Physics 325
Spring 2011
Discussion 6
February 28, 2011
A particle of mass
μ
and initial angular momentum
‘
is moving in a central force field
F
(
r
) =
kr,
with
k >
0
. Starting from the central force equation
u
00
+
u
=

μ
‘
2
u
2
F
1
u
,
where
u
0
=
du/dθ
and
u
= 1
/r
, we want to solve for the particle’s path using the following
steps.
(a) If
u
0
= 0
at
u
= 1
/s
, solve for
(
u
0
)
2
as a function of
u
. Here,
s
is given by
s
=
‘
2
μk
1
/
4
(b) Assuming
u
0
≥
0
, integrate once more to solve for
u
as a function of
θ
. Use initial
condition
r
2
sin
α
=
s
2
at
θ
= 0
, for a given
α
.
Hint:
Make the substitution
u
2
=
a
2
sin
φ
, where
a
is some constant chosen appro
priately.
(c) Using
x
=
r
cos
θ
and
y
=
r
sin
θ
, show that the previous result can be rewritten as
s
2
= (
x
2

y
2
) sin
α
+ 2
xy
cos
α.
(d) Let
σ
2
+
λ
2
= 1
and
sin
α
=
σ
2

λ
2
,
cos
α
= 2
σλ.
(We need
σ
2
+
λ
2
= 1
so that
sin
2
α
+ cos
2
α
= 1
.)
Show that part (c)’s result can also be written as
s
2
= ˜
x
2

˜
y
2
,
where
˜
x
=
σx
+
λy,
˜
y
=
λx

σy.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Staff
 Physics, mechanics, Angular Momentum, Force, Mass, Momentum, The Central

Click to edit the document details