Three Body Problem for Constant Angular
Velocity
P. Coulton∗
Department of Mathematics
Eastern Illinois University
Charleston, Il 61920
∗
Corresponding author, E-mail: cfprc@eiu.edu
1
Abstract
•
•
•
We give the conditions for a periodic
orbit condition to hold for three distinct masses and show that every distribution 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 m1 , m2 and m3 . Assume
that the particle m1 rests at the point (r1, 0)
on the x-axis and that the particle m2 lies on
the circle of radius r2 at a positive angle θ
from the x-axis.
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 m1 , m2 and m3 . Assume
that the particle m1 rests at the point (r, 0)
on the x-axis and that the particle m2 lies on
the circle of radius r at a positive angle θ
from the x-axis.
Assume that the particle m3 lies on the
circle of radius r3 at a negative angle φ from
the x-axis. The angle between m2 and m1 is
2π − θ − φ.
5
The vectors between each pair of particles is
a12 = (r2 cos θ − r1 )r̂1 + r2 sin θr̂2 ,
a13 = (r3 cos φ − r1 )r̂1 + r3 sin φr̂3 ,
a23 = (r3 cos(θ + φ) − r2 )r̂2 + r3 sin(θ + φ)r̂3 .
6
The vectors between each pair of particles satisfy
a12 = (r2 cos θ − r1 )r̂1 + r2 sin θr̂2 ,
a13 = (r3 cos φ − r1 )r̂1 + r3 sin φr̂3 ,
a23 = (r3 cos(θ + φ) − r2 )r̂2 + r3 sin(θ + φ)r̂3 .
The distance between the particles is given
by
q
a12 =
a13 =
a23 =
q
q
(r1 − r2 cos(θ))2 + r22 sin2 (θ)
(r1 − r3 cos(φ))2 + r32 sin2 (φ),
(r2 − r3 cos(φ + θ))2 + r32 sin2 (φ + θ),
7
If
r̂ij =
aij
.
aij
Then by Newton’s gravitational law
F1 =
Km1 m2
Km1 m3
r̂12 +
r̂13 ,
3
a12
a313
F2 = −
Km2 m3
Km1 m2
r̂12 +
r̂23 ,
3
a12
a323
F3 = −
Km2 m3
Km1 m3
r̂13 −
r̂23 .
3
a13
a323
8
If
r̂ij =
aij
.
aij
Then by Newton’s gravitational law
F1 =
Km1 m2
Km1 m3
r̂12 +
r̂13 ,
3
a12
a313
F2 = −
Km2 m3
Km1 m2
r̂12 +
r̂23 ,
3
a12
a323
F3 = −
Km2 m3
Km1 m3
r̂13 −
r̂23 .
3
a13
a323
Now observe that if we place the m1 particle on the positive x-axis, then the ĵ component of F1 must vanish if the force is radial,
therefore
m2 r2 sin θ
m3 r3 sin φ
=
,
3
a12
a313
and
F1 = |F1 | = Km1 [
m2 (r2 cos θ − r1 ) m3 (r3 cos θ − r1 )
+
].
a312
a313
9
Apply the same reasoning for each particle in turn. We obtain the radial mass
relations:
m3 r3 sin φ
m2 r2 sin θ
=
,
a312
a313
m2 r2 sin(θ + φ)
m1 r1 sin φ
=
,
3
a23
a313
m3 r3 sin(θ + φ)
m1 r1 sin θ
=
.
3
a23
a312
10
Continuing this observation the force equations become :
F1 = |F1 | = Km1 [
m2 (r2 cos θ − r1 ) m3 (r3 cos θ − r1 )
+
].
a312
a313
F2 = |F2 | = Km2 [
m1 (r1 cos θ − r2 ) m3 (r3 cos(θ + φ) − r2 )
+
],
a312
a323
F3 = |F3 | = Km3 [
m2 (r2 cos(θ + φ) − r3 ) m1 (r1 cos φ − r3 )
+
].
a312
a313
11
Continuing this observation the force equations become :
F1 = |F1 | = Km1 [
m2 (r2 cos θ − r1 ) m3 (r3 cos θ − r1 )
+
].
a312
a313
F2 = |F2 | = Km2 [
m1 (r1 cos θ − r2 ) m3 (r3 cos(θ + φ) − r2 )
+
],
a312
a323
F3 = |F3 | = Km3 [
m2 (r2 cos(θ + φ) − r3 ) m1 (r1 cos φ − r3 )
+
].
a312
a313
The angular velocity equations are:
[
v1 2 K m2 (r1 − r2 cos θ) m3 (r1 − r3 cos θ)
] = [
+
],
r1
r1
a312
a313
[
v2 2 K m1 (r2 − r1 cos θ) m3 (r2 − r3 cos(θ + φ))
] = [
+
],
r2
r2
a312
a323
[
v3 2 K m2 (r3 − r2 cos(θ + φ)) m1 (r3 − r1 cos φ)
+
].
] = [
r3
r3
a312
a313
12
The problem involves solving the non-linear
system of equations:
m3 r3 sin φ
m2 r2 sin θ
=
,
a312
a313
m1 r1 sin φ
m2 r2 sin(θ + φ)
=
,
3
a23
a313
m3 r3 sin(θ + φ)
m1 r1 sin θ
=
.
3
a23
a312
K m2 (r1 − r2 cos θ) m3 (r1 − r3 cos θ)
vi
+
],
[ ]2 = ω 2 = [
ri
r1
a312
a313
=
K m1 (r2 − r1 cos θ) m3 (r2 − r3 cos(θ + φ))
+
],
[
r2
a312
a323
=
K m2 (r3 − r2 cos(θ + φ)) m1 (r3 − r1 cos φ)
[
+
].
r3
a323
a313
13
•
•
WOLOG we may choose
to obtain 4 equations
m1 = 1, r1 = 1
m2 r2 sin θ
m3 r3 sin φ
=
,
2
3/2
(1 + r2 − 2r2 cos θ)
(1 + r32 − 2r3 cos φ)3/2
(r22
[
sin φ
m2 r2 sin(θ + φ)
=
,
2
3/2
+ − 2r3 r2 cos(φ + θ))
(1 + r3 − 2r3 cos φ)3/2
r32
m2 (1 − r2 cos θ)
m3 (1 − r3 cos θ)
+
]=
2
3/2
(1 + r2 − 2r2 cos θ)
(1 + r32 − 2r3 cos θ)3/2
=
=
m3 (r2 − r3 cos(θ + φ))
1
m1 (r2 − cos θ)
+ 2
],
[
2
3/2
r2 (1 + r2 − 2r2 cos θ)
(r2 + r32 − 2r3 r2 cos(φ + θ))3/2
1
m2 (r3 − r2 cos(θ + φ))
m1 (r3 − cos φ)
[ 2
].
+
2
3/2
r3 (r2 + r3 − 2r3 r2 cos(φ + θ))
(1 + r32 − 2r3 cos φ)3/2
14
•
•
the third radial equation is now redundant
This implies two degrees of freedom in
the solution.
m2 r2 sin θ
m3 r3 sin φ
=
,
2
3/2
(1 + r2 − 2r2 cos θ)
(1 + r32 − 2r3 cos θ)3/2
(r22
sin φ
m2 r2 sin(θ + φ)
=
,
2
3/2
+ − 2r3 r2 cos(φ + θ))
(1 + r3 − 2r3 cos φ)3/2
(r22
m3 r3 sin(θ + φ)
sin θ
=
,
2
3/2
+ − 2r3 r2 cos(φ + θ))
(1 + r3 − 2r3 cos θ)3/2
r32
r32
15
•
•
4 equations and 6 unknowns.
two degrees of freedom.
m2 r2 sin θ
m3 r3 sin φ
=
,
2
3/2
(1 + r2 − 2r2 cos θ)
(1 + r32 − 2r3 cos θ)3/2
(r22
[
sin φ
m2 r2 sin(θ + φ)
=
,
2
3/2
+ − 2r3 r2 cos(φ + θ))
(1 + r3 − 2r3 cos θ)3/2
r32
m2 (1 − r2 cos θ)
m3 (1 − r3 cos θ)
+
]=
2
3/2
(1 + r2 − 2r2 cos θ)
(1 + r32 − 2r3 cos θ)3/2
=
1
m1 (r2 − cos θ)
m3 (r2 − r3 cos(θ + φ))
[
+ 2
],
2
3/2
r2 (1 + r2 − 2r2 cos θ)
(r2 + r32 − 2r3 r2 cos(φ + θ))3/2
=
m2 (r3 − r2 cos(θ + φ))
1
m1 (r3 − cos φ)
[ 2
].
+
2
r3 (r2 + r3 − 2r3 r2 cos(φ + θ))3/2 (1 + r32 − 2r3 cos θ)3/2
16
If r1 = r2 = r3 , then we have an equilateral
triangle with rotation about the center point.
The masses are equal and
v=
s
and
where
cles.
√
v
ω= =
r
3r = d
Km
,
d
s
3Km
.
d3
is the distance between parti-
17
2. The Centriod Case
Let
M = m 1 + m2 + m3
and assume that
m1 r1 + m2 r2 + m3 r3 = 0.
We have
F1 =
m1 m2 (−r1 + r2 ) m1 m3 (−r1 + r3 )
+
a3
a3
m1 m2 (−r2 + r1 ) m2 m3 (−r2 + r3 )
+
a3
a3
m1 m3 (−r3 + r1 ) m3 m2 (−r3 + r2 )
F3 =
+
a3
a3
F2 =
18
Fig.1 Centroid configuration
19
Replacing −m2 r2−m3 r3 by m1 r1 in the first gravitational force equation we obtain
F1 = −
m1 m2 r1 m1 m3 r1 m21 r1
−
− 3
a3
a3
a
=−
M m1 r1
.
a3
20
Now the gravitational equations become
F1 = −
M m1 r1
.
a3
21
Now the gravitational equations become
F1 = −
M m1 r1
.
a3
F2 = −
M m2 r2
a3
22
Now the gravitational equations become
F1 = −
M m1 r1
.
a3
M m2 r2
a3
M m3 r3
F3 = −
.
a3
F2 = −
23
The centripetal force equations take the
form
mi ω 2 r i =
24
M mi ri
,
a3
The centripetal force equations take the
form
M mi ri
,
a3
mi ω 2 r i =
Which implies that
ω=
s
KM
,
a3
This is a complete solution.
25
We now consider the case such that
• the distance between m1 and m2 is a,
26
We now consider the case such that
• the distance between m1 and m2 is a,
• between m1 and m3 is η 1/3 a,
27
We now consider the case such that
• the distance between m1 and m2 is a,
• between m1 and m3 is η 1/3 a
• and the distance between m2 and m3 is
28
ν 1/3 a.
Following the derivation for the centroid
above we obtain
F1 =
m1 m2 (−r1 + r2 ) m1 m3 (−r1 + r3 )
+
a3
ηa3
29
Following the derivation for the centroid
above we obtain
F1 =
m1 m2 (−r1 + r2 ) m1 m3 (−r1 + r3 )
+
a3
ηa3
F2 =
m1 m2 (−r2 + r1 ) m2 m3 (−r2 + r3 )
+
a3
νa3
30
Following the derivation for the centroid
above we obtain
F1 =
m1 m2 (−r1 + r2 ) m1 m3 (−r1 + r3 )
+
a3
ηa3
m1 m2 (−r2 + r1 ) m2 m3 (−r2 + r3 )
+
a3
νa3
m1 m3 (−r3 + r1 ) m3 m2 (−r3 + r2 )
F3 =
+
ηa3
νa3
F2 =
31
The condition that the forces be directed
to the center is given by
m2 r 2 +
m3
r3 = −τ1 r1 ,
η
32
The condition that the forces be directed
to the center is given by
m2 r 2 +
m3
r3 = −τ1 r1 ,
η
m1 r 1 +
m3
r3 = −τ2 r2 ,
ν
33
The condition that the forces be directed
to the center is given by
m2 r 2 +
m3
r3 = −τ1 r1 ,
η
m3
r3 = −τ2 r2 ,
ν
m2
m1
r1 +
r2 = −τ3 r3 .
η
ν
m1 r 1 +
By elimentary linear analysis these equations must be multiples when the points are
non-linear.
34
Therefore,
τ1 r1 + m 2 r2 +
m3
m3
r3 = t(m1 r1 + τ2 r2 +
r3 )
η
ν
= s(
35
m1
m2
r1 +
r2 + τ3 r3 ).
η
ν
Therefore,
τ1 r1 + m 2 r2 +
m3
m3
r3 = t(m1 r1 + τ2 r2 +
r3 )
η
ν
= s(
m1
m2
r1 +
r2 + τ3 r3 ).
η
ν
Apparently (by the first identity)
t=
ν
,
η
and
τ1 =
ν
m1 ,
η
36
and
τ2 =
η
m2
ν
Therefore,
τ1 r1 + m 2 r2 +
m3
m3
r3 = t(m1 r1 + τ2 r2 +
r3 )
η
ν
= s(
m1
m2
r1 +
r2 + τ3 r3 ).
η
ν
Apparently (by the first identity)
t=
ν
,
η
and
τ1 =
ν
m1 ,
η
and
τ2 =
η
m2
ν
and by the second identity
s = ν,
and
τ3 =
1
m3 .
νη
The centripetal force equations become
37
Therefore,
τ1 r1 + m 2 r2 +
m3
m3
r3 = t(m1 r1 + τ2 r2 +
r3 )
η
ν
= s(
m1
m2
r1 +
r2 + τ3 r3 ).
η
ν
The centripital force equations become
m1 ω 2 r 1 = (
m1 m2 m1 m3 m21 ν
+
+
)r1 ,
a3
ηa3
ηa3
38
Therefore,
τ1 r1 + m 2 r2 +
m3
m3
r3 = t(m1 r1 + τ2 r2 +
r3 )
η
ν
= s(
m1
m2
r1 +
r2 + τ3 r3 ).
η
ν
The centripital force equations become
m1 ω 2 r 1 = (
m1 m2 m1 m3 m21 ν
+
+
)r1 ,
a3
ηa3
ηa3
m2 ω 2 r 2 = (
m1 m2 m2 m3 m22 η
+
+
)r2 ,
a3
νa3
νa3
39
Therefore,
τ1 r1 + m 2 r2 +
m3
m3
r3 = t(m1 r1 + τ2 r2 +
r3 )
η
ν
= s(
m1
m2
r1 +
r2 + τ3 r3 ).
η
ν
The centripital force equations become
m1 ω 2 r 1 = (
m1 m2 m1 m3 m21 ν
+
+
)r1 ,
a3
ηa3
ηa3
m1 m2 m2 m3 m22 η
+
+
)r2 ,
a3
νa3
νa3
m1 m3 m2 m3
m23
m3 ω 2 r 3 = (
+
+
)r3 .
ηa3
νa3
ηνa3
m2 ω 2 r 2 = (
40
We simplify this to obtain the relations:
m2 +
m3 m1 ν
m3 m2 η
+
= m1 +
+
η
η
ν
ν
41
We simplify this to obtain the relations:
m2 +
m3 m1 ν
m3 m2 η
+
= m1 +
+
η
η
ν
ν
=
m1 m2 m3
+
+
.
η
ν
ην
42
We obtain the relations
ηm2 (ν − η) + m3 (ν − η) + νm1 (ν − η) = 0,
and
43
We obtain the relations
ηm2 (ν − η) + m3 (ν − η) + νm1 (ν − η) = 0,
and
m3 (ν − 1) + ηm2 (ν − 1) + νm1 (ν − 1) = 0.
44
We obtain the relations
ηm2 (ν − η) + m3 (ν − η) + νm1 (ν − η) = 0,
and
m3 (ν − 1) + ηm2 (ν − 1) + νm1 (ν − 1) = 0.
We observe that the only solution is
η = ν = 1.
This implies that every solution with constant angular velocity about a fixed point
is an equilateral triangle solution.
45
3. The Isoceles-Mass Case
Fig.1 Mass configuration
46
Take m1 = 1, and m2 = m3 = m. The condition
that the masses lie on an equilateral triangle reduces to
8r 3 (1 − cos2 θ)3/2 = (1 + r 2 − 2r cos θ)3/2 .
47
Take m1 = 1, and m2 = m3 = m. The condition
that the masses lie on an equilateral triangle reduces to
8r 3 (1 − cos2 θ)3/2 = (1 + r 2 − 2r cos θ)3/2 .
Solving for
cos θ
we obtain
cos θ =
1−
√
12r 2 − 3
.
4r
48
Take m1 = 1, and m2 = m3 = m. The condition
that the masses lie on an equilateral triangle reduces to
8r 3 (1 − cos2 θ)3/2 = (1 + r 2 − 2r cos θ)3/2 .
Solving for
cos θ
we obtain
cos θ =
1−
√
12r 2 − 3
.
4r
Using the equilateral condition and replacing cos θ in the radial mass relation we have
m=−
1
2
=√ 2
2r cos θ
12r − 3 − 1
49
Finally,
m=−
ω=
v
u
u
t
1
2
=√ 2
2r cos θ
12r − 3 − 1
√
√
2 2K(3 + 12r 2 − 3)
√
√
( 12r 2 − 3 − 1)(1 + 2r 2 + 12r 2 − 3)3/2
50
Finally,
m=−
ω=
v
u
u
t
1
2
=√ 2
2r cos θ
12r − 3 − 1
√
√
2 2K(3 + 12r 2 − 3)
√
√
( 12r 2 − 3 − 1)(1 + 2r 2 + 12r 2 − 3)3/2
We observe the following relations:
lim
√ m = ∞,
r→
1/3
51
Finally,
m=−
ω=
v
u
u
t
1
2
=√ 2
2r cos θ
12r − 3 − 1
√
√
2 2K(3 + 12r 2 − 3)
√
√
( 12r 2 − 3 − 1)(1 + 2r 2 + 12r 2 − 3)3/2
We observe the following relations:
lim
√ m = ∞,
r→
1/3
lim
√ cos θ = 0,
r→
1/3
52
Finally,
m=−
ω=
v
u
u
t
1
2
=√ 2
2r cos θ
12r − 3 − 1
√
√
2 2K(3 + 12r 2 − 3)
√
√
( 12r 2 − 3 − 1)(1 + 2r 2 + 12r 2 − 3)3/2
We observe the following relations:
lim
√ m = ∞,
r→
1/3
lim
√ cos θ = 0,
r→
1/3
lim cos θ =
r→∞
53
√
3/2.
4. The Scalene-Mass Case
Fig.1 Mass configuration
54
We recall that the solution requires that
the masses be equidistant at a distance, say
a.
• The distance from the centriod to m1 is
r1 = r .
55
We recall that the solution requires that
the masses be equidistant at a distance, say
a.
• The distance from the centriod to m1 is
r1 = r .
• Define the angle ψ to be the angle between the centroid and the line joining
m1 and m2 .
56
We recall that the solution requires that
the masses be equidistant at a distance, say
a.
• The distance from the centriod to m1 is
r1 = r .
• Define the angle ψ to be the angle between the centroid and the line joining
m1 and m2 .
• r2 /r1 = α,
57
We recall that the solution requires that
the masses be equidistant at a distance, say
a.
• The distance from the centriod to m1 is
r1 = r .
• Define the angle ψ to be the angle between the centroid and the line joining
m1 and m2 .
• r2 /r1 = α,
• and r3 /r1 = β .
58
Fig.1 Mass configuration
Let the centroid be at the origin and consider the following configuration:
r1 = r î,
59
Fig.1 Mass configuration
Let the centroid be at the origin and consider the following configuration:
r1 = r î,
r2 = (αr − a cos ψ)î + a sin ψ ĵ,
60
Fig.1 Mass configuration
Let the centroid be at the origin and consider the following configuration:
r1 = r î,
r2 = (αr − a cos ψ)î + a sin ψ ĵ,
r3 = (βr − a cos(π/3 − ψ))î + a sin(π/3 − ψ)ĵ,
61
The centroid equations lead to
m1 r + m2 (αr − a cos ψ) + m3 (βr − a cos(π/3 − ψ)) = 0,
and
m2 sin ψ − m3 sin(π/3 − ψ) = 0.
We obtain the mass relations
m3 =
and
2m2 sin ψ
√
,
cos ψ − 3 sin ψ
m1 = m2 [a cos ψ − αr +
2 sin ψ(βr − a cos(π/3 − ψ))
√
].
cos ψ − 3 sin ψ
62
In addition we have a triangle determined
by the sides r1 = r, r2 = αr and a. By the law of
cosines we have
α=
and
β=
1q 2
a + r 2 − 2ar cos ψ
r
1q 2
a + r 2 − 2ar cos(π/3 − ψ).
r
63
Finally the interior angles are given by
θ = arcsin(
and
φ = arcsin(
a sin ψ
),
αr
a sin(π/3 − ψ)
)
βr
ω=
s
KM
a3
64
This completes the classification of the
three body scalene mass distribution in terms
of r and ψ for constant angular momentum.
65
5. The Colinear Case
Fig.4 Mass configuration
66
We assume that:
• The masses lie in a common line;
67
We assume that:
• The masses lie in a common line;
•
The masses satisfy
M >> m >> µ;
68
We assume that:
• The masses lie in a common line;
•
The masses satisfy
M >> m >> µ;
•
The angular velocity is constant;
69
We assume that:
• The masses lie in a common line;
•
The masses satisfy
•
The angular velocity is constant;
M lies essentially at the centroid and
axis of rotation.
•
M >> m >> µ;
70
In addition, for simplicity we take the distance between M and m as 1, and K = 1. Now
we have
Fm = M m = mω 2 ,
71
In addition, for simplicity we take the distance between M and m as 1, and K = 1. Now
we have
Fm = M m = mω 2 ,
or
ω 2 = M,
72
In addition, for simplicity we take the distance between M and m as 1, and K = 1. Now
we have
Fm = M m = mω 2 ,
or
ω 2 = M,
and
Fµ =
mµ
ω2µ
− 2 = µω 2 (1 − x).
2
(1 − x)
x
We wish to solve for x.
73
Gives
m
ω2
− 2 = ω 2 (1 − x).
Fµ =
2
(1 − x)
x
ω 2 x2 − m(1 − x)2 = ω 2 x2 (1 − x)3 .
74
Gives
or
m
ω2
− 2 = ω 2 (1 − x).
Fµ =
2
(1 − x)
x
ω 2 x2 − m(1 − x)2 = ω 2 x2 (1 − x)3 .
ω 2 x2 − m + 2mx − mx2 = ω 2 (x2 − 3x3 + 3x4 − x5 ),
75
Gives
or
m
ω2
− 2 = ω 2 (1 − x).
Fµ =
2
(1 − x)
x
ω 2 x2 − m(1 − x)2 = ω 2 x2 (1 − x)3 .
ω 2 x2 − m + 2mx − mx2 = ω 2 (x2 − 3x3 + 3x4 − x5 ),
which reduces to (replace
ω2
by
M)
M x5 − 3M x4 + 3M x3 − mx2 + 2mx − m = 0.
76
Consider the case: M = 103 , m = 1 which closely
approximates the earth and sun. Then our
equation
6
M x5 − 3M x4 + 3M x3 − mx2 + 2mx − m = 0.
77
Consider the case: M = 103 , m = 1 which closely
approximates the earth and sun. Then our
equation
6
M x5 − 3M x4 + 3M x3 − mx2 + 2mx − m = 0.
becomes
106 5
x − 106 x4 + 106 x3 − x2 + 2x − 1 = 0.
3
78
Consider the case: M = 103 , m = 1 which closely
approximates the earth and sun. Then our
equation
6
M x5 − 3M x4 + 3M x3 − mx2 + 2mx − m = 0.
becomes
106 5
x − 106 x4 + 106 x3 − x2 + 2x − 1 = 0.
3
which has approximate solution
10−2 ,
2
10−4
− 10−2 + 1 − 10−4 + −2 − 1 ≈ 0.
3
10
79
or
Consider the case: M = 103 , m = 1 which closely
approximates the earth and sun. Then our
equation
6
M x5 − 3M x4 + 3M x3 − mx2 + 2mx − m = 0.
becomes
106 5
x − 106 x4 + 106 x3 − x2 + 2x − 1 = 0.
3
which has approximate solution
10−2 ,
or
2
10−4
− 10−2 + 1 − 10−4 + −2 − 1 ≈ 0.
3
10
Thus x ≈ 1, 000, 000 miles. This point is essentially independent of µ, the mass of the satellite.
80
Another interesting case
Fig.5 Mass configuration
81
In this case we have
M = ω2
82
and
In this case we have
M = ω2
and
Mµ
mµ
+
= µω 2 (1 + x)
2
2
(1 + x)
x
83
Then
m
M
+ 2 = M (1 + x),
2
(1 + x)
x
and
84
Then
and
or
m
M
+ 2 = M (1 + x),
2
(1 + x)
x
M x2 + m(1 + x)2 = M x2 (1 + x)3 ,
m + 2mx + x2 = 3M x3 + 3M x4 + M x5 ,
85
Then
m
M
+ 2 = M (1 + x),
2
(1 + x)
x
and
M x2 + m(1 + x)2 = M x2 (1 + x)3 ,
or
Let M
comes
m + 2mx + x2 = 3M x3 + 3M x4 + M x5 ,
=
106
3
and
m = 1
then our equation be106 5
x ,
3
solution x = 10−2,
1 + 2x + x2 = 106 x3 + 106 x4 +
which has approximate
86
Then
m
M
+ 2 = M (1 + x),
2
(1 + x)
x
and
M x2 + m(1 + x)2 = M x2 (1 + x)3 ,
or
Let M
comes
m + 2mx + x2 = 3M x3 + 3M x4 + M x5 ,
=
106
3
and
m = 1
then our equation be106 5
x ,
3
solution x = 10−2,
1 + 2x + x2 = 106 x3 + 106 x4 +
which has approximate
1+
2
1
1
10−4
+
=1+
+
,
100 10000
100
3
87
or
Let M
comes
=
106
3
and
m=1
then our equation be106 5
x ,
3
solution x = 10−2,
1 + 2x + x2 = 106 x3 + 106 x4 +
which has approximate
1+
or
2
1
1
10−4
+
=1+
+
,
100 10000
100
3
Once again x ≈ 1, 000, 000 miles. This completes
our analysis of the colinear case in the plane.
88