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Chapter 6
Page 1
CHAPTER 6  Gravitation and Newton’s Synthesis
1.
Because the spacecraft is 2 Earth radii above the surface, it is 3 Earth radii from the center.
The gravitational force
on the spacecraft is
F
=
GM
E
M
/
r
2
= (6.67
×
10
–11
N
·
m
2
/kg
2
)(5.98
×
10
24
kg)(1400 kg)/[3(6.38
×
10
6
m)]
2
=
1.52
×
10
3
N
.
2.
The acceleration due to gravity on the surface of a planet is
g
=
F
/
M
=
GM
planet
/
r
2
.
For the Moon we have
g
Moon
= (6.67
×
10
–11
N
·
m
2
/kg
2
)(7.35
×
10
22
kg)/(1.74
×
10
6
m)
2
=
1.62 m/s
2
.
3.
The acceleration due to gravity on the surface of a planet is
g
=
F
/
M
=
GM
planet
/
r
2
.
If we form the ratio of the two accelerations, we have
g
planet
/
g
Earth
= (
M
planet
/
M
Earth
)/(
r
planet
/
r
Earth
)
2
,
or
g
planet
=
g
Earth
(
M
planet
/
M
Earth
)/(
r
planet
/
r
Earth
)
2
= (9.80 m/s
2
)(1)/(2.5)
2
=
1.6 m/s
2
.
4.
The acceleration due to gravity on the surface of a planet is
g
=
F
/
M
=
GM
planet
/
r
2
.
If we form the ratio of the two accelerations, we have
g
planet
/
g
Earth
= (
M
planet
/
M
Earth
)/(
r
planet
/
r
Earth
)
2
,
or
g
planet
=
g
Earth
(
M
planet
/
M
Earth
)/(
r
planet
/
r
Earth
)
2
= (9.80 m/s
2
)(3.0)/(1)
2
=
29 m/s
2
.
5.
The acceleration due to gravity at a distance
r
from the center of the Earth is
g
=
F
/
M
=
Gm
Earth
/
r
2
.
If we form the ratio of the two accelerations for the different distances, we have
g
h
/
g
surface
= [(
r
Earth
)/(
r
Earth
+
h
)]
2
= [(6400 km)/(6400 km + 300 km)]
2
which gives
g
h
= 0.91
g
surface
.
6.
The acceleration due to gravity at a distance
r
from the center of the Earth is
g
=
F
/
M
=
Gm
Earth
/
r
2
.
If we form the ratio of the two accelerations for the different distances, we have
g
/
g
surface
= [(
r
Earth
)/(
r
Earth
+
h
)]
2
;
(
a
)
g
= (9.80 m/s
2
)[(6400 km)/(6400 km + 3.20 km)]
2
=
9.80 m/s
2
.
(
b
)
g
= (9.80 m/s
2
)[(6400 km)/(6400 km + 3200 km)]
2
=
4.36 m/s
2
.
7.
We choose the coordinate system shown in the figure and
find the force on the mass in the lower left corner.
Because the masses are equal, for the magnitudes of the
forces from the other corners we have
F
1
=
F
3
=
Gmm
/
r
1
2
= (6.67
×
10
–11
N
·
m
2
/kg
2
)(8.5 kg)(8.5 kg)/(0.70 m)
2
= 9.83
×
10
–9
N;
F
2
=
Gmm
/
r
2
2
= (6.67
×
10
–11
N
·
m
2
/kg
2
)(8.5 kg)(8.5 kg)/[(0.70 m)/cos 45°]
2
= 4.92
×
10
–9
N.
From the symmetry of the forces we see that the resultant will be
along the diagonal.
The resultant force is
F
= 2
F
1
cos 45° +
F
2
= 2(9.83
×
10
–9
N) cos 45° + 4.92
×
10
–9
N =
1.9
×
10
–8
N toward center of the square
.
8.
For the magnitude of each attractive
gravitational force, we have
F
1
F
2
F
3
4
2
1
x
y
L
3
L
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Page 2
F
V
=
Gm
E
m
V
/
r
V
2
=
Gf
V
m
E
2
/
r
V
2
= (6.67
×
10
–11
N
·
m
2
/kg
2
)(0.815)(5.98
×
10
24
kg)
2
/[(108 – 150)
×
10
9
m]
2
= 1.10
×
10
18
N;
F
J
=
Gm
E
m
J
/
r
J
2
=
Gf
J
m
E
2
/
r
J
2
= (6.67
×
10
–11
N
·
m
2
/kg
2
)(318)(5.98
×
10
24
kg)
2
/[(778 – 150)
×
10
9
m]
2
= 1.92
×
10
18
N;
F
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This note was uploaded on 04/08/2008 for the course PHY 53 taught by Professor Mueller during the Fall '08 term at Duke.
 Fall '08
 Mueller
 Force

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