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(a)
1
1
1
1
2
dV
r
V
r
dr
r
≈
D
(b)
()
11
1
2
2 60 1%
120%!
dr
r
dV
rr
V
===
D
The approximation of a differential has broken down – a correct result can be obtained by
calculating finite differences, but the implication is clear – a 1% error in boost causes
rocket to miss the moon by a huge factor 
∼
2!
6.16
From section 6.5,
0.967
ε
=
and
mi.
6
55 10
r
=×
D
From equations 6.5.21a&b,
1
1
1
+
=
−
D
6
1
6
15
5
1
0
2
1
1
0.967
93 10
r
mi
AU
ar
r
mi
×
=+
= =
×
−−
×
D
D
1
U
17.92
aA
=
From equation 6.6.5,
3
2
ca
τ
=
33
22
1
17.92
yr AU
AU
−
=⋅
×
3
2
75.9
yr
=
From equation 6.5.21a and 6.5.19 …
2
00
ml
rk
r
α
=−
=
−
2
1
mrv
k
=
DD
−
and
kG
M
m
=
1
2
1
GM
v
r
D
D
From Example 6.5.3 we can translate the factor GM into the more convenient
2
ee
GM
a v
=
… with
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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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