Problem
Set 4: Problem 4.
Problem:
During a flyby of Earth, the speed of a spacecraft is
V
as it reaches its minimum altitude of
1
6
R
above the surface at Point O, where
R
is Earth’s radius. At Point B the spacecraft’s altitude is
4
3
R
.
(a) Verify that the spacecraft’s trajectory is parabolic and determine its velocity at Point O as a
function of
R
, the universal constant of gravitation,
G
, and Earth’s mass,
M
.
(b) Using the energyconservation principle, compute
V
A
. Express your answer as a function of
V
.
Solution: (a)
In general, the shape of the trajectory is
1
r
=
GM
h
2
(1 +
6
cos
θ
)
At Point O,
r
=
7
6
R
and
θ
=0
o
,sotha
t
6
7
R
=
GM
h
2
(1 +
6
·
1) =
GM
h
2
(1 +
6
)
At Point B,
r
=
7
3
R
and
θ
=90
o
, which gives
3
7
R
=
GM
h
2
(1 +
6
·
0) =
GM
h
2
Combining these two equations yields
6
7
R
=
w
3
7
R
W
(1 +
6
)=
⇒
2=1+
6
Therefore, we conclude that
6
=1
wherefore the orbit is indeed parabolic. Consequently, its velocity at Point O, which is the perigee, is the
escape velocity, i.e.,
V
=
5
2
GM
r
O
=
5
12
7
GM
R
wherewehaveusedthefac
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 '06
 Shiflett
 Gm Ro Ro, GM GM, mvO rA rO

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