Chapter 11 Gravity

Gravity
Picture the Problem Let the zero of gravitational potential energy be at infinity
and let m represent the mass of the object. We’ll use conservation of energy to
relate the initial potential energy of the object-Earth system to the final potential
and kinetic energies.
Use conservation of energy to relate
the initial potential energy of the
system to its energy as the object is
about to strike Earth:
Express the potential energy of the
object-Earth system when the object
is at a distance r from the surface of
Earth:
Substitute in equation (1) to obtain:
Solving for v yields:
Substitute numerical values and evaluate v:
v =
Gravitational Orbits
2
231
K f − Ki
+ U f −U
i = 0
or, because Ki = 0,
K ( RE
) + U ( RE
) −U
( RE
+ h)
= 0 (1)
where h is the initial height above
Earth’s surface.
U
() r
mv
GM
= −
r
GM
E
m
m
GM
m
1 2 E
E
2 − + =
RE
RE
+ h
v =
=
⎛ GM
2 ⎜
⎝ RE
E
GM ⎞ E − ⎟
RE
+ h ⎠
⎛ h ⎞
2gR
⎜
⎟
E
⎝ RE
+ h ⎠
2
6
6
( 9.81m/s
)( 6.37×
10 m)(
4.
0×
10 m)
= 6.
9km/s
6.37 × 10
6
m + 4.
0×
10
59 •• Many satellites orbit Earth with maximum altitudes of
1000 km or less. Geosynchronous satellites, however, orbit at an altitude of
35 790 km above Earth’s surface. How much more energy is required to launch a
500-kg satellite into a geosynchronous orbit than into an orbit 1000 km above the
surface of Earth?
Picture the Problem We can express the energy difference between these two
orbits in terms of the total energy of a satellite at each elevation. The application
of Newton’s second law to the force acting on a satellite will allow us to express
6
m
0