Problem: Orbits with a 13 central potential
The equation of motion for the onedimensional radial problem is
μ
··
r
=
f
ef
(
r
)=
−
dV
dr
+
l
2
μr
3
,
(1)
where
l
is the constant angular momentum,
μr
2
·
θ
,and
μ
is the reduced mass of
the two particle system. The 13 potential is de
f
ned as
V
(
r
)=
−
k
r
−
a
3
r
3
,
(2)
where
k
and
a
are positive contants. With
a
=0
, we recover the unperturbed
Kepler problem. The attractive
r
−
3
term arises from general relativistic correc
tions to Newtonian dynamics. It causes the Keplerian orbits to precess. The
radial equation of motion can be written explicitly as
μ
··
r
=
f
ef
(
r
)=
−
k
r
2
−
a
r
4
+
l
2
μr
3
=
−
dV
0
dr
,
(3)
where
V
0
is the e
f
ective potential for this problem,
V
0
=
V
(
r
)+
l
2
2
μr
2
.
(4)
(a) Circular Orbits:
The circular orbit condition is that the e
f
ective force
must vanish when
r
=
r
c
,
f
ef
(
r
c
)=
−
[
dV
0
(
r
c
)
/dr
c
]=0
,where
r
c
is the radius
of the circular orbit,
f
ef
(
r
c
)=
−
k
r
2
c
−
a
r
4
c
+
l
2
μr
3
c
=0
.
(5)
This equation can be simpli
f
ed and solved as follows. First, introduce a scaled
radial coordinate
x
c
=
r
c
/r
0
,
(6)
where
r
0
is the radius of the circular orbit when
a
=0
,
r
0
=
l
2
/μk.
(7)
Then Eq.(5) reduces to
x
−
3
c
−
x
−
2
c
−
cx
−
4
c
=0
,
(8)
where
c
is the positive, dimensionless constant
c
=
a/
(
kr
2
0
)
.
(9)
Then, after multiplying Eq.(8) by
x
4
c
, we obtain a quadratic equation for the
variable
x
c
,
x
2
c
−
x
c
+
c
=0
,
(10)
1
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View Full Documentwith solutions given by the quadratic formula
2
x
c
=1
±
√
1
−
4
c.
(11)
Real roots are possible only for
c
≤
1
/
4
. The larger root always corresponds
to a stable circular orbit (see the subsequent analysis and plot of
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 Fall '10
 Wilemski
 mechanics, Angular Momentum, Mass, Momentum, Orbits, Quadratic equation, circular orbit, Kepler's laws of planetary motion, Celestial mechanics, Elliptic orbit

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