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This is the same expression as derived in Problem 6.3 for a particle dropped into a
hole drilled through the earth.
T
1.4
≈
hours.
6.8
The Earth’s orbit about the Sun is counterclockwise as seen from, say, the north
star. It’s coordinates on approach at the latus rectum are
( ) ( )
,,
xy
a
ε
α
=
−
.
The easiest way to solve this problem is to note that
1
60
=
is small. The
orbit is almost circular!
2
2
S
GM m
mv
rr
∴
=
and
2
S
GM
r
=
v
with
ra
b
=≈≈
when
0
≈
4
310
S
GM
m
v
s
≈=
⋅
More exactly
cos
rv
v
l
β
×=
=
GG
, but
2
ml
k
=
(equation 6.5.19)
Since
S
kG
Mm
=
2
S
l
GM
=
,
hence
()
1
2
cos
S
G
M
αβ
==
lv
Or
1
2
1
cos
S
GM
v
=
The angle
can be calculated as follows:
22
1
yx
ba
+=
(
s
e
e
a
p
p
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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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