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Integrate LHS by parts
2
0
00
11
1
mr r
mr dt
F rdt
ττ
τ
⋅−
= ⋅
∫∫
±±
The first term is zero – since the quantity has the same value at 0 and
.
Thus
2
TF
=−
⋅
r
where
denote time average of the quantity within brackets.
but
dV
k
rF
r
r
V
dr
r
V
−⋅ =
=
=
=
−
⋅∇
hence
2
TV
but
22
VV
ETV
V
=+=
−+=
hence
2
V
=
E
but
2
k
E
constant
a
=
and
0
1
2
k
EE
d
t
E
a
==
=
−
∫
so
2
k
E
a
= −
Thus:
k
V
a
as before and therefore
1
k
a
=
6.21
The energy of the initial orbit is
2
1
kk
mv
E
ra
−==
−
(1)
2
21
k
v
mr a
Since
(
1
a
)
ε
=+
at apogee, the speed
, at apogee is
1
v
()
( )
2
1
1
v
ma
a
m
a
−
=
++
To place satellite in circular orbit, we need to boost its speed to
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.
 Fall '11
 JohnAnderson
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