Example 10-7 A Satellite for Satellite Television
The broadcasting satellites used in a satellite television system orbit Earth’s equator with a period of exactly 24 hours,
the same as Earth’s rotation period. As a result, these satellites are geostationary: They always remain over the same spot
on the equator, and so they always appear to be in the same position in the sky as seen from anywhere on Earth’s surface.
(A satellite TV receiver “dish” is aimed to receive the signal broadcast from one of these satellites.) (a) At what distance
from Earth’s center must a television satellite orbit? (b) What must be its orbital speed? (c) If a television ­satellite has a
mass of 3.50 * 103 kg, how much energy must it be given to place it in orbit?
Set Up
A geostationary satellite has T = 24 h; we’ll
use this information and Equation 10-12 to
find the radius r of the orbit. Once we find
the value of r, we’ll use Equation 10-11 to
find the orbital speed of the satellite.
Equation 10-13 tells us the total
mechanical energy of the spacecraft in orbit,
and Equation 10-4 will tell us the total
mechanical energy when the spacecraft is at
rest on Earth’s surface before being launched.
The difference between these two values is
the energy that must be given to the satellite.
Relationship between period and
radius for a circular orbit:
4p 2
r 3
T2 =
GmEarth
(10-12)
Speed of an Earth satellite in a
circular orbit:
v =
GmEarth
r
B
(10-11)
Earth
Total mechanical energy for a
circular orbit:
GmEarth m
E = 2r
Gravitational potential energy:
Gm1 m2
Ugrav = r
Solve
(a) To use Equation 10-12 to find the orbital
radius r, first convert the orbital period T to
seconds.
T = 24 h
r=?
(10-13)
(10-4)
Period of the satellite orbit:
T = 24 ha
60 min
60 s
ba
b = 8.64 * 104 s
1h
1 min
Solve Equation 10-12 for r3:
r3 =
=
GmEarth T 2
4p 2
16.67 * 10-11 N # m2 >kg 2 2 15.97 * 1024 kg2 18.64 * 104 s2 2
= 7.53 * 1022
4p 2
N#m #s
= 7.53 * 1022 m3
kg
2
2
(We used 1 N = 1 kg # m>s 2.)
Take the cube root to find r:
3 3
3
r = 2
r = 2
7.53 * 1022 m3
= 4.22 * 107 m = 4.22 * 104 km
This is 6.63 times Earth’s radius (REarth = 6.37 * 103 km).
(b) Then use Equation 10-11 to find the
orbital speed of the satellite.
Orbital speed:
v =
=
=
=
GmEarth
r
B
C
B
16.67 * 10-11 N # m2 >kg 2 2 15.97 * 1024 kg2
4.22 * 107 m
9.43 * 106
9.43 * 106
C
kg # m m
N#m
=
9.43 * 106
kg
B
s 2 kg
m2
s2
= 3.07 * 103 m>s = 1.11 * 104 km>h = 6.87 * 103 mi>h
(Again, we used 1 N = 1 kg # m>s 2.)
(c) Find the total
mechanical energy
in orbit, the total
mechanical energy
on Earth’s surface, and
the difference between
these values for the
satellite of mass
m = 3.50 * 103 kg.
When the satellite is in orbit, the total mechanical energy is
E in orbit = = -
GmEarth m
2r
Ein orbit = –1.65 x 1010 J
16.67 * 10-11 N # m2 >kg 2 2 15.97 * 1024 kg2 13.50 * 103 kg2
= 21.65 * 1010 N # m
Eon surface = –2.19 x 1011 J
214.22 * 107 m2
= 21.65 * 1010 J
Earth
With the spacecraft at rest on Earth’s surface, the total mechanical
energy is just the gravitational potential energy:
E on surface = Ugrav, on surface = = -
GmEarth m
r
16.67 * 10-11 N # m2 >kg 2 2 15.97 * 1024 kg2 13.50 * 103 kg2
6.37 * 106 m
= 22.19 * 1011 N # m = 22.19 * 1011 J
The amount of energy that must be given to the satellite to put it
into orbit is
Ein orbit 2 Eon surface = (21.65 * 1010 J) 2 (22.19 * 1011 J)
= 2.02 * 1011 J
Reflect
Our answer to part (a) shows that geostationary broadcast satellites orbit at a tremendous distance from Earth.
Part (c) shows that it takes a tremendous amount of energy to put them there.
You can check the result for part (b) by showing that the satellite’s orbital speed v equals 2pr (the circumference of
the circular orbit of radius r) divided by T (the time to complete one orbit). Does it?