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(a) The ratio of potential energies is
/
1
.
/2
BB
A
AA
B
UG
m
M
r
r
m
M
r
r
−
==
=
−
(b) Using Eq. 1338, the ratio of kinetic energies is
1
.
2
A
B
KG
m
M
rr
m
M
=
(c) From Eq. 1340, it is clear that the satellite with the largest value of
r
has the smallest
value of 
E
 (since
r
is in the denominator). And since the values of
E
are negative, then
the smallest value of 
E
 corresponds to the largest energy
E
. Thus, satellite
B
has the
largest energy.
(d) The difference is
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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

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