1
The two-body problem
Here we discuss the motion of two particles under their mutual gravitational
in¢uence. The forces acting on the two bodies are:
GM 1 M 2
F~ i = ¡
(~x i ¡ ~x j )
R3
where R is the separation of the masses, i; j = 1; 2 and j 6= i: Since no external
forces act on the system, the center of mass moves at constant velocity, so we
may choose our origin at the CM. The position of the CM is at a distance s
from the larger mass (call it M 1 ) where
s=
M2
R
M1 + M2
Thus the polar coordinates of the two bodies with respect to this origin are
r1 = s;
r2 = R ¡ s =
M1
M1
R=
r1
M1 + M2
M2
To simplify, we introduce the reduced mass ¹ :
¹=
so that
r1 =
M1 M 2
M1 + M2
¹
¹
R and r2 =
R
M1
M2
Now we write the equation of motion for each body:
M i~ai = ¡
GM 1 M 2
r^
R2
To solve these equations, ¤rst note that the torque about the CM is
~¿ = ~r £ F~ = 0
1
and thus the angular momentum of each body about the CM remains constant.
Mi` = M i r2i µ_ i = Li = constant
and we can solve for µ_ i
Li
= µ_ i
M i r2i
(1)
The velocity and acceleration in polar coordinates are
~vi
=
~ai
=
d
dri
d
dri
d
(rir^) =
r^ + ri r^ =
r^ + ri µ_ i r^ = r_i ^r + ri µ_ i ^µ
(2)
dt
dt
dt
dt
dµ
µ
¶
³
´
¢¢
¢¢
2
2
¢¢
¢¢
rir^ + r_iµ_ i^µ + r_i µ_ i ^µ + riµ i^µ ¡ riµ_ i ^r = ri ¡ riµ_ i r^ + 2 r_iµ_ i + riµ i ^µ
The theta component is zero, which con¤rms our previous result that L is constant. The r-component gives
¢¢
ri
Mj2
2
GM j
GM j
¡ riµ_ i = ¡ 2 = ¡ 2
R
ri (M 1 + M 2 )2
Substituting in for µ_ i; we have
¢¢
ri
¡ ri
µ
Li
Mir2i
¶2
¢¢
= ri ¡
M j2
L2i
GM j
G¹2 M j
=
¡
2 2
2
2 = ¡
M i ri
ri (M1 + M 2)
r2i M i2
Next note that we expect the path to be described by a relation r = r (µ) ; so
we rewrite using
dr _
dr Li 1
r_i =
µi =
dµ
dµ M i r2i
and hence
¢¢
ri
=
=
µ
¶2
µ ¶2
dr_i _
d2 r L i 1
2 dr
Li Li 1
µi = 2
¡
dµ
Mi r2i
r3i dµ
M i M i r2i
dµ
µ
¶2
µ ¶2 µ
¶2
d 2 r Li 1
2 dr
Li 1
¡
ri dµ
M i r2i
dµ2 M i r2i
2
Then the equation of motion becomes:
d2 r
dµ 2
µ
Li 1
Mi ri2
¶2
2
¡
ri
µ
dr
dµ
¶2 µ
Li 1
M i ri2
¶2
d2 r 1
2
2 2 ¡
ri
dµ ri
¡ ri
µ
dr
dµ
µ
¶2
¶2
=
¡
1
1
2 ¡
ri
ri
=
¡G¹2
Li
M iri2
G¹2 M j
r2i Mi2
Mj
L2i
De¤ning u i = 1=ri ; we have
u 0i = ¡
and
r00i =
1 0
r = ¡u2i r0i
ri2 i
2
1
2
(u0i ) ¡ 2 u0i0
3
ui
ui
The equation of motion becomes:
u2i
·
¸
µ 0 ¶2
2
1 00
ui
0 2
3
(u
)
¡
u
¡
2u
¡ ui
i
u 3i i
u2i i
u 2i
= ¡G¹2
u 00i
= ¡ui +
Mj
L2i
G¹2 M j
L2i
(3)
Now note the dimensions of the physical quantities are:
·
¸
GM
L
L3
= 2 ) [GM ] = 2
2
R
T
T
and the angular momentum
[L] = M
and so
·
G¹2 Mj
L2i
¸
L2
T
·
¸
µ
¶2
Mj
L3
T
1
= [G¹] ¹ 2 = 2 ¢ M 2 ¢
=
Li
T
M L2
L
as required.
Equation (3) is easily solved:
µ
¶
d2
G¹2 M j
u
¡
i
L2i
dµ2
G¹2 M j
ui ¡
L2i
=
µ
¶
G¹2 Mj
¡ ui ¡
L2i
=
A cos (µ ¡ µ 0 )
Thus the solution is
ri =
1
=
ui
1
G¹ 2M
L2
i
j
+ A cos (µ ¡ µ 0 )
3
=
L2i =G¹2 M j
1 + B cos (µ ¡ µ0 )
(4)
From this we see that r is a maximum when µ ¡ µ 0 = ¼ and a minimum when
µ = µ0 : The path is a conic section with the major axis de¤ned by µ = µ 0 : We
may choose µ 0 = 0 for convenience.
Since the center of mass remains ¤xed, we have µ2 = ¼ + µ 1 and
M 1 r1
M1 L21 =G¹2
M2 1 + B1 cos µ1
for all µ 1 : Thus, writing ® =
M1
;
M2
=
=
M 2 r2
M2 L22 =G¹2
M1 1 ¡ B 2 cos µ1
the mass ratio, we have
® 2 (1 ¡ B2 cos µ 1 ) =
L22
(1 + B1 cos µ 1 )
L21
This implies that
L2 = ®L1
(5)
and B2 = ¡B1 = ¡B: We still need to ¤nd B:
The total energy is also conserved, and is most easily found at rmax or rmin ;
since at these points r_ = 0
ri; max = a (1 + e) =
L2i
L1 L2
=
(1 ¡ B i) G¹2 M j
(1 ¡ Bi ) G¹2M i
and
ri;min = a (1 ¡ e) =
Thus
ai =
and
(1 ¡
L2i
2
B ) G¹2 M
e=
L2i
(1 + Bi) G¹2 M j
=
j
L1 L2
(1 ¡ B 2 ) G¹2 M i
ri;max ¡ ri;min
=B
2ai
4
(6)
(7)
Then when particle 1 is at r1;max ; r2 is at r2; max
³
´2 1
³
´2
GM1 M2
1
E = ¡
+ M1 r1µ_ 1 + M 2 r2 µ_ 2
R
2
2
µ
¶2
µ
¶2
GM1 M2
1
L1
1
L2
= ¡
¹ + M1
+ M2
M 1 r1
2
M 1 r1
2
M2 r2
µ
¶
µ
¶
¤ M 1 (1 ¡ B) G¹2 2 M 2 (1 ¡ B) G¹2 2
GM1 M2 ¹ £
2
= ¡
(1 ¡ B) G¹ +
+
L1 L2
2
L2
2
L1
=
¡
G2 M1 M2 ¹3
G 2 ¹4 (1 ¡ B)2 M 1
G 2 ¹4 (1 ¡ B)2 M 2
(1 ¡ B) +
+
L1 L2
2
L22
2
L21
=
¡
G2 M1 M2 ¹3
G 2 ¹4 (1 ¡ B) M 2
G 2 ¹4 (1 ¡ B) M 1
(1 ¡ B) +
+
L1 L2
2
L1 L2
2
L1 L2
=
¡
2
2
2
=
Thus
G2 M1 M2 ¹3
G 2 ¹4 (1 ¡ B)
(1 ¡ B) +
(M 1 + M 2 )
L1 L2
2
L1 L2
i
¢
G 2 M 1 M 2 ¹3 h
G 2 M 1 M 2 ¹3 ¡
2
(1 ¡ B) ¡ 2 (1 ¡ B) = ¡
1 ¡ B2
2L1 L2
2L1 L2
1 ¡ B2 +
and so
B=e=
(8)
2L1 L2 E
=0
G 2 M1 M 2 ¹3
s
1+
2L1 L2 E
G2 M 1 M 2 ¹3
(9)
If the system is bound (E < 0) the eccentricity is less than 1 and the path is an
ellipse.
In the limit M2 À M1 ; L2 = ®L1 ¿ L1 and ¹ ' M 1 ; so
s
2L2 E
e ' 1 + 2 12 3
G M2 M1
which agrees with equation E3.4 in Lea and Burke.
We may summarize our results for the path as:
r1;2
=
L1 L2
(1 § e cos µ i) G¹2 M i
(10)
=
a1:2 (1 ¡ e)
(1 § e cos µ i)
(11)
2
Putting our result for e back into equation (7), we have
ai
=
=
³
L1 L2
´
1L 2E
¡ G22L
M 1M2¹ 3
GMj ¹
¡2E
5
G¹2 M i
(12)
1.1
Period
_ Thus the period is
The time needed to travel through an angle dµ is dt = dµ= µ:
T
=
=
=
=
Z
2¼
0
Mi
Li
dµ
µ_
Z 2¼
0
r2i dµ
¡
¢ !2
a1 1 ¡ e2
dµ
1 + e cos µ
0
Z 2¼ µ
¶2
¢
M1 2 ¡
1
2 2
a 1¡e
dµ
L1 1
1 + e cos µ
0
M1
L1
Z
2¼
Ã
We can do the integral by using the unit circle in the complex plane:
I
1
dz
I =
£
¡
¢¤
2
iz
1 + e2 z + 1z
u nit c irc le
I
4
z
=
dz
e2 i
[2z=e + z2 + 1]2
u n it c irc le
The poles are at
1
zp = ¡ §
e
0.2
0.4
x
r
1
¡1
e2
0.6
0.8
1
0
-0.2
-0.4
-0.6
-0.8
-1
zp versus e
Since e < 1; both roots are real, but only one is inside the circle.
6
The
residue is
2
d
(z ¡ z p+ ) z
2
z!zp+ dz [(z ¡ z
p+) (z ¡ zp¡ )]
lim
d
z
2
z!zp+ dz (z ¡ z
p¡ )
1
2z
lim
¡
2
3
z!zp+ (z ¡ z
(z ¡ zp¡ )
p¡ )
z ¡ zp¡ ¡ 2z
lim
3
z!zp+ (z ¡ z
p¡ )
¡ (zp¡ + zp+)
=
lim
=
=
=
(zp+ ¡ zp¡ )3
=
I
=
=
¡
23
¡
¡2=e
1
e2
¡1
¢3=2
=
e2
4 (1 ¡ e2 )3=2
4
e2
2¼i
2
3=2
e i
4 (1 ¡ e 2 )
2¼
3=2
(1 ¡ e 2 )
Thus the period is
T
=
=
=
=
T
=
=
¢2
M1 2 ¡
2¼
a 1 ¡ e2
3=2
L1 1
(1 ¡ e 2 )
¢1=2
M1 2 ¡
2¼
a 1 ¡ e2
L1
µ
¶2 µ
¶1=2
M 1 GM 2 ¹
2L1 L2 E
2¼
¡ 2
L1
¡2E
G M 1 M 2 ¹3
µ
¶ 3=2 µ
¶ 1=2
GM 2 ¹
L2 M 1
2¼
¡2E
L1 G¹2
µ
¶ 3=2 µ
¶ 1=2
GM 2 ¹
M 12
2¼
¡2E
G¹2 M 2
µ
¶
1=2
M 12
3=2
2¼a1
G¹2 M2
(13)
(14)
In this form we can see the dependence on a. In the limit M 2 À M 1 ; we have
¹ ' M1 and
2¼
T 'p
a3=2
GM 2
and we recover Keplerzs Law.
7
Now letzs show the symmetry in the expression for T :
µ
T
1.2
GM 2 ¹
¡2E
¶3=2 µ
M 12
G¹2 M 2
= 2¼
µ
¶r
GM 2 M 1
¹
= ¼
¡2
¡2E
E
¶ 1=2
(15)
Observations
We can observe the period of the system easily, but other quantities are harder.
Letzs rewrite the system energy in terms of the masses and observables, using
equation 12:
GM j ¹
E = ¡2
ai
Thus
T
¶r
GM 2 M 1
¹ai
ai
4
GM j ¹
GM j ¹
s
M2 M 1 3=2
1
= 2¼
a
Mj ¹ i
GM j
r
M1 3=2
1
= 2¼
a
¹ 1
GM 2
= ¼
µ
If we have a visual binary and we know the distance to the system, we will have
a value for a1 ; and so we will be able to evaluate the following function of the
masses
M 1 + M2
3=2
M2
If we have both a1 and a2 ; then we can evaluate both masses.
For spectroscopic binaries we have only the velocities of one or both stars
along the line of sight.
8
~v i ¢ ^l =
=
r_ir^ ¢ ^l + riµ_ i^µ i ¢ ^l
r_i sin µ sin ® + ri µ_ i cos µ sin ®
where
r_i
d
L1 L2
µ_ i
dµ (1 § e cos µ i) G¹2 M i
Li
L 1L 2 (¨2) e sin µ i
2
Mi ri (1 § e cos µi )2 G¹2 M i
£
¤2
(1 § e cos µ i) G¹2 M i
Li L1 L2 (¨2) e sin µi
Mi (1 § e cos µ i)2 G¹2 M i
(L1 L2 )2
=
=
=
r_1:2
=
and
ri µ_ i =
¨2
G¹2
e sin µ i
L2;1
Li
Li (1 § e cos µi ) G¹2 Mi
(1 § e cos µ i) G¹2
=
=
M i ri
Mi
L1 L2
Lj
Thus
~v i ¢ ^l =
=
·
¸
G¹2
(1 § e cos µ i) G¹2
¨2
e sin µ i sin µ +
cos µ sin ®
Lj
Lj
G¹2
[¨2e sin µ i sin µ + (1 § e cos µ i) cos µ] sin ®
Lj
If we can observe both stars we can obtain the mass ratio,. since
r_1
L1
M2
=
=
r_2
L2
M1
If we can observe only one, say #1, then we observe the quantity
¹2
e sin ®
L2
(16)
We can usually obtain the eccentricity by ¤tting a model to the observed curve
of velocity versus time (sin µi (t)): We still need to express L2 in terms of observables. So use equation (8) in equation (15) to get:
¡
¢
2
M1M2¹ 3
¡ G 2L
1 ¡ B2
1 L2
T
r
·
¸3=2
¹
2L1 L2
2 G2 M 1 M 2 ¹3 (1 ¡ e2 )
·
¸3=2
M 2 M1 1
L22
= 2¼
G 2 ¹4 M 12 (1 ¡ e2 )
= ¼GM 2 M 1
9
Thus
L2
·
¸1=3
¢
G 2 M 12 ¹4 T ¡
2 3=2
1¡ e
2¼ M2
¢1=2
G 2=3 M 1 T 1=3 ¹ ¡
1 ¡ e2
1=3
2¼ (M1 + M 2)
=
=
and therefore we can measure (eqn 16)
¹2
e sin ®
L2
1=3
¹ (M 1 + M 2 )
e sin ®
M 1T 1=3 (1 ¡ e2 )
·
¸
M2
2¼
e
sin ®
p
T 1 ¡ e2
(M 1 + M 2 )2=3
= 2¼
=
which is the same function of the masses that we found for visual binaries.
2
The three-body problem
We cannot solve this problem in general, so we look at the special case in which
the third body is much less massive than the ¤rst two. Wezll also specialize to
the case in which the ¤rst two bodies orbit each other in a circular path, so that
the angular velocity µ_ is constant. Then we can work in the rotating frame,
and the ¤rst two bodies are at rest in this frame. In this frame there are two
¤ctitious forces: the centrifugal force
F~ ce n t = ¡m~! £ (~! £ ~r )
and the Coriolis force
F~ c oriolis = ¡2m~! £ ~v
Expanding the triple cross product, the centrifugal force may be written as
µ
¶
m! 2 r2
2
~
~
Fc e nt = m! ~r = ¡r ¡
2
where
m! 2r 2
2
is the e¤ective potential energy associated with this force. The Coriolis force
may be neglected to ¤rst approximation if
Vef f = ¡
v ¿ !r
and we shall do so for now. Thus we may write the Lagrangian for a particle
moving in the rotating frame as
1
GM 1 m
GM 2 m
m! 2 r2
L = mv 2 +
+
+
2
r1
r2
2
10
where r1 and r2 are distances from the two masses and r is the distance from the
center of mass. We may simplify by putting the origin at the CM, and x¡axis
along the line joining the two bodies. The distance between the two massive
bodies is R; Then the two bodies are at positions
x 1 = ¡R
M2
M1
and x 2 = R
M1 + M 2
M1 + M2
and
1
L = mv 2 + q
2
GM1 m
2
(x ¡ x 1 ) + y 2
where
!
=
=
2¼
=
T
GM 2 m
+q
2
(x ¡ x 2) + y 2
3=2
2¼a1
p
³
2¼
M12
G ¹2M 2
G (M 1 + M 2 )
R 3=2
Now write all coordinates in dimensionless form:
potential is
VR
¡
= r³
GmM2
®
x+
1
1+®
´2
+
y2
+ r³
¡
¢
m! 2 x2 + y 2
+
2
´1=2
i.e. x ! x=R: Then the
1
x¡
®
1+®
´2
+ y2
¡
¢
(1 + ®) x 2 + y 2
+
2
~ =0
The small third body will be in equilibrium at points where rV
Plot of equipotentials with M 1 =M 2 = ® = 2
11
(17)
2
y1
-2
0
-1
1x
2
-1
-2
Green 4.5, Black- 5, magenta 5.15, blue 6, red 7.5,
VR
¡
= r³
GmM2
®
x+
1
1+®
´2
+ y2
+ r³
1
x¡
®
1+®
´2
+ y2
¡
¢
(1 + ®) x 2 + y 2
+
2
³
´
³
´
1
®
® x + 1+®
x ¡ 1+®
@V
= ¡·
¸3=2 ¡ ·³
¸3=2 + x (1 + ®)
³
´2
´2
@x
1
®
2
2
x + 1+® + y
x ¡ 1+® + y
(18)
and
@V
= ¡·
³
@y
x+
®y
´2
1
1+®
+
y2
¸3=2 ¡ ·³
x¡
®
1+®
y
´2
+
y2
¸3=2 + y (1 + ®)
So @V =@y = 0 at y = 0 and then we also have @V =@x = 0 at:
For x > ®=(1 + ®)
¡³
®
x+
1
1+®
´2 ¡ ³
1
x¡
®
1+®
12
´2 + x (1 + ®) = 0
(19)
For ¡ 1+1 ® < x <
®
1+® ,
¡³
®
x+
1
1+®
1
and for x < ¡ 1+®
+³
®
x+
1
1+®
´2 + ³
´2 + ³
1
x¡
®
1+®
1
x¡
®
1+®
´2 + x (1 + ®) = 0
´2 + x (1 + ®) = 0
The di¤erent values arise from the fact that the numerators in equation (18)
are distances and thus must be positive. Thus the points of stability are at:
¡
¢2 ¡
¢2
¡
¢2 ¡
¢2
¡2 x ¡ 23 ¡ x + 13 +3x x + 13
x ¡ 23 = 0, Solution is: fx = 1: 249g
¡
¢
¡
¢
¡
¢
¡
¢
2
2
2
2
¡2 x ¡ 23 + x + 13 +3x x + 13
x ¡ 23 = 0, Solution is: fx = 0:237 42g
¡
¢
¡
¢
¡
¢
¡
¢
2
2
2
2
+2 x ¡ 23 + x + 13 +3x x + 13
x ¡ 23 = 0, Solution is: fx = ¡1: 136 4g
These points agree with the values read from the plot of the equipotential
surfaces.
5
-1.5
-1
-0.5
0.5
x
1
1.5
0
-5
-10
-15
For ® ¿ 1 :
¡³
®
1
x + 1+®
¡
´2 ¡ ³
®
1
x¡
1
´2 + x (1 + ®)
2 + x (1 + ®)
(x + 1 ¡ ®)
(x ¡ ®)
®
1 ³
®´
¡
¡ 2 1+2
+ x + x®
2
x
x
(x + 1)
2
¡
®
1+ ®
= 0
= 0
= 0
To zeroth order in ® the solution is x = 1: Then let x = 1 + "; where " is of
13
order ®:
¡
®
1
µ
®
1+2
1+ "
¶
+1+"+ ® =
2 ¡
2
(2 + ")
(1 + ")
®
¡ (1 ¡ ") ¡ (1 ¡ 2") (1 + 2®) + 1 + " + ® =
4
5
¡ ® + 3" =
4
" =
0
0
0
5
®
12
5
Thus L2 is at 1 + 12
": To ¤nd the next point, we change one sign.
¡³
®
x+
1
1+®
´2 + ³
1
x¡
®
1+®
´2 + x (1 + ®) = 0
The zeroth order solution (® = 0) is x = ¡1: Again we look for a solution
x = ¡1 + ": We have to be very careful with the ¤rst term, so letzs go to second
order in ® this time.
2
2
¡® (x (1 + ®) ¡ ®) + (x (1 + ®) + 1) +
2
x
2
2
(x (1 + ®) + 1) (x (1 + ®) ¡ ®)
(1 + ®)
2
2
¡® (1 + ®) (x + ® (x ¡ 1)) + (1 + ®) (x + 1 + x®) + x (x + 1 + x®) (x + ® (x ¡ 1))
0
=
=
2
= 0
= 0
³
´
¡
¢
¡® (1 + ®) x 2 + 2®x (x ¡ 1) + (1 + ®) (x + 1) 2 + 2x (x + 1) ® + x 2 ®2
³
´³
´
2
2
+x (x + 1) + 2x (x + 1) ® + ® 2 x2 x 2 + 2®x (x ¡ 1) + (x ¡ 1) ® 2
¡® (1 + ®) x 2 ¡ 2® 2 x (x ¡ 1) + (1 + ®) (x + 1) 2 + 2x (x + 1) ® (1 + ®) + x 2® 2
+4x4 ® + 6x5 ® 2 + 4x5 ® + x 3 + 2x4 + x 5 ¡ 2®x 3 + x®2 ¡ 6x 3 ®2 ¡ 2x 2 ®
0 = +2x4 +x 5 +x 3 +x2 +2x+1+4x®+4x 5 ®¡2®x 3 +4x 4 ®2 +6x5 ® 2 ¡6x 3 ®2 +5x® 2
Now set x = ¡1 + "
0
=
=
³
´
1 + 2 (" ¡ 1) + (" ¡ 1)2 + (" ¡ 1)3 + 2 (" ¡ 1)4 + (" ¡ 1)5 ¡ 2® (" ¡ 1)3 ¡ 2 (" ¡ 1) ¡ 2 (" ¡ 1)5
³
´
3
4
2
+ (" ¡ 1) ® 2 4 (" ¡ 1) + 6 (" ¡ 1) ¡ 6 (" ¡ 1) + 5
¡
¢
¡
¢
3"3 ¡ 3"4 + "5 ¡ 2® 3 ¡ 9" + 17"2 ¡ 19"3 + 10" 4 ¡ 2" 5 + ®2 ¡1 + " ¡ 18"2 + 38" 3 ¡ 26"4 + 6"5
As expected the terms in "0 ® 0 have vanished. Gathering terms in powers of
"
0 = ¡® 2 ¡ 6® + ®" (18 + ®) + ¢ ¢ ¢
14
Thus
"
³
®+6
1
®´
=
(® + 6) 1 ¡
18 + ®
18
18
1
1
+
®
3
27
=
=
and
x = ¡1 +
1
1
2
®
+ ®=¡ +
3
27
3
27
Now for the last one:
³
x+
®
1
1+®
´2 + ³
1
®
x ¡ 1+®
´2 + x (1 + ®) = 0
As before, the zeroth order solution is ¡1and we have to be especially careful
¤nding the correction. The only di¤erence is the sign of the leading term. Thus
³
´
¡
¢
2
0 = +® (1 + ®) x 2 + 2®x (x ¡ 1) + (1 + ®) (x + 1) + 2x (x + 1) ® + x 2 ®2
³
´³
´
+x (x + 1)2 + 2x (x + 1) ® + ® 2 x2 x 2 + 2®x (x ¡ 1) + (x ¡ 1)2 ® 2
and
These points are on either side of the small mass.
The other values of y are given by
¡·
³
x+
1
1+®
®
´2
+
y2
¸3=2 ¡ ·³
x¡
®
1+®
1
´2
+
y2
¸3=2 + (1 + ®) = 0
Together with the x¡relation
³
´
³
´
1
®
® x + 1+®
x ¡ 1+®
¡·
¸3=2 ¡ ·³
¸3=2 + x (1 + ®) = 0
³
´2
´2
1
®
x + 1+ ® + y 2
x ¡ 1+® + y 2
Letzs look again at the special cases ® = 2 and ® ¿ 1: First ® = 2 :
¡h
¡
(black curve) and
x+
2
¢
2
1
3
+y
i 3=2 ¡ h¡
2
x¡
1
¢
2 2
3
+
y2
i3=2 + 3 = 0
¡
¢
¡
¢
2 x + 13
x ¡ 23
¡h
i3=2 ¡ h¡
i3=2 + 3x = 0
¡
¢2
¢2
x + 13 + y 2
x ¡ 23 + y 2
15
(red curve)
1.5
1
y
0.5
-1
-0.5
0.5 x
1
-0.5
-1
-1.5
There are two intersections, symmetrically located about the x¡axis.
The points that we have found are the ¤ve Lagrange points.
For ® ¿ 1 : The two equations are
¡h
and
¡h
®
1
i 3=2 ¡ h
i3=2 + (1 + ®) = 0
(x + 1 ¡ ®)2 + y 2
(x ¡ ®) 2 + y 2
®x
x¡®
i 3=2 ¡ h
i3=2 + x (1 + ®) = 0
2
2
2
2
(x + 1 ¡ ®) + y
(x ¡ ®) + y
Keeping ¤rst order in ® :
¡h
®
2
(x + 1) ¡ 2® (x + 1) +
®
y2
i3=2 ¡
1
[x 2
1
+ (1 + ®) =
0
¸
3®x
1+ 2
+ (1 + ®) =
x + y2
0
¡ 2®x + y 2 ]3=2
·
¡h
i3=2 ¡ 2
2
(x + y 2 ) 3=2
(x + 1) + y 2
8
9
>
>
<
=
1
1
3x
1¡
+
®
1
¡
¡
h
i
3=2
3=2
5=2
>
(x2 + y 2 )
:
(x 2 + y 2 ) >
;
(x + 1)2 + y 2
16
=
0(20)
The zeroth order result is r = 1: For ® ¿ 1; the second equation becomes:
·
¸
®x
x¡ ®
3®x
¡h
1+ 2
+ x (1 + ®) =
i 3=2 ¡ 2
x + y2
2
[x + y 2 ]3=2
(x + 1) + y 2
0
®x
x¡ ®
3®x 2
¡h
¡
+ x (1 + ®) = 0
i3=2 ¡ 2
3=2
2]
2 + y 2 )5=2
2
[x
+
y
(x
2
(x + 1) + y
0
1
Ã
!
2
1
x
1
3x
B
C
x 1¡
+ ® @x ¡ h
+
¡
A =(21)0
i
3=2
3=2
3=2
2 + y2 ]
2 + y 2 )5=2
2
[x2 + y 2 ]
[x
(x
(x + 1) + y 2
which gives r = 1 or x = 0 as the zeroth order solution. Then let r = 1 + " in
equation (20) to get
0
1
1¡
1
1
(1 + ") cos µ C
B
+ ® @1 ¡ h
A =
i3=2 ¡ 3
5
(1 + ")
(1 + ")
(1 + ") 2 + 2 (1 + ") cos µ + 1
Ã
!
1
3" + ® 1 ¡
¡ 3 cos µ
=
3=2
[1 + 2 cos µ]
3
and so
®
"=¡
3
Ã
1¡
1
[2 + 2 cos µ]
3=2
¡ 3 cos µ
!
0
0
(22)
Now we expand equation (21) to ¤rst order in ® and " :
Ã
!
cos µ
2
3" cos µ + ® cos µ ¡
+ 1 ¡ 3 cos µ = 0
2 3=2 [1 + cos µ]3=2
and then insert relation (22) for " :
Ã
!
Ã
!
1
cos µ
2
¡® cos µ 1 ¡
¡ 3 cos µ + ® cos µ ¡
+ 1 ¡ 3 cos µ
3=2
3=2
[2 + 2 cos µ]
[2 + 2 cos µ]
cos µ
¡p
+1
3=2
2 (1 + cos µ)
cos µ
cos 2 µ
=
p
2 (1 + cos µ)3=2
3
= 2 (1 + cos µ) = 2 + 6 cos µ + 6 cos2 µ + 2 cos3 µ
Thus cosµ satis¤es the equation
0
= 2 + 6 cos µ + 5 cos2 µ + 2 cos3 µ
¡
¢
= (2 cos µ + 1) cos2 µ + 2 cos µ + 2
17
= 0
= 0
The real solution is cos µ = ¡1=2; or µ = 120 ± . Finally we put this result into
equation (22) to get
Ã
!
®
1
3
" = ¡
1¡
+
3=2
3
2
[2 ¡ 1]
®
= ¡
2
Thus the last two Lagrange points are at
r=1¡
®
;
2
µ = §120 ±
These points are 60± ahead of and behind the smaller mass.
2.1
Stability of the Lagrange points.
We can see from the ¤gure on page 12 that L4 and L5 are potential minima,
and thus stable, while L1 ; L2 and L3 are saddle points and thus unstable.
To analyze the stability, we need to include the Coriolis force. We will need
to include a velocity dependent potential. Then Lagrangezs equations take the
form
d @L (~x ; ~v ) @L (~x; ~v )
¡
=0
dt @vi
@x i
Letzs reacall what happens in E&M. The magnetic force is
~
F~ = q~v £ B
where
~ =r
~ £A
~
B
so
F~
=
=
³
´
~ £A
~
q~v £ r
·
³
´ ¸
@
~
~
q vj
Aj + ~v ¢ r A
@xi
18
(23)
We de¤ne the velocity-dependent potential as
~ ¢ ~v
¡q A
so that
@V
= ¡qAi
@vi
and
d
dt
µ
¶ µ
¶
@V
@
@
@
¡
=
+ vj
qAi = vj
qAi
@v i
@t
@x j
@x j
d
dt
µ
then
¡
@V
@vi
¶
+
@V
@x i
=
=
@
@
qAi + q
v j Aj
@x j
@xi
@
@
vj
qAi + qvj
Aj
@x j
@x i
vj
in agreement with equation (23).
Now in our case the Coriolis force is
F~ c or = ¡m~! £ ~v = m~v £ ~!
where
³ !r ´
^µ = r
~ £A
~
2
(cf Lea problem 1.5). Thus our potential is now
VR
¡
= r³
GmM2
~ £
~! = r
®
x+
1
1+®
´2
+
y2
+ r³
1
x¡
19
®
1+®
´2
+ y2
¡
¢ p
(1 + ®) x2 + y 2
1 + ®r ^
+
¡ p
µ¢~v
2
2 GM 2