Integrate LHS by parts
2
0
0
0
1
1
1
mr
r
mr dt
F rdt
τ
τ
τ
τ
τ
τ
⋅
−
=
⋅
∫
∫
±
±
The first term is zero – since the quantity has the same value at 0 and
τ
.
Thus
2
T
F
= −
⋅
r
where
denote time average of the quantity within brackets.
but
dV
k
r
F
r
r
V
dr
r
V
−
⋅
=
=
=
= −
⋅∇
hence
2
T
V
= −
but
2
2
V
V
E
T
V
V
=
+
= −
+
=
hence
2
V
=
E
but
2
k
E
constant
a
= −
=
and
0
1
2
k
E
Edt
E
a
τ
τ
=
=
= −
∫
so
2
k
E
a
= −
Thus:
k
V
a
= −
as before and therefore
1
2
2
k
T
V
a
= −
=
6.21
The energy of the initial orbit is
2
1
2
2
k
k
mv
E
r
a
−
=
= −
(1)
2
2
1
k
v
m
r
a
=
−
Since
(
1
a
r
a
)
ε
=
+
at apogee, the speed
, at apogee is
1
v
(
)
(
)
(
)
2
1
1
2
1
1
1
k
k
v
m
a
a
ma
ε
ε
ε
−
=
−
=
+
+
To place satellite in circular orbit, we need to boost its speed to
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 Fall '11
 JohnAnderson
 Work, Celestial mechanics, Semimajor axis, Hyperbolic trajectory, denote time average

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