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290 CHAPTER 9 Gravitation

The total mechanical energy is the sum of the potential energy and the kinetic energy.
Since we are assuming that the mass M is stationary, the kinetic energy is entirely due to
the motion of the mass m, and the Law of Conservation of Energy takes the form

GMm
2
1
Law of Conservation of Energy E K U mv [constant] (9.21)
2 r
If the only force acting on the body is the gravitational force (no rocket engine or
other external force!), then this total energy remains constant during the motion. For
instance, the energy (9.21) is constant for a planet orbiting the Sun, and for a satellite or
a spacecraft (with rocket engines shut off) orbiting the Earth. As we saw in Chapter 8,
examination of the energy reveals some general features of the motion. Equation (9.21)
shows how the orbiting body trades distance (“height”) for speed; it implies that if r
1 2
decreases, v must increase, so that the sum of the two terms 2 mv and GMm/r
remains constant. Conversely, if r increases, v must decrease.
Let us now investigate the possible orbits around, say, the Sun from the point of
view of their energy. For a circular orbit, we saw in Eq. (9.10) that the orbital speed is
GM S
v (9.22)
B r
and so the kinetic energy is
GM m
S
2
1
K mv (9.23)
2
2r
Hence the total energy is
GM m GM m GM m
S
S
S
1
2
E K U mv
2
r 2r r
or
GM m
S
E (9.24)
2r
Consequently, the total energy for a circular orbit is negative and is exactly one-half of
the potential energy.
The 1300-kg Syncom communications satellite was placed in
Concepts EXAMPLE 9 7
in its high-altitude geosynchronous orbit of radius 4.23 10 m
Context
in two steps. First the satellite was carried by the Space Shuttle to a low-altitude
6
circular orbit of radius 6.65 10 m; there it was released from the cargo bay of
the Space Shuttle, and it used its own booster rocket to lift itself to the high-
altitude circular orbit. What is the increase of the total mechanical energy during
this change of orbit?

SOLUTION: The total mechanical energy is exactly one-half of the potential energy
[Eq. (9.24)]. For an Earth orbit, we replace M in Eq. (9.24) by M . For the low-
S E
altitude circular orbit of radius r , the total energy is E GM m 2r , and for
1 1 E 1
the high-altitude circular orbit of radius r , the total energy is E GM m 2r .
2 2 E 2
So the change of the energy is
GM m 1 1
E
1
E E ¢ ≤
2
2 r 2 r 1