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```equationâ ÎU = mg Îy gives a 1.0% error in the change in the potential energy. What
is this height?
Answer Because the surface equation assumes a constant value for g, it will give a
ÎU value that is larger than the value given by the general equation, Equation 13.13.
Set up a ratio reflecting a
1.0% error:
Substitute the expressions
for each of these changes
ïU surface
ï½ 1.010
ïU general
gr r
mg ïy
ï½ i f ï½ 1.010
GM E m ï¨ ïy ri rf ï© GM E
ÎU:
Substitute for ri, rf, and g
ï¨GM
E
GM E
from Equation 13.5:
Solve for Îy:
ï½
RE ï« ïy
ïy
ï½ 1ï«
ï½ 1.010
RE
RE
Îy = 0.010RE = 0.010(6.37 Ã 106 m) = 6.37 Ã 104 m =
63.7 km
Example 13.7 Changing the Orbit of a Satellite
A space transportation vehicle releases a 470-kg communications satellite while in an
orbit 280 km above the surface of the Earth. A rocket engine on the satellite boosts it
into a geosynchronous orbit. How much energy does the engine have to provide?
SOLUTION
Conceptualize Notice that the height of 280 km is much lower than that for a
geosynchronous satellite, 36 000 km, as mentioned in Example 13.5. Therefore,
energy must be expended to raise the satellite to this much higher position.
Categorize This example is a substitution problem.
Find the initial radius of ri = RE + 280 km = 6.65 Ã 106 m
the satelliteâs orbit
when it is still in the
```
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