Chapter 13 – Student Solutions Manual
1. The magnitude of the force of one particle on the other is given by
F
=
Gm
1
m
2
/
r
2
,
where
m
1
and
m
2
are the masses,
r
is their separation, and
G
is the universal gravitational
constant. We solve for
r
:
(
)
(
)(
)
11
2
2
1
2
12
6.67
10
N m / kg
5.2kg
2.4kg
19m
2.3
10
N
Gm m
r
F
−
−
×
⋅
=
=
=
×
7. At the point where the forces balance
, where
M
2
1
/
e
s
GM m r
GM m r
=
2
2
/
e
is the mass of
Earth,
M
s
is the mass of the Sun,
m
is the mass of the space probe,
r
1
is the distance from
the center of Earth to the probe, and
r
2
is the distance from the center of the Sun to the
probe. We substitute
r
2
=
d
−
r
1
, where
d
is the distance from the center of Earth to the
center of the Sun, to find
(
)
2
2
1
1
=
.
e
s
M
M
r
d
r
−
Taking the positive square root of both sides, we solve for
r
1
. A little algebra yields
(
)
9
24
8
1
30
24
150
10
m
5.98
10
kg
=
=
= 2.60
+
1.99
10
kg +
5.98
10
kg
e
s
e
d
M
r
M
M
×
×
×
×
×
10
m.
Values for
M
e
,
M
s
, and
d
can be found in Appendix C.
17. The acceleration due to gravity is given by
a
g
= GM/r
2
, 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
a
g
= GM /
(
R + h
)
2
. We solve for
h
and obtain
/
g
h
GM
a
=
R
−
.
According to Appendix C,
R
= 6.37
×
10
6
m and
M
= 5.98
×
10
24
kg,
so
(
)(
)
(
)
11
3
2
24
6
6
2
6.67
10
m / s
kg
5.98
10
kg
6.37
10 m
2.6
10 m.
4.9m / s
h
−
×
⋅
×
=
−
×
=
×
29. (a) The density of a uniform sphere is given by
ρ
= 3
M
/4
π
R
3
, where
M
is its mass and
R
is its radius. The ratio of the density of Mars to the density of Earth is
3
3
4
3
3
0.65
10
km
=
= 0.11
= 0.74.
3.45
10
km
M
M
E
E
E
M
M
R
M
R
ρ
ρ
⎛
⎞
×
⎜
⎟
×
⎝
⎠

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