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MA 36600 LECTURE NOTES: MONDAY, JANUARY 26
This is the implicit solution for the initial value problem. (The lefthand side is the change in kinetic energy,
whereas the righthand side is the change in potential energy.)
This expression states how the distance
x
=
x
(
t
) is related to the velocity
v
=
v
(
t
). As
t
→∞
, we want
x
(
t
)
→∞
and
v
(
t
)
≥
0; this means the rocket can travel inFnitely far away from the Earth and never reverse
its velocity. This gives the inequality
−
1
2
mv
2
0
≤
lim
t
→∞
°
1
2
mv
(
t
)
2
−
1
2
mv
2
0
±
=l
im
t
→∞
°
mgR
2
x
(
t
)+
R
−
mgR
±
=
−
mgR
which reduces to
1
2
mv
2
0
≥
mgR
=
⇒
v
0
≥
²
2
gR.
The expression
v
e
=
²
2
gR
= 11
.
19 km
/
sec = 6
.
95 miles/sec
is the escape velocity.
Is it possible to reach the escape velocity in a commercial jet plane, or even in the space shuttle? The
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.
 Spring '09
 EdrayGoins

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