Problem GPS 331 (2
nd
ed. 3.26)
Start with Eq.(3.97) (in both 2
nd
and 3
rd
editions), relating the scattering
angle
Θ
(
s
)
to the interaction potential
V
(
r
)
and the impact parameter
s
,
Θ
(
s
) =
π
−
2
s
Z
u
m
0
¡
1
−
V
(1
/u
)
/E
−
s
2
u
2
¢
−
1
/
2
du.
(1)
For the repulsive central force
f
=
kr
−
3
, we
fi
nd easily that
V
(1
/u
) = (
k/
2)
u
2
.
(2)
With Eq.(2), Eq.(1) becomes
Θ
(
s
) =
π
−
2
s
Z
u
m
0
¡
1
−
(
k/
2
E
+
s
2
)
u
2
¢
−
1
/
2
du,
(3)
which we can integrate as soon as we determine the value of
u
m
= (
r
m
)
−
1
.
The distance of closest approach (
periapsis
)
r
m
is the value of
r
at the turning
point of the orbit, i.e., where
·
r
= 0
. This value is easily found from the energy
equation evaluated with
·
r
= 0
and
r
=
r
m
:
E
=
k
2
r
2
m
+
l
2
2
μr
2
m
.
(4)
Upon replacing the angular momentum
l
by the impact parameter,
l
=
s
√
2
μE
,
and solving for
r
m
, we fnd
r
m
= (
u
m
)
−
1
=
p
s
2
+
k/
2
E,
(5)
which allows us to further simplify Eq.(3) to read
Θ
(
s
) =
π
−
2
s
Z
u
m
0
¡
1
−
(
u/u
m
)
2
¢
−
1
/
2
du.
(6)
With the variable substitution,
cos
φ
=
u/u
m
,
(7)
Eq.(6) then becomes
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 Fall '10
 Wilemski
 mechanics, Angular Momentum, SCATTERING, Replacements, Eq.

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