3 Angular Momentum, or Areal Velocity, Integral
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7.3 Angular Momentum, or Areal Velocity, Integral
147
Fig. 7.2 The angular
momentum vector shown in a
Cartesian coordinate system
The Cartesian equations for the angular momentum are
3
X
mi Œyi zPi
zi yP i D a1
(7.17a)
mi Œzi xP i
xi zPi D a2
(7.17b)
mi Œxi yP i
yi xP i D a3
(7.17c)
iD1
3
X
iD1
3
X
iD1
The brackets in these equations are the projections of the areal velocities of the
different bodies upon the three coordinate planes. They are the integrals of area,
1
ŒyPz
2
1
ŒzPx
2
1
ŒxPy
2
1
c1
2
1
xPz D c2
2
1
yPx D c3
2
zPy D
(7.18a)
(7.18b)
(7.18c)
q
and h D c21 C c22 C c23 . These expressions have a counterpart in two-body motion
(see Sect. 5.5).
148
7 The n-Body Problem
Let Ai1 ; Ai2 ; Ai3 denote the projections of the areas swept out by the radius vector
of the mass mi upon the yz; xz; and xy planes, respectively. Equations (7.17) may be
written as
3
X
mi AP i1 D a1
(7.19a)
mi AP i2 D a2
(7.19b)
mi AP i3 D a3
(7.19c)
mi Ai1 D a1 t C b1
(7.20a)
mi Ai2 D a2 t C b2
(7.20b)
mi Ai3 D a3 t C b3
(7.20c)
iD1
3
X
iD1
3
X
iD1
Integrated, these yield
3
X
iD1
3
X
iD1
3
X
iD1
Consequently, the sums of the products of the masses and the projections of
the areas swept out by the corresponding radius vectors increase uniformly with
time. As the three bodies move, their position and velocity vectors are oriented
so the vector L has a fixed direction in space and a constant magnitude, .a21 C
a22 C a23 /1=2 . L is directed along a line that is the invariable line. A plane, which
is perpendicular to invariable line, and passes through the center of mass of the
system, is the invariable plane (see also Sect. 3.5.1). This term was introduced by
Laplace; the Laplacian plane is used for natural satellite systems, i.e. around Jupiter
and Saturn. The invariable plane has the following properties:
(a) The angular momentum about any line in the plane is zero.
(b) The angular momentum about a line normal to the plane is a maximum.
In our notation, the direction numbers of the normal to the invariable plane are
a1 ; a2 ; a3 .
The orbital angular momentum of Jupiter and Saturn is nearly 87 % of the whole
solar system. As a consequence, and because the two planets’ orbital planes are
nearly in the ecliptic, the vector L is directed only about 1ı 350 from the pole of the
ecliptic. The invariable plane lies between the orbita l planes of these two planets.
Since the masses, positions, and velocities of planets of the solar system are well
7.4 Integral of Energy
149
known, the constants a1 ; a2 ; a3 can be determined with considerable accuracy. So,
the solar system’s invariable plane is relatively well determined.
What is stated here about the three-body system can be generalized to the n-body
problem, and will be shown later (McCuskey 1963, pp. 94–97).
7.4 Integral of Energy
The energy integral of the equations of motion in the three-body problem is
analogous to 12 mv 2 C V.r/ D E under central force motion (see Sect. 4.4).
Take the scalar product of Eqs. (7.13) by rP 1 ; rP 2 ; rP 3 , respectively, and add the
results. Then
3
X
mi rP i vP i D k
2
Ä
m1 m2
2
12
iD1
C
with rP 1
rP 2 D
P ij D
P 12 ; rP 1
d
.
dt
ij
m2 m3
2
23
uO 12 .Pr1
uO 23 .Pr2
rP 3 D
m1 m3
2
13
uO 13 .Pr1
rP 3 /
P 13 ; rP 2
uO ij / D Pij uO ij C
rP 2 / C
ij
rP 3 /
(7.21)
rP 3 D
P 23 and
uPO ij ; i; j D 1; 2; 3; i Ô j
(7.22)
Since uO ij is a unit vector, uPO ij uO ij D 0 and uO ij uO ij D 1. Then, using this in Eq. (7.21)
3
X
mi rP i vP i D k2
Ä
m1 m2
2
12
iD1
P12 C
m1 m3
2
13
P13 C
m2 m3
2
23
P23
which can be written
#
" 3
Ä
d 1X
m1 m3
m2 m3
d m1 m2
2
mi rPi D k2
C
C
dt 2 iD1
dt
12
13
23
(7.23)
(7.24)
By definition, the kinetic energy of the system is
3
TD
3
1X
1X
mi rP 2i D
mi vi2
2 iD1
2 iD1
(7.25)
150
7 The n-Body Problem
where vi is the speed of the ith mass. The potential energy of the system is defined as
V D k2
Ä
m1 m2
12
C
m1 m3
13
C
m2 m3
23
(7.26)
Integration of Eq. (7.24) yields
T C V D E D constant
(7.27)
Equation (7.27), the energy integral, shows the conservation of energy for the system
of three masses. This result can be generalized to n bodies.
7.5 Stationary Solutions of the Three-Body Problem
Two special solutions of the three-body problem, designated stationary solutions,
were discovered in 1772 by Lagrange. Assume three mass points, m1 ; m2 ; m3 ,
are projected in one plane. A stationary solution is one where the geometric
configuration of the three masses is invariant with time. If the masses move such that
their mutual distances remain unchanged, the configuration simply rotates around
the center of mass in a plane. Alternatively, a contraction, which does not alter the
pattern of the points, may take place.
First, consider the special case where the three masses revolve with constant
angular speed around the center of mass in coplanar circular orbits. Their position
vectors relative to the center of mass C are r1 D r1 uO 1 ; r2 D r2 uO 2 ; r3 D r3 uO 3 .
uO 1 ; uO 2 ; uO 3 are the unit vectors in the directions r1 ; r2 ; r3 , respectively, as shown in
Fig. 7.3. For circular motion, r1 ; r2 ; r3 are constant.
Fig. 7.3 Coplanar orbits in
the three-body problem
7.5 Stationary Solutions of the Three-Body Problem
151
The total planar acceleration of each mass point is given by (recall Eq. (4.6))
R uO Â
rÂP 2 / uO r C .2PrÂP C rÂ/
a D .Rr
(7.28)
where uO r and uO Â are unit vectors in the radial and transverse directions, respectively.
For the special case of the three-body problem considered here, for each mass rP D
rR D 0; ÂR D 0; ÂP D n, the constant speed of the revolution about C. Thus, the
accelerations become
ai D ri n2 uO r ; i D 1; 2; 3
(7.29)
With these values in Eqs. (7.1)–(7.3), we have
n2 r1 uO 1 D k2
n2 r2 uO 2 D k2
n2 r3 uO 3 D k2
Ä
Ä
Ä
m2
3
12
m3
3
23
.r2 uO 2
r1 uO 1 / C
m3
.r3 uO 3
r2 uO 2 /
m1
m2
3
23
.r3 uO 3
3
13
3
12
r2 uO 2 /
.r3 uO 3
r1 uO 1 /
(7.30a)
.r2 uO 2
r1 uO 1 /
(7.30b)
m1
3
13
.r3 uO 3
r1 uO 1 /
(7.30c)
However, the origin is selected so
m1 r1 uO 1 C m2 r2 uO 2 C m3 r3 uO 3 D 0
(7.31)
If we multiply Eq. (7.30a), by m1 and (7.30b) by m2 and add, we derive an
equation, which upon substitution from Eq. (7.31) yields Eq. (7.30c). Thus, we can
use Eq. (7.31) in place of Eq. (7.30c) in the unit vectors uO 1 ; uO 2 ; uO 3 . Rewrite the
system of equations in the form
Â
n2 C
k 2 m1
3
12
k 2 m2
3
12
Â
r1 uO 1 C
C
k 2 m3
3
13
n2 C
Ã
r1 uO 1
k 2 m3
3
23
k 2 m2
k 2 m3
r3 uO 3 D 0
(7.32a)
r3 uO 3 D 0
(7.32b)
m1 r1 uO 1 C m2 r2 uO 2 C m3 r3 uO 3 D 0
(7.32c)
C
3
12
k 2 m1
3
12
r2 uO 2
Ã
r2 uO 2
3
13
k 2 m3
3
23
These conditions must be fulfilled for the three mass points to move with uniform
angular speed in circular orbits around the center of mass in a plane.
If we have a rectangular coordinate system, origin at C, rotating in the counterclockwise direction with a constant angular speed, then the unit vectors uO 1 ; uO 2 ; uO 3
are fixed in position. The angles in the equations
ui D cos Âi Oi C sin Âi Oj; i D 1; 2; 3
(7.33)
152
7 The n-Body Problem
denote the orientations relative to these Cartesian coordinates. Taking the scalar
products of Oi and Eqs. (7.32), after simplification
n 2 x1 C
n 2 x2 C
k 2 m2
3
12
k 2 m1
3
12
.x1
x2 / C
.x2
x1 / C
k 2 m3
.x1
x3 / D 0
(7.34)
.x2
x3 / D 0
(7.35)
m1 x 1 C m2 x 2 C m3 x 3 D 0
(7.36)
3
13
k 2 m3
3
23
where
xi D ri cos Âi ; i D 1; 2; 3
(7.37)
Taking the scalar products of Oj and the same Eqs. (7.32), we have
n 2 y1 C
n 2 y2 C
k 2 m2
3
12
k 2 m1
3
12
.y1
y2 / C
.y2
y1 / C
k 2 m3
.y1
y3 / D 0
(7.38)
.y2
y3 / D 0
(7.39)
m1 y 1 C m2 y 2 C m3 y 3 D 0
(7.40)
3
13
k 2 m3
3
23
where
yi D ri sin Âi ; i D 1; 2; 3
(7.41)
Equations (7.34)–(7.36) and (7.38)–(7.40) are a system of six simultaneous
equations in the unknowns .xi ; yi /; i D 1; 2; 3. Consider a case where the masses
are at the vertices of an equilateral triangle. Then, 12 D 23 D 13 D at all times.
If the scale of the length is adjusted so that is unity at all times, the equations may
be written with k2 D 1 by proper choice of time units,
. n2 C m2 C m3 /x1
m2 x 2
m3 x 3 D 0
(7.42a)
m1 x1 C . n2 C m1 C m3 /x2
m3 x 3 D 0
(7.42b)
m1 x 1 C m2 x 2 C m3 x 3 D 0
(7.42c)
. n2 C m2 C m3 /y1
m2 y 2
m3 y 3 D 0
(7.43a)
m1 y1 C . n2 C m1 C m3 /y2
m3 y 3 D 0
(7.43b)
m1 y 1 C m2 y 2 C m3 y 3 D 0
(7.43c)
7.5 Stationary Solutions of the Three-Body Problem
153
These have a nontrivial solution if
3
2
M 0 m2 m3 0
0
0
6 m M0 m 0
0
0 7
1
3
7
6
7
6
0
0 7
6 m1 m2 m3 0
det 6
7D0
0
6 0
m2 m3 7
0
0 M
7
6
4 0
0
0
m1 M 0 m3 5
0
0
0 m1 m2 m3
(7.44)
where M 0 D . n2 C m2 C m3 /.
The solution xi D 0; yi D 0; i D 1; 2; 3 is inconsistent with the requirement that
D 1. Let M D m1 C m2 C m3 denote the total mass of the system. By the usual
reduction rules for determinants, we have Eq. (7.44)
m23 .M
n 2 /4 D 0
(7.45)
This is satisfied when n2 D M. If we left k2 in the equations and the equal values of
D ij ; i; j D 1; 2; 3I i Ô j, this condition would be
n2 D
k2 M
(7.46)
3
Equation (7.46) is dimensionally consistent and specifies that n is in radians per unit
of time.
With n2 D M in Eqs. (7.42) and (7.43), any two pairs of coordinates .xi ; yi / can
arbitrarily be chosen, the scale adjusted, and the third pair determined such that
D 1. Therefore, there is a solution to the equations in this special problem of three
bodies.
There is a second solution for Eqs. (7.34)–(7.36) and (7.38)–(7.40). If y1 D y2 D
y3 D 0, then all three mass points lie on the x-axis, and Eqs. (7.38)–(7.40) are
satisfied. Arrange the masses on the x-axis as shown in Fig. 7.4.
Denote the distance between x3 and x2 by and adjust the scale so the distance
between x2 and x1 is unity. Due to the inequalities, x1 < x2 < x3 , Eqs. (7.34)–(7.36)
become
n2 x1 C k2 m2 C k2 m3 .1 C /
2
D0
(7.47a)
2
D0
(7.47b)
m1 x1 C m2 .1 C x1 / C m3 .1 C x1 C / D 0
(7.47c)
2
n .1 C x1 /
2
2
k m1 C k m3
We eliminate n2 and x1 from these to obtain
.m1 C m2 /
5
.2m2 C 3m3 /
C .3m1 C 2m2 /
4
C .3m1 C m2 /
.m2 C m3 / D 0
3
.m2 C m3 /
2
(7.48)
154
7 The n-Body Problem
3( 3)
2( 2)
Fig. 7.4 Three masses arranged on the x-axis
This 5th degree equation in , originally due to Lagrange, has only one real positive
root, because there is only one sign change in the coefficients. With this arrangement
of the masses along the x axis, only one positive solution for exists. When is
found, m3 is uniquely located, because m1 and m2 are arbitrarily located to set the
distance scale, x2 x1 D 1. If the location of the masses is interchanged, a similar
unique solution is obtained for each arrangement. The solution for Eq. (7.48) can be
found by numerical, or iterative, means.
Lagrange found two distinct solutions to the three-body problem:
(i) The equilateral triangle solution
(ii) The straight-line solution
These solutions are valid for any masses moving with uniform angular speed in
coplanar circular orbits around their center of mass. Let us consider under what
dynamical conditions such motions take place.
Since each mass moves uniformly in a circular path around the center of mass,
C, the areal velocity in each orbit is constant. So, the resultant force, acting on each
mass, must pass through C.
From Eq. (7.29) the acceleration of each mass is ri n2 uO i ; i D 1; 2; 3. So, the
force is mi ri n2 uO i . The resultant force on each mass is directly proportional to the
distance from C, and is directed toward that point.
The initial velocity vectors, required for circular motion of the system, are
perpendicular to the initial position vectors of the respective masses. For each mass,
mi , the velocity, vi D ri uPO i , so the velocity is proportional to the distance of mi from
C and is perpendicular to uO i . The motion in these cases is a rotation of the system
around the center of mass, C.
If the initial velocity vectors are not perpendicular to the position vectors of
the respective mass points, the configuration will change with time. If each vector,
vi ; i D 1; 2; 3, makes the same angle with respect to its corresponding ri , then the
configuration will expand or contract such that the problem is still solvable. Each
mass will follow a conic section, the distances between the bodies varying with time,
7.6 Generalization to n Bodies
155
Fig. 7.5 A particular
solution of the three-body
problem, wherein each mass
follows a conic section,
resulting in a rotating
equilateral triangle
2
1
but such that their mutual distances remain in the same ratio. Thus, if 12 ; 23 ; 13
are initial distances at time t, then the distances become ˛ 12 ; ˛ 23 ; ˛ 13 , where ˛
is the proportional factor. The distances of m1 ; m2 ; m3 from the center of mass at
any time are r1 D ˛r01 ; r2 D ˛r02 ; r3 D ˛r03 , where r0i ; i D 1; 2; 3 are the initial
position vector lengths. The shape of the configuration remains invariant with time.
˛ may be a function of time. Each mass traverses a conic section with a resultant
pattern as shown in Fig. 7.5, with each triangle being equilateral.
The vertices of the equilateral triangle solution and the points on the straight line
solution of the three-body problem are the Lagrangian points (see also Chap. 1).
These are fixed in a planar coordinate system rotating with constant angular
speed around the center of mass. A body, situated and initially at rest at one of
the Lagrangian points, will remain so unless disturbed by external forces. The
gravitational and centrifugal forces balance each other at these points (McCuskey
1963, pp. 102–108).
7.6 Generalization to n Bodies
Previously, important characteristics of three body motions, which can be generalized to the n-body problem, have been developed. These characteristics are
summarized as follows:
(a) The system potential energy is
V D k2
n
X
mi mj
i;jD1
ij
; iÔj
(7.49)
156
7 The n-Body Problem
This depends on the relative distances between the mass points. This value is
independent of location of the origin of the coordinate system and any arbitrary
angular rotation of the axes of the system.
(b) Because of this independence, the integrals yielding the motion of the center
of mass and areas, follow. There are nine constants of integration, six define
the center of mass position at any instant, and three define the orientation of
the angular momentum vector. A fixed origin and orientation of axes cannot
be defined in space, so these nine constants cannot be determined from the
observations.
(c) There is a tenth constant, E, the energy integral. These are ten constants of
integration for the n-body problem. From the equations of motion, 6n constants
are needed to solve the problem completely. For the three-body problem we
have only 10 of 18 constants needed.
Other integrals of the equations of motion, other constants, have been sought
by many. These efforts have not been successful. These integrals, such as the
energy integral, are relations between coordinates and velocities. They are called
the classical integrals of the three-body problem.
Bruns and Poincaré have shown that these 10 integrals are the only independent
integrals, and all others are combinations of these 10.
Thus, the coordinates and velocity components of the three mass points as
functions of time cannot be solved for, when given their values at some initial epoch.
This is also true for the n-body problem. A closed form solution of the n-body
problem is impossible. Specifically, given the coordinates and velocities of n mass
points, which are moving under mutual gravitational attractions, the motion for any
arbitrary succeeding time interval can not be predicted. Whether masses will collide
or escape in an arbitrary finite time interval cannot be predicted. Only some special
cases of the three-body problem can be addressed (McCuskey 1963, pp. 98–99).
7.7 Equations of Relative Motion
Since the motions of celestial bodies cannot be referred to a fixed coordinate system
in space, the relative motions of n 1 bodies with respect to the nth body can be
established. In the solar system, the Sun is the origin of coordinates, and the plane of
the ecliptic or equator is the fundamental reference plane. The other bodies motions,
planets, comets, and asteroids, are given relative to the Sun. Now, we generalize the
three-body problem to the n body problem.
Let m1 , shown in Fig. 7.6, be chosen as the origin. Taking the first three equations
from the three-body problem
m1 rR 1 D k
2
Ä
m1 m2
3
12
12
C
m1 m2
3
13
13
(7.50)
7.7 Equations of Relative Motion
157
1
Fig. 7.6 Relative motion in
the three-body problem
12
1
2
13
2
23
3
m2 rR 2 D k2
m3 rR 3 D k2
Ä
m2 m3
Ä
3
23
m3 m2
3
23
m2 m1
23
3
12
m3 m1
23
3
13
(7.51)
12
(7.52)
13
So we will subtract Eq. (7.50) from Eqs. (7.51) and (7.52), and after some simplification
Ä
k2 .m1 C m2 /
1
1
2
R 12 D
uO 12 C k m3 2 uO 23
uO 13
(7.53)
2
2
12
R 13 D
2
k .m1 C m3 /
2
13
uO 13
k 2 m2
Ä
23
1
2
23
13
uO 23 C
1
2
12
uO 12
(7.54)
These are the equations of motion of m2 relative to m1 and m3 relative to m1 . The
right hand side of the first equation has three terms which represent, respectively:
(i) the acceleration of m2 due to m1 ; (ii) the acceleration of m2 due to m3 ; and (iii) the
reactive, or negative, acceleration of m1 due to m3 . A similar explanation can be
given for the terms of Eq. (7.51).
If m3 D 0, Eq. (7.53) describes the two-body motion of m2 around m1 . Likewise,
if m2 D 0, Eq. (7.54) gives the two-body motion of m3 with respect to m1 .
Let m1 be the dominant mass of the three masses and at the origin of a Cartesian
coordinate system, as shown in Fig. 7.7.