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81. Energy conservation for this situation may be expressed as follows:
22
11
2 2
1
2
12
GmM
GmM
KU KU
m
v
m
v
rr
+=+
¡
−=−
where
M
= 5.98
×
10
24
kg,
r
1
=
R
= 6.37
×
10
6
m and
v
1
= 10000 m/s. Setting
v
2
= 0 to
find the maximum of its trajectory, we solve the above equation (noting that
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Energy

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