy
equation becomes −GM2/ri = −GM2/rf + Mv2. The solution for v is
 ⎛1 1⎞ ⎛ 1 1 ⎞
v = GM ⎜ − ⎟ = (6.67 × 10−11 m3 / s 2 ⋅ kg) (1030 kg) ⎜ − 10 ⎟
⎜r ri ⎟⎠ ⎝ 2 × 10 m 10 m ⎠
5
⎝ f
= 1.8 × 107 m/s.

45. Let N be the number of stars in the galaxy, M be the mass of the Sun, and r be the
radius of the galaxy. The total mass in the galaxy is N M and the magnitude of the
gravitational force acting on the Sun is F = GNM2/r2. The force points toward the galactic
center. The magnitude of the Sun’s acceleration is a = v2/R, where v is its speed. If T is
the period of the Sun’s motion around the galactic center then v = 2πR/T and a = 4π2R/T2.
Newton’s second law yields GNM2/R2 = 4π2MR/T2. The solution for N is

4π 2 R 3
N = .
GT 2 M

The period is 2.5 × 108 y, which is 7.88 × 1015 s, so

4π 2 (2.2 × 1020 m)3

N = = 5.1 × 1010.
(6.67 × 10−11 m3 / s 2 ⋅ kg) (7.88 × 1015 s) 2 (2.0 × 1030 kg)

47. (a) The greatest distance between the satellite and Earth’s center (the apogee distance)
is Ra = (6.37 × 106 m + 360 × 103 m) = 6.73 × 106 m. The least distance (perigee distance)
is Rp = (6.37 × 106 m + 180 × 103 m) = 6.55 × 106 m. Here 6.37 × 106 m is the radius of
Earth. From Fig. 13-13, we see that the semi-major axis is

Ra + R p 6.73 × 106 m + 6.55 × 106 m

a= = = 6.64 × 106 m.
2 2

(b) The apogee and perigee distances are related to the eccentricity e by Ra = a(1 + e) and
Rp = a(1 − e). Add to obtain Ra + Rp = 2a and a = (Ra + Rp)/2. Subtract to obtain Ra − Rp
= 2ae. Thus,

Ra − R p Ra − R p 6.73 × 106 m − 6.55 × 106 m

e= = = = 0.0136.
2a Ra + R p 6.73 × 106 m + 6.55 × 106 m

61. (a) We use the law of periods: T2 = (4π2/GM)r3, where M is the mass of the Sun (1.99
× 1030 kg) and r is the radius of the orbit. The radius of the orbit is twice the radius of
Earth’s orbit: r = 2re = 2(150 × 109 m) = 300 × 109 m. Thus,

4π 2 r 3 4π 2 (300 × 109 m)3

T = = −11
= 8.96 × 107 s.
GM (6.67 × 10 m / s ⋅ kg) (1.99 × 10 kg)
3 2 30
Dividing by (365 d/y) (24 h/d) (60 min/h) (60 s/min), we obtain T = 2.8 y.

(b) The kinetic energy of any asteroid or planet in a circular orbit of radius r is given by
K = GMm/2r, where m is the mass of the asteroid or planet. We note that it is
proportional to m and inversely proportional to r. The ratio of the kinetic energy of the
asteroid to the kinetic energy of Earth is K/Ke = (m/me) (re/r). We substitute m = 2.0 ×
10−4me and r = 2re to obtain K/Ke = 1.0 × 10−4.

75. (a) Using Kepler’s law of periods, we obtain

⎛ 4π 2 ⎞ 3
T = ⎜ ⎟ r = 2.15 × 10 s .
4

⎝ GM ⎠

(b) The speed is constant (before she fires the thrusters), so vo = 2πr/T = 1.23 × 104 m/s.

(c) A two percent reduction in the previous value gives v = 0.98vo = 1.20 × 104 m/s.

(d) The kinetic energy is K = ½mv2 = 2.17 × 1011 J.

(e) The potential energy is U = −GmM/r = −4.53 × 1011 J.

(f) Adding these two results gives E = K + U = −2.35 × 1011 J.

(g) Using Eq. 13-42, we find the semi-major axis to be

−GMm
a= = 4.04 × 107 m .
2E

(h) Using Kepler’s law of periods for elliptical orbits (using a instead of r) we find the
new period is

⎛ 4π 2 ⎞ 3
T′ = ⎜ ⎟ a = 2.03 × 10 s .
4

⎝ GM ⎠

This is smaller than our result for part (a) by T − T´ = 1.22 × 103 s.

(i) Elliptical orbit has a smaller period.

79. We use F = Gmsmm/r2, where ms is the mass of the satellite, mm is the mass of the
meteor, and r is the distance between their centers. The distance between centers is r = R
+ d = 15 m + 3 m = 18 m. Here R is the radius of the satellite and d is the distance from
its surface to the center of the meteor. Thus,
 F=
( 6.67 ×10 −11
N ⋅ m 2 / kg 2 ) ( 20kg )( 7.0kg )
= 2.9 × 10−11 N.
(18m )
2

83. (a) We write the centripetal acceleration (which is the same for each, since they have
identical mass) as rω2 where ω is the unknown angular speed. Thus,

G (M ) (M ) GM 2
= = Mrω 2
( 2r )
2 2
4r

which gives ω = 1 2 MG / r 3 = 2.2 × 10−7 rad/s.

(b) To barely escape means to have total energy equal to zero (see discussion prior to Eq.
13-28). If m is the mass of the meteoroid, then

1 2 GmM GmM 4GM

mv − − =0 ⇒ v= = 8.9 × 104 m/s .
2 r r r

87. We apply the work-energy theorem to the object in question. It starts from a point at
the surface of the Earth with zero initial speed and arrives at the center of the Earth with
final speed vf. The corresponding increase in its kinetic energy, ½mvf2, is equal to the
work done on it by Earth’s gravity: ∫ F dr = ∫ (− Kr )dr (using the notation of that Sample
Problem referred to in the problem statement). Thus,

1 2 0 0 1
2
mv f = ∫ R
F dr = ∫
R
(− Kr ) dr =
2
KR 2

where R is the radius of Earth. Solving for the final speed, we obtain vf = R K / m . We
note that the acceleration of gravity ag = g = 9.8 m/s2 on the surface of Earth is given by
ag = GM/R2 = G(4πR3/3)ρ/R2, where ρ is Earth’s average density. This permits us to write
K/m = 4πGρ/3 = g/R. Consequently,

K g
vf = R =R = gR = (9.8 m/s 2 ) (6.37 × 106 m) = 7.9 × 103 m/s .
m R

93. The magnitude of the net gravitational force on one of the smaller stars (of mass m) is

GMm Gmm Gm ⎛ m⎞
+ = 2 ⎜ M + ⎟.
( 2r ) ⎝ 4⎠
2
r2 r

This supplies the centripetal force needed for the motion of the star:
 Gm ⎛ m⎞ v2 2pr
⎜ M + ⎟ = m where v = .
r2 ⎝ 4⎠ r T

Plugging in for speed v, we arrive at an equation for period T:

2π r 3 2
T = .
G ( M + m / 4)