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Chapter 13
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
:
(
29
(
29
(
29
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


=
=
=
2. We use subscripts s, e, and m for the Sun, Earth and Moon, respectively.
2
2
2
s
m
sm
sm
s
em
e
m
em
e
sm
em
Gm m
F
r
m
r
Gm m
F
m
r
r
=
=
Plugging in the numerical values (say, from Appendix C) we find
2
30
8
24
11
1.99 10
3.82 10
2.16.
5.98 10
1.50 10
×
×
=
×
×
3. The gravitational force between the two parts is
(
29
(
29
2
2
2
=
=
Gm M
m
G
F
mM
m
r
r


which we differentiate with respect to
m
and set equal to zero:
(
29
2
= 0 =
2
= 2
dF
G
M
m
M
m
dm
r

⇒
which leads to the result
m
/
M
= 1/2.
4. Using
F
=
GmM/r
2
, we find that the topmost mass pulls upward on the one at the origin
with 1.9
×
10

8
N, and the rightmost mass pulls rightward on the one at the origin with
1.0
×
10

8
N. Thus, the (
x, y
) components of the net force, which can be converted to
polar components (here we use magnitudeangle notation), are
(
29
(
29
8
8
8
net
= 1.04 10 ,1.85 10
2.13 10
60.6 .
F



а
r
581
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View Full DocumentCHAPTER 13
(a) The magnitude of the force is 2.13
×
10

8
N.
(b) The direction of the force relative to the +
x
axis is 60.6
. .
5. At the point where the forces balance
2
2
1
2
/
/
e
s
GM m r
GM m r
=
, where
M
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
(
29
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
(
29
9
24
8
1
30
24
150 10 m
5.98 10 kg
=
=
= 2.60 10 m.
+
1.99 10 kg + 5.98 10 kg
e
s
e
d M
r
M
M
G
G
G
Values for
M
e
,
M
s
, and
d
can be found in Appendix C.
6. The gravitational forces on
m
5
from the two 5.00g masses
m
1
and
m
4
cancel each other.
Contributions to the net force on
m
5
come from the remaining two masses:
(
29
(
29
(
29
(
29
11
2
2
3
3
3
net
2
1
14
6.67 10
N m /kg
2.50 10 kg 3.00 10 kg 1.00 10 kg
=
2 10 m
=1.67 10
N.
F







״
The force is directed along the diagonal between
m
2
and
m
3
, towards
m
2
. In unitvector
notation, we have
14
14
net
net
ˆ
ˆ
ˆ
ˆ
(cos45 i
sin 45 j)
(1.18 10
N)i
(1.18 10
N) j
F
F


=¬
+
=
E
+
r
7. We require the magnitude of force (given by Eq. 131) exerted by particle
C
on
A
be
equal to that exerted by
B
on
A
.
Thus,
=
.
We substitute in
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 Fall '08
 Many
 Physics, Force, Mass

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