1
Physics 221
Fall 2008
Homework #10 Solutions
Ch. 12:46; Ch. 13
Due Tues, Nov 4, 2008
10.1 A 500 kg satellite is placed in a circular orbit above the earth’s equator such that the satellite
always stays directly above the same point on the equator.
This is called a geosynchronous orbit.
Some communications and other satellites are placed in such orbits in order to provide
communications coverage over a large fixed area of the earth’s surface.
(a)
What is the period of the orbit?
By definition, the period of the orbit is
T
=
24 h
=
86400 s .
(b)
What is the radius of the orbit?
How far above the earth’s surface is the satellite, in km and in
miles?
From Eq. (12.12), we have
r
=
Gm
E
T
2
4
±
2
²
³
´
µ
¶
·
1/3
=
(6.67
¸
10
¹
11
N m
2
/kg
2
)(5.97
¸
10
24
kg)(8.64
¸
10
4
s)
2
4
2
º
»
¼
½
¾
¿
=
4.22
±
10
7
m
=
4.22
±
10
4
km
=
2.62
±
10
4
mi
.
Since the radius of the earth is 6.38
±
10
6
m, the radius of the orbit is 6.61 times the radius of
the earth.
The height of the satellite above the earth’s surface is 3.58
±
10
4
km = 2.23
±
10
4
mi.
(c)
What is the speed of the satellite?
v
=
2
r
T
=
2
(4.22
²
10
7
m)
8.64
²
10
4
s
=
3.07 km/s
.
(d) What are the kinetic energy and the potential energy of the satellite?
What is the total
mechanical energy of the satellite?
K
=
1
2
mv
2
=
1
2
(500 kg)(3.07
±
10
3
m/s)
2
=
2.36
±
10
9
J
.
U
=
±
Gm
E
m
r
=
±
(6.67
²
10
±
11
N m
2
/kg
2
)(5.97
²
10
24
kg)(500 kg)
4.22
²
10
7
m
=
±
4.72
²
10
9
J
E
=
K
+
U
=
2.36
±
10
9
J
²
4.72
±
10
9
J
=
²
2.36
±
10
9
J
.
We see that the magnitude of the mechanical energy is equal to the kinetic energy, and the
magnitude of the potential energy is twice the kinetic energy.
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(e)
At what speed would a small object have to be launched from the satellite in order to escape
earth’s gravity, assuming that the satellite is not moving?
Compare this escape speed with the
speed at which the object would have to be launched from the earth’s surface in order to
escape earth’s gravity, again assuming that the earth is not moving.
Taking into account the
motion of the satellite, does the escape speed from the satellite depend on the direction that
the object is launched from the satellite?
Explain.
Using Example 12.5 on p. 392 in the text, the escape speed of the object from the satellite is
v
=
2
Gm
E
r
=
2(6.67
±
10
²
11
N m
2
/kg
2
)(5.97
±
10
24
kg)
4.22
±
10
7
m
=
4.34 km/s
.
The escape speed from the earth’s surface is given in Example 12.5 as 11.2 km/s, about a
factor of 2.6 larger than the escape speed from the satellite, as verified by the following:
v
escape, earth
v
escape, satellite
=
r
satellite
r
earth
=
42.2
±
10
3
km
6.38
±
10
3
km
=
2.57
.
Since the escape speed is measured with respect to the earth, the speed does depend on which
direction the object is launched from the satellite, since the satellite itself is moving at 3.07
km/s.
If the object were launched perpendicular to the direction of motion of the satellite,
then the escape speed is the same as calculated above.
However, if the object is launched in
the direction that the satellite is moving, the object would have to be launched only at (4.34 –
3.07) km/s = 1.27 km/s.
If the object were launched in the opposite direction to the direction
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