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CHAPTER 13
SOLUTION FOR PROBLEM 5
At the point where the forces balance GMe m/r12 = GMs 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. Substitute
r2 = d − r1 , where d is the distance from the center of Earth to the center of the Sun, to find
Me
Ms
=
.
2
(d − r1 )2
r1
Take the positive square root of both sides, then solve for r1 . A little algebra yields
0
√
d Me
(150 × 109 m) 5.98 × 1024 kg
√
0
= 2.6 × 108 m .
=0
r1 = √
Ms + Me
1.99 × 1030 kg + 5.98 × 1024 kg
Values for Me , Ms , and d can be found in Appendix C.
CHAPTER 13
SOLUTION FOR PROBLEM 13
(a) The magnitude of the force on a particle with mass m at the surface of Earth is given by
F = GM m/R2 , where M is the total mass of Earth and R is Earth’s radius. The acceleration
due to gravity is
2
ag =
GM (6.67 × 10−11 m3 /s · kg)(5.98 × 1024 kg)
F
2
=
=
= 9.83 m/s .
m
R2
(6.37 × 106 m)2
(b) Now ag = GM/R2 , where M is the total mass contained in the core and mantle together and
R is the outer radius of the mantle (6.345 × 106 m, according to Fig. 13–36). The total mass is
M = 1.93 × 1024 kg + 4.01 × 1024 kg = 5.94 × 1024 kg. The first term is the mass of the core and
the second is the mass of the mantle. Thus
2
(6.67 × 10−11 m3 /s · kg)(5.94 × 1024 kg)
2
ag =
= 9.84 m/s .
6
2
(6.345 × 10 m)
(c) A point 25.0 km below the surface is at the mantle-crust interface and is on the surface of
a sphere with a radius of R = 6.345 × 106 m. Since the mass is now assumed to be uniformly
distributed the mass within this sphere can be found by multiplying the mass per unit volume by
the volume of the sphere: M = (R3 /Re3 )Me , where Me is the total mass of Earth and Re is the
radius of Earth. Thus
]3
}
6.345 × 106 m
(5.98 × 1024 kg) = 5.91 × 1024 kg .
M=
6.37 × 106 m
The acceleration due to gravity is
2
ag =
GM (6.67 × 10−11 m3 /s · kg)(5.91 × 1024 kg)
2
=
= 9.79 m/s .
R2
(6.345 × 106 m)2
CHAPTER 13
SOLUTION FOR PROBLEM 27
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 = GN M 2 /r 2 . 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 = 4π2 R/T 2 . Newton’s second law yields
GN M 2 /R2 = 4π 2 M R/T 2 . The solution for N is
N=
4π 2 R3
.
GT 2 M
The period is 2.5 × 108 y, which is 7.88 × 1015 s, so
N=
(6.67 ×
4π 2 (2.2 × 1020 m)3
2
10−11 m3 /s
· kg)(7.88 ×
1015
s)2 (2.0
×
1030 kg)
= 5.1 × 1010 .
CHAPTER 13
SOLUTION FOR PROBLEM 37
(a) Use the law of periods: T 2 = (4π 2 /GM )r 3 , 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
5
4π 2 r 3
T =
GM
4π 2 (300 × 109 m)3
= 8.96 × 107 s .
=
2
−11
3
30
(6.67 × 10
m /s · kg)(1.99 × 10 kg)
Divide by (365 d/y)(24 h/d)(60 min/h)(60 s/min) to 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 = GM m/2r, where m is the mass of the asteroid or planet. Notice 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). Substitute m = 2.0 × 10−4 me and r = 2re to obtain
K/Ke = 1.0 × 10−4 .
CHAPTER 13
HINT FOR PROBLEM 7
Symmetry tells us that the horizontal component of the net force on the central sphere√is zero.
A little geometry shows that the distance from a vertex to the center of the triangle is 3L/4,
where L is the length of a triangle side. Furthermore, the angle between a side and the line from
a vertex to the center is 30◦ and the sine of this angle is 1/2. Thus vertical component of the
force on the central sphere is
GM m4
2Gmm4 (1/2)
− √
.
Fnet, y = √
2
( 3L/4)
( 3L/4)2
Set this expression equal to zero and solve for M . Notice that the algebraic result for M does
not depend on the value of m4 .
J
ans: (a) m; (b) 0
o
CHAPTER 13
HINT FOR PROBLEM 25
The period T and orbit radius r are related by the law of periods: T2 = (4π 2 /GM )r 3 , where M
is the mass of Mars. Convert the given period to seconds and solve for M .
J
ans: 6.5 × 1023 kg
o
CHAPTER 13
HINT FOR PROBLEM 39
The period T is given by the Kepler’s law of periods: T2 = (4π 2 /GM )r 3 . The speed is given
by v = 2πr/T , the kinetic energy by K = 12 M v 2 , and the magnitude of the angular momentum
by mrv.
J
ans: (a) r 3/2 ; (b) 1/r; (c)
√
√ o
r; (d) 1/ r