Three Body Problem for Constant Angular Velocity

88 Pages·2012·0.49 MB·English

Three Body Problem for Constant Angular
Velocity
P. Coulton
∗
Department of Mathematics
Eastern Illinois University
Charleston, Il 61920
Corresponding author, E-mail: [email protected]
∗
1
Abstract
We give the conditions for a periodic
• orbit condition to hold for three dis-
tinct masses and show that every distri-
bution has a periodic orbit solution.
We show that the only solution is an
• equilateral arrangement with rotation
about the center of mass.
We also study related n-body problems.
•
2
Assume that we have a configuration of
three bodies rotating at a uniform angular
speed about a common fixed point.
Fig.1 Mass configuration
3
Assume that we have a configuration of
three bodies rotating at a uniform angular
speed about a common fixed point.
Let the masses be , and . Assume
m m m
1 2 3
that the particle rests at the point
m (r ,0)
1 1
on the -axis and that the particle lies on
x m
2
the circle of radius at a positive angle
r θ
2
from the -axis.
x
4
Assume that we have a configuration of
three bodies rotating at a uniform angular
speed about a common fixed point.
Let the masses be , and . Assume
m m m
1 2 3
that the particle rests at the point
m (r,0)
1
on the -axis and that the particle lies on
x m
2
the circle of radius at a positive angle
r θ
from the -axis.
x
Assume that the particle lies on the
m
3
circle of radius at a negative angle from
r φ
3
the x-axis. The angle between and is
m m
2 1
.
2π θ φ
− −
5
The vectors between each pair of parti-
cles is
a = (r cosθ r )ˆr +r sinθˆr ,
12 2 1 1 2 2
−
a = (r cosφ r )ˆr +r sinφˆr ,
13 3 1 1 3 3
−
a = (r cos(θ+φ) r )ˆr +r sin(θ+φ)ˆr .
23 3 2 2 3 3
−
6
The vectors between each pair of parti-
cles satisfy
a = (r cosθ r )ˆr +r sinθˆr ,
12 2 1 1 2 2
−
a = (r cosφ r )ˆr +r sinφˆr ,
13 3 1 1 3 3
−
a = (r cos(θ+φ) r )ˆr +r sin(θ+φ)ˆr .
23 3 2 2 3 3
−
The distance between the particles is given
by
a = (r r cos(θ))2 +r2sin2(θ)
12 1 − 2 2
(cid:113)
a = (r r cos(φ))2 +r2sin2(φ),
13 1 − 3 3
(cid:113)
a = (r r cos(φ+θ))2 +r2sin2(φ+θ),
23 2 − 3 3
(cid:113)
7
If
a
ij
ˆr = .
ij
a
ij
Then by Newton’s gravitational law
Km m Km m
1 2 1 3
F = ˆr + ˆr ,
1 a3 12 a3 13
12 13
Km m Km m
1 2 2 3
F = ˆr + ˆr ,
2 − a3 12 a3 23
12 23
Km m Km m
1 3 2 3
F = ˆr ˆr .
3 − a3 13 − a3 23
13 23
8
If
a
ij
ˆr = .
ij
a
ij
Then by Newton’s gravitational law
Km m Km m
1 2 1 3
F = ˆr + ˆr ,
1 a3 12 a3 13
12 13
Km m Km m
1 2 2 3
F = ˆr + ˆr ,
2 − a3 12 a3 23
12 23
Km m Km m
1 3 2 3
F = ˆr ˆr .
3 − a3 13 − a3 23
13 23
Now observe that if we place the parti-
m
1
cle on the positive -axis, then the ˆ compo-
x j
nent of must vanish if the force is radial,
F
1
therefore
m r sinθ m r sinφ
2 2 3 3
= ,
a3 a3
12 13
and
m (r cosθ r ) m (r cosθ r )
2 2 1 3 3 1
F = F = Km [ − + − ].
1 | 1| 1 a3 a3
12 13
9
Apply the same reasoning for each par-
ticle in turn. We obtain the radial mass
relations:
m r sinθ m r sinφ
2 2 3 3
= ,
a3 a3
12 13
m r sin(θ +φ) m r sinφ
2 2 1 1
= ,
a3 a3
23 13
m r sin(θ+φ) m r sinθ
3 3 1 1
= .
a3 a3
23 12
10