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A Geocentric Solution to the 3-body Problem
1
A GEOCENTRIC SOLUTION TO THE
THREE-BODY PROBLEM
Prepared by
Gerard Bouw, Ph.D.
Introduction
In 2013 some humanistic, self-professed scientists proposed that, for the 3-body problem,
the theory of geocentricity should yield a different solution than what is observed. In
anticipation of that challenge Bob Sungenis invited me to derive a geocentric solution
within the framework of the theory of geocentricity. Our purpose was to derive a viable
geocentric framework for the 3-body problem. We were successful and this paper is the
fruit of our labors.
The Approach
The version of the 3-body problem we consider here is, as the name suggests, of three
bodies in mutual orbit about each other. Classically, the 3-body problem assumes that
two of the bodies are massive and the third body’s mass can be regarded as zero. Unlike
most modern dynamic derivations posted on the web which know the solutions ahead of
time, the approach in this paper is completely general and so could be used to locate any
additional critical points besides the five Lagrangian points. Geocentrically, the 3 bodies
are: the sun, the earth, and a third body, of negligible mass, that appear at rest relative to
those two bodies. Fundamentally, we are looking to answer this question: “At what
locations is the orbital speed of the negligible-mass body zero with respect to both the
sun and earth?” As a point of terminology, the negligible body is technically called the
“infinitesimal mass” or “the test mass.”
Figure 1 presents the geocentric geometry of the problem. The earth is located at point E
and the sun is located at point S. The angular velocity vector,  D represents the diurnal
rotation of the firmament about the axis that runs from the center of the earth through its
poles. The angular velocity vector,  A, is inclined at 23.5° to  D, which is the angular
velocity of the earth-sun line about the earth in the course of the year. The geometry is
such that the sun is located at its southernmost point, namely the first day of winter, that
is, roughly 21 December. Thus the line from E to S goes clockwise around the point E as
seen from  A, with a period of one year. Likewise, as seen from the tip of  D, the entire
page rotates counterclockwise with a period of one sidereal day (23 hours 56 minutes).
We shall simplify where we can legitimately do so. In an earlier paper we have shown
specifically for the diurnal (daily) rotation case that the system holds together; that the
firmament’s gravitational field, which we call inertia, holds the stellar universe together
even though some may think that the orbital speeds would be immense. But rotation is
not the same as revolution, which is to say that if the entire universe rotates as one inertial
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A Geocentric Solution to the 3-body Problem
field, the presence of that rotation cannot be felt by the atomic matter of the universe, or
by photons traveling through it.
We shall shortly demonstrate that neither the daily rotation of the firmament nor the
yearly orbital motion of the sun’s entire about the earth can disrupt the motions we
observe. When we are finished with that section, we can simplify our problem to that
presented in Figure 4 (pg. 7).
Figure 1: The Geocentric Situation
Derivation Of The Geocentric Equations For A Daily-Rotating Universe
By definition, physics deals with matter in motion. Mathematics is the language of
choice, used by physicists to describe motion. Usually physicists are well behaved in
their use of math, but at times, they introduce fudge factors to bridge what theory
demands and experiment lacks. Even then, the fudging is quite obvious from the names
given the fudge factors such as “guillotine factor,” for instance. But there are times when
reputations and careers are at stake and at those times, the fudging becomes quite subtle,
even mean-spirited at times.
A Geocentric Solution to the 3-body Problem
3
The mathematical language used to describe the gravitational forces of orbiting bodies,
and the behavior of spinning bodies is a case in point. When confronted by the mass of
evidence for the geocentric universe, physicists resort to sleight of hand to keep the earth
in orbit about the sun when all fundamental experimental results reveal earth to be
standing still in the firmament. In this case, they multiply one side of the generalized
equation of motion by the number one. Before multiplying by one, the equation is said to
be kinematic, describing the accelerations and velocities of the bodies but not taking the
masses of the bodies into consideration. For instance, consider this equation that
describes the velocity, v, of a body in circular motion with a rotational speed of  at a
distance R from the center of the circle:
v =   R.
(1)
This equation is said to be kinematic and even though it perfectly describes the velocity
and behavior of a body’s rotational and orbital motions, it is said to be unphysical.
Now suppose that we multiply the left-hand side of the equation by one, namely, by the
mass, m, divided by itself, i.e., m/m. This is equivalent to multiplying both sides of the
equation by the mass, m. Our equation (1) now looks as follows:
m v = m  R.
(2)
This is said to be a dynamic description, that is to say, somehow this equation is more
“physical,” more “real,” than the kinematic equation (1) even though we can obviously
cancel out the m’s and simplify equation (2) back to equation (1). To hide this sleight of
hand, equation (2)’s left hand side is replaced by a single variable, p, called momentum.1
Thus equation (3), which is the same as equation (2) is rewritten as”
p = m  R.
(3)
Since momentum is a dynamic concept, the mass is hidden and no physicist will cancel
its appearance on the right-hand side of the equation with its hidden counterpart in p.
But two can play at that game. Let us assume that God created the firmament with a
built-in set of reaction rules. These rules dictate the behaviors of accelerating bodies and
the set of all such reactions we group together under in the name of inertia.
Deriving the Geocentric Equations From First Principles
As seen from earth, its coordinates determine a star’s location. Just as our coordinates on
earth are specified by longitude and latitude, so a star’s coordinates are given by its right
ascension and declination. A star’s longitude is specified by its right ascension and its
latitude is measured by its declination north or south of the plane of earth’s equator.
Since the star’s coordinates are fixed to the celestial sphere, to model the rotation of the
firmament—carrying the star with it—we only need the star’s declination (see Figure 1).
1
Technically, it is more correct to say that p is the angular momentum, but that is irrelevant to the
argument at hand.
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A Geocentric Solution to the 3-body Problem
Figure 2: The Geocentric View of the Daily Rotation
The following is a derivation of the dynamical equations for the universe rotating about
the earth in a daily rotation. In the derivation, we use the following notation:
F is the net gravitational force exerted on the star;
a is the net acceleration experienced by the star in its daily path about earth;
R is the shortest distance from the axis of rotation to the star;
D is the distance from earth to the star;
v is the velocity of the star;
m is the star’s mass;
 is the declination (celestial latitude) of the star as measured from the equator; and
 is the rotation rate of the firmament about the rotational axis that passes through
the north and south poles of earth: imagine it measured in degrees per second
although technically we use radians per second.2
The usual objection against geocentricity is that the earth is not massive enough to have
the universe “orbit” it once a day. In reality, neither the mass of the earth nor the mass of
the star enter into the force that holds the universe together during its rotation.
Acceleration is defined as a change in velocity per unit time. We can write this as:
d 2R
a 2
dt
2
There are 2 radians in the circumference of a circle, so a radian is roughly 57 degrees.
(4)
A Geocentric Solution to the 3-body Problem
5
Here, R is the distance to a moving object and t is time. This can be rewritten more
explicitly as:
a
d dR
dt dt
(5)
where dR/dt is the velocity, v, of the moving object, the star in our case. This equation
says, “Acceleration is the rate of change in velocity.”
But we’re not trying to model the speed and acceleration of an automobile here but that
of a distant star rotating about the earth once every 23 hours and 56 minutes. We must
thus add the rotational velocity (Equation (1)) into the mix. This requires us to rewrite
equation (5) as:
d  dR

(6)
a 
   R
dt  dt

where  is the angular velocity (measured in degrees per second, for instance) and R is
the distance of the star from the axis of rotation.
Distributing the derivative (d/dt) through the terms in parentheses of Equation (6) gives
us equation (7):
d 2 R d
dR
(7)
a 2 
 R  2 
     R  .
dt
dt
dt
Here the first term on the right-hand side of equation (7) is any acceleration that may be
 d

imparted to the earth (the central point). The second term, 
 R  , is the Euler force,
 dt

which is not of interest here since it only kicks in if the length of the day changes
significantly over the course of a day. The third term (starting with the 2), is the Coriolis
force and the last term [R)] is the centrifugal force.
The Coriolis and centrifugal forces dominate the motion of the sun, planets, and stars in a
geocentric system. We shall thus ignore the Euler and local acceleration terms of
equation (7) and work only with:
a  2  v      R 
(8)
where v is the orbital speed of the star. Since the firmament rotates and not the earth, the
sign of v is in the opposite direction to the heliocentric system, and is thus negative. The
v in equation (8) is thus replaced by    R .
After expanding v, equation (8) is now:
a  2    R       R  ;
or
a      R  .
Distributing the cross-product through the term in parentheses gives us:
(9)
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A Geocentric Solution to the 3-body Problem
a     R   R    .
(10)
Now the star is not located on the equator but at declination , whence
 ·R = D  sin().
Our final equation for the geocentric system is thus:
a  2 R  Dˆ sin  .
(11)
Here ̂ is a unit vector pointing along the rotation axis, that is, in the direction of 
which is perpendicular to the equator in general and here in the plane of the star’s circle
in Figure 1. This keeps the acceleration experienced by the star confined to the star’s
latitude, swept out by R and noted as the “Star’s daily path” in Figure 1.
Let’s Examine Our Results Thus Far
Equation (11) has two components, two vectors. They are pictured in Figure 2 where
they are shown as acceleration vectors. To make them dynamic, multiply each by the
star’s mass. The acceleration pictured by the sine term is aligned along the rotation
vector,  , and serves to keep the star’s rotational
plane from “falling” up or down the rotational axis.
The second component is the cosine term. That
acceleration pulls the star towards the axis of
rotation. If multiplied by the star’s mass it becomes
a centripetal (non-fictitious) force, meaning that it
pulls the star towards the axis of rotation. The net
result of these two accelerations is to keep the star
in its place in the inertial field of the universe which
is the gravitational field of the firmament.
Of course, equation (11) is kinematic, not dynamic
and we have to show the geocentric model is
dynamically correct. To do that, all we have to do
is to multiply both sides by the star’s mass, m:
F  ma m2 R  Dˆ sin 
(12)
Figure 3: Accelerations (Forces)
Although we assumed the body was a star in
deriving (12), it could just as well have been the sun, moon, any planet, artificial satellite,
or star circling the earth’s polar axis. Yet some will ask, “What about the speed of light?
Won’t the distant planets and stars orbit the earth way above the speed of light?”
The answer is, “No.” The speed of light is determined by the firmament. It is the
firmament that rotates on the polar axis once a day and so photons, which are transmitted
by the firmament, also participate in the daily rotation. Light, will also obey the above
A Geocentric Solution to the 3-body Problem
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equations superimposed on its own motion. To object that it still exceeds the speed of
light we answer that the speed of light speed limit does not apply for rotation. In this
case it is equivalent to claiming that when a supersonic jet flies faster than the speed of
sound, you could not talk to the person in front of you because you were flying faster
than the speed of sound. But the air in the plane, too, was “flying” faster than the speed
of sound, so you can talk to the person in the seat in front of you because the soundbearing medium was carried with you, even as the light-bearing medium is carried along
with the sun, moon, and stars in the daily rotation of the firmament.
Conclusion
We have shown that the physics of the geocentric universe accounts perfectly for what
we see and measure of the daily rotation whether that rotation is of the earth within the
universe or the universe around the earth. In the final analysis, proofs based on
dynamical equations are not proofs of anything; nor are they proofs against the geocentric
universe.
By the same approach, we could show that the yearly orbital motion of the sun and
planets can be represented in the geocentric framework of the firmament. But for this
paper we shall only point out that every object in the universe obeys equation (12). Since
S, m, and even the c.m. in Figure 1 will all obey equation (12) for the rotational case, we
can simplify Figure 1 to Figure 4:
Figure 4: The Simplified Vector Diagram
.
Here: S is the sun;
E is the earth;
m is the infinitesimal mass (a.k.a. the test mass);
c.m. is the center of mass of the local system;
r1 is the vector representing the distance from sun to the c.m. (which is inside the
sun);
r2 is the vector representing the distance of the center of mass from the earth;
r is the distance of the test mass m from the c.m.
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A Geocentric Solution to the 3-body Problem
There remains for us to derive the generalized equations of motion describing the path of
the test mass, m. Then we need to solve the generalized equations for locations where
m’s velocity is zero relative to both sun and earth.
Equations of Motion for the Infinitesimal Body
As a result of deriving Figure 4 the way we did by using the geocentric equations of
diurnal forces acting on distant masses, we can assume that the two finite masses revolve
in circles around their common center of mass. Since the test mass is assumed
infinitesimally small, it does not change the location of the earth-sun center of mass (i.e.,
the barycenter).
To further simplify our approach, let us assume the unit of mass such that the sum of the
3 masses = 1. This allows us to set the mass of one body equal to 1 - and the other
equal to. In our case,  is the mass of the earth and 1 -  is the mass of the sun. Here
we select the notation such that  is less than or equal to ½.
Let the unit of length be the earth-sun distance, that is, the distance from E to S = 1.
Likewise, we select the unit of time to be such that k2 = 1, where k is such that
m
s k 2 F .
Set the origin of our coordinate system at the center of mass and let the -plane be
determined by their mutual rotation (i.e., the ecliptic plane). Set the coordinates of the
bodies so that 1 -  and  and the test mass be (and (
respectively, and
r1 
   
r2 
  

1
   1    2
2
2     2 2   2 
2
2
Then the differential equations of motion for m are:
  1     2 
d 2
 1   

'
2
dt
r13
r23
  1     2 
d 2
 1   

'
2
dt
r13
r23
d 2


 1    3   3 .
2
dt
r1
r2
The choice of units makes the mean angular motion of the finite bodies be:
(13)
A Geocentric Solution to the 3-body Problem


  k


9
1       
a
3
2
 = 1,

(14)
where a = r1 + r2  1.
If we now change coordinate systems so that the origin is still at the c.m. and the rotation
is still in the -plane in the direction that the bodies move with uniform angular velocity
unity see (14). The coordinates of the new system are defined by the equations:
  x cos t  y sin t ,
  x sin t  y cos t ,
 z
(15)
and similar equations for letters with subscripts 1 and 2.
Taking the second derivatives of (15) we obtain and substituting into equation (13)
yields:
d 2 y

d 2 x

dy
dx
 2  2  x  cos t   2  2  y  sin t
dt
dt
 dt

 dt



x  x1  x  x 2 
y  y1
 y  y 2 
  1   

 cos t  1    3  
 sin t ,
3
3
r1
r2 
r1
r23 


d 2 y

d 2 x

dy
dx

2

x
sin
t

 2

 2  2  y  cos t
dt
dt
 dt

 dt

(16)


x  x1  x  x 2 
 y  y1   y  y 2  
  1   


 sin t  1   
 cos t ,
3
3
r1
r2 
r13
r23 


d 2z
z
z
 1    3   3 .
2
dt
r1
r2
Multiplying the first two equations of the three by cos t and sin t respectively, and then by
–sin t and cos t, and adding; the results are:
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A Geocentric Solution to the 3-body Problem
x  x1  x  x2 
d 2x
dy
2
 x  1   

,
2
dt
dt
r13
r23
 y  y1   y  y2 
d2y
dx
2
 y  1   

,
2
dt
dt
r13
r23
d 2z
z
z
 1    3   3 .
2
dt
r1
r2
Assuming the x-axis as fixed to the centers of the finite bodies (earth and sun, for
instance), then y1=0 and y2=0 and the equations become:
x  x1  x  x2 
d 2x
dy
2
 x  1   

,
2
dt
dt
r13
r23
d2y
dx
y
y
2
 y  1    3   3 ,
2
dt
dt
r1
r2
(17)
d 2z
z
z
 1    3   3 .
2
dt
r1
r2
We now have the differential equations describing the motion of the test mass, m with
respect to axes rotating in such a way that the finite bodies always lie on the x-axis
(earth-sun). The unique property of these expressions is that they do not explicitly
include time. When we started our analysis in equations (13); were
functions of time. 
Equations 17 can be integrated by Jacobi’s Integral if we let
U
1 2
1      ;
x  y2 
2
r1
r2


(18)
so that equations 17 can be rewritten as:
d 2x
dy U
2

,
2
dt
dt
x
d2y
dx U
2 
,
2
dt
y
dt
(19)
d 2 z U

.
z
dt 2
Multiplying these equations by 2(dx/dt), 2(dy/dt) and 2(dz/dt) respectively, and then
summing, the resulting equation can be integrated because U is a function of x, y, and z
alone. Doing so gives:
A Geocentric Solution to the 3-body Problem
2
11
2
2
 dx   dy   dz 
2
         v  2U  C
 dt   dt   dt 
where C is the constant of integration.
By definition this gives
1     2   C
 dx   dy   dz 
2
2
       x  y 2
r1
r2
 dt   dt   dt 
2
2
2
(20)
Since our solution to the problem is of 6th order instead of the usual approach which is of
5th order, we need five more equations to solve the problem. If we confine the motion of
the test mass to the xy-plane, we only need three more equations. Finding one equation,
we can find the remaining two by Jacobi’s last multiplier; so we actually need to find one
more equation.
Equation (20) is a relation between the square of the velocity and the coordinates of m
referenced by the rotating axes. The constant C can be found by initial conditions, so
(20) specifies the velocity of the test mass at all points of the rotating space; and
conversely, given a velocity, (20) gives the locus of points where only m can be. In
particular, if the velocity is set to zero in (20), it will define the surfaces at which the
velocity is zero. On the one side of these surfaces, the velocity will be real and on the
other side, imaginary. We can regard this as saying that it is possible for the body to
move on one side of the surface and impossible to move on the other side.
The equation of the zero surfaces of relative velocity is:
x2  y2  2
r1 
r2 
1     2 
r1
r2
 C,
x  x   y  z ,
x  x   y  z .
2
2
2
1
2
2
(21)
2
2
Since only the squares of y and z occur, the surfaces defined by (21) are symmetrical
with respect to the xy and xz-planes, and, for the case that = 0.5, with respect to the yzplane also. The surfaces for  0.5 can be regarded as deformations of those for = 0.5.
From the geometry of z it follows that a line parallel to the z-axis pierces the surfaces in
two or none real points. Also, the surfaces are contained within a cylinder whose axis is
parallel to the z-axis and whose radius is C , to which certain folds are asymptotic at
z2 = ; for, as z2 increases, the equation approaches as a limit:
x2 + y2 = C.
12
A Geocentric Solution to the 3-body Problem
The equation of the curves of intersection of the surfaces with the xy-plane is obtainable
by setting z = 0 in equation (21):
x2  y2 
21   
x  x 
2
1
 y2

2
 x  x 
2
2
 y2

 C.
(22)
There are two cases we can use to approximate (22).
Case 1: x and y are large.
If x and y are large, the 3rd and 4th terms in (22) are negligible and the equation can
be written as:
x2  y2  C  
where  = the sum of terms 3 and 4, which are negligible in this case. This is the
equation of a circle whose radius is C    . The larger the value of C, the greater
the values of x and y which satisfy the equation. The smaller , the more circular the
curve and the more nearly it reaches the asymptotic cylinder.
Case 2: x and y are small.
For small values of x and y, the first two terms of equation (22) become relatively
unimportant and the equation may be rewritten as:
1  
  C 
r1
r2
These curves plot the locus of points of equal potential energy for the two centers of
force, 1- and . For large values of C they consist of closed ovals around each of
the bodies E and S. For small values of C these ovals unite between the bodies
forming a dumbbell shaped figure in which the ends are of different size except
when  = 0.5. And for still smaller values of C, the handle of the dumbbell enlarges
until the figure becomes an oval enclosing both of the bodies.
It thus follows that the approximate forms of the curves in which the surfaces intersect
the xy-plane are given in Figure 5. The curves C1, C2, C3, C4, and C5 are in order of
decreasing values of the constant C. (Figures 5, 6, and 7 were not computed numerically
but are intended to show qualitatively the characteristics of the curves.)
A Geocentric Solution to the 3-body Problem
Figure 5: xy-Plane Contours
13
14
A Geocentric Solution to the 3-body Problem
Figure 6: xz-Plane Contours
A Geocentric Solution to the 3-body Problem
Figure 7: yz-Plane Contours
15
16
A Geocentric Solution to the 3-body Problem
The equation of the curves of intersection of the surfaces and the xz-plane is obtained by
setting y = 0 in equation (21):
x2 
21   
x  x 
2
1
z
2
2

 x  x 
2
2
z
2

C
(23)
Again we have two cases:
Case 1: large values of x and z
For large values of x and z the 2nd and 3rd terms are negligible and may be written as:
x2 = C - 
which is the equation of a symmetrical pair of straight lines parallel to the z-axis.
The larger the value of C, the larger the value of x which, for a given value of z,
satisfies the equation and, therefore, the smaller is . Hence, the larger C, the closer
the lines are to the asymptotic cylinder.
Case 2: small x and z.
For small values of x and z, the first term in (23) becomes negligible and the
equation can be written:
1   C
  
r1
r2 2
Hence the forms of the curves in the xz-plane are qualitatively like those in Figure 6.
Again, C1, …, C5 are in order of decreasing values of the integration constant, C.
We can likewise compute the curves for the yz=plane by setting x=0 in equation (21).
y2 
x
21   
2
1
 y2  z2


x
2
2
2
 y2  z2

C
(24)
Case 1: y, z large
Following the same reasoning as done for the earlier cases, for large y and z, we can
write
y 2 = C - ,
which is near the asymptotic cylinder.
Case 2: small y and z
For small values of y and z, (24) may be written as
21   
 C 
r1
A Geocentric Solution to the 3-body Problem
17
which is the equation of a circle which becomes larger as C decreases. Hence the
forms of the curves in the yz-plane are qualitatively as given in Figure 7.
From these 3 figures it is easy to infer their forms for the different values of the
integration constant. They may be roughly described as consisting of, for large values of
C, a closed fold approximately spherical in form around each of the finite bodies, and of
curtains hanging from the asymptotic cylinder symmetrically with respect to the xyplane. For smaller values of C, the folds expand and coalesce (Figure 5, curve C3); for
still smaller values if C the united folds coalesce with the curtains, the first points of
cntact being in every case in the xy-plane; and for sufficiently small values of C the
surfaces consist of two parts symmetrical with respect to the xy-plane but not intersecting
it. (Figure 6, curve C5 and Figure 7 curve C6.)
Now that we know the forms of the surfaces, we have to find where the space motion is
real and where it is imaginary.
The square of the velocity is:
v2  x2  y2 
21    2 

C
r1
r2
Assume C is so large that the ovals and curtains all are separate. The motion will be real
in those portions of relative space for which the right member of the velocity equation is
positive. If it is positive in one point in a closed fold, it is positive in every other point
within it because the function changes sign only at a surface of zero relative velocity.
From the velocity equation that x and y can be taken so large that the right member will
be positive, regardless of how great C may be. Therefore, the motion is real outside the
curtains. It is also clear that a point can be chosen so near to either 1- or  (earth or
sun), that is, either r1 or r2 may be taken so small that the right expression will be positive
however great C may be. Therefore, the motion is real within the folds around the finite
bodies.
If the value of C is so large that the folds around the finite bodies were closed, and if the
infinitesimal body should be within one of these folds at the origin of time, it would
always remain there since it could not cross a surface of zero velocity.
If the sun’s motion about the earth is taken to be circular, and the mass of the moon
infinitesimal, we find that the contour of C3, is 40 times larger than the orbit of the moon.
This is so large that the fold around it and the earth is closed with the moon within it.
Therefore, the moon cannot escape earth’s gravity.
Points on the surfaces can be found by determining the curves in the xy-plane and then
finding by approximations the values of z which satisfy equation 20. Specifically, the
curves in the xy-plane are of interest because the first points of contact, as the various
18
A Geocentric Solution to the 3-body Problem
folds coalesce, occur in this plane, and, indeed, on the x-axis as are evident from the
symmetries of the surfaces.
The equation of the curves in the xy-plane is:
x2  y2 
21   
x  x 
2
1
y
2

2
 x  x 
2
2
 y2

C
If this equation is rationalized and cleared of fractions the result is a polynomial of the
16th degree in x and y. When the value of one of the variables is taken arbitrarily the
corresponding values of the other can be found.
This problem presents great practical difficulties because of the high degree of the
equation (16th order), and these difficulties are exacerbated by the presence of extraneous
(imaginary) solutions which are introduced by the process of rationalization.
Transforming to polar coordinates can significantly reduce the degree of the equation.
That is, points on the curves can be defined by giving their distances from two fixed
points on the x-axis. We could not use this method if the curves were not symmetrical
with respect to the axis on which the poles lie.
Let the centers of the bodies 1- and  be taken as poles; the distances from these poles
are r1 and r2 respectively. To complete the transformation it is necessary to express x2+y2
in terms of these quantities.
Figure 8: Transformed x-y axes
Let P be a point on one of the curves; then OA = x, AP=y, and, since O is the center of
mass of 1- and , O  1   , and O1       .
It follows that:
y 2  r12   x     r12  x 2  2 x   2
2
y 2  r22  x  1     r22  x 2  21   x  1    .
On eliminating the first power of x from these equations and solving for x2 + y2, we find
that:
2
2
A Geocentric Solution to the 3-body Problem
19
x 2  y 2  1   r12  r22   1   .
As a consequence of this equation, equation (22) becomes:




r1 

1   r12  2    r22 
2
  C   1     C 
r2 
(25)
If an arbitrary value of r2 is assumed, r1 can be computed from this equation: the points of
intersection of the circles around 1- and  as centers, with the computed and assumed
values respectively of r1 and r2 as radii will be points on the curves. As a result, we may
let equation (25) be written in the form:
r13  ar1  b  0,
a
C


1  1 
 2 2
r2  ,
r2 

(26)
b  2.
Since b=2 is positive, there is at least one real negative root of the first part of (26)
whatever value a may have. But the only value of r1, which has any meaning in this
problem, is real and positive; hence, the condition for real positive roots must be
considered.

2
It follows from (25) that C  is always greater than  r22   for all real, positive values
r2 

of r1 and r2; therefore, a is always negative.
From the Theory of Equations we know that a cubic equation of this form (top line of
(26)) has three distinct real roots if
27b 2  4a 3  0 ; or, since b=2, if
a + 3 < 0.
Given this inequality, we can find the cubic roots as:
(27)
20
A Geocentric Solution to the 3-body Problem
sin  
b 27

,  ,
3
2 a
2
r11  2
a

sin ,
3
3
r12  2
a   
sin  60  ,
3
3

r13  2
(28)
a   
sin  60  .
3
3

where r11, r12, r13 are the three roots of the cubic.
The limit of the inequality (27), or, in terms of he original quantities, is,
r23  a r2  b  0
C  31   
a   
,


b   2.
(29)
The solution of this equation gives the extreme values of r2 for which (26) has real roots.
Therefore, in actual computation equation (29) should first be solved for r21, and r22. The
values of r2 to be substituted in (26) should be chosen at convenient intervals between
these roots.
Equation (29) will not have real, positive roots for all values of a´, the condition for real,
positive roots being:
a  3  0 ;
the limiting value of which is, in the original quantities,

C  31   

 3,


whence C´ = 3.
Therefore, C´ must be equal to, or greater than, 3 in order that the curves shall have real
points in the xy-plane. For C´=3, the curves are just vanishing from the xy-plane and it
follows that equation (25) is satisfied by r1 = 1, and r2 = 1; i.e., the surfaces vanish from
the xy-plane at the points which form equilateral triangles with 1- and .
From the overall form of the surfaces that the pairs of points which appear as C decreases
are all in the xy-plane. Therefore, it is sufficient here to consider the equation of the
curves in the xy-plane.
A Geocentric Solution to the 3-body Problem
21
There are three pairs of points on the x-axis which appear when the ovals around the
finite bodies touch each other and when they touch the exterior curve enclosing the both
of them. Two more appear as the surfaces vanish form the xy-plane, at the two points
making equilateral triangles with the finite bodies. These points are critical points of
their respective contours and they are connected with important dynamical properties of
the system.
Let the equation of a contour be written as:
F  x, y   x 2  y 2 
21    2

 C  0.
r1
r2
(30)
Differentiating we get the conditions for the twin pairs:
x  x1    x  x2   0
1 F
 x  1   
2 x
r13
r23
1 F
y
y
 y  1    3   3  0
2 y
r1
r2
(31)
The left members of these equations are the same as the right members of equations (17)
for z=0. The terms to the left of the equal signs in (31) are proportional to the direction
cosines of the normal at all ordinary points of the curves; and, since dx/dt and dy/dt are
zero at the surfaces of zero velocity, it follows from (17) that the directions of
acceleration, i.e., the lines of effective force are orthogonal to he surfaces of zero
velocity.
Thus, if the infinitesimal body is placed on a surface of zero relative velocity it will start
moving in the direction of the normal. But at the twin points, the sense (direction and
amount) of the normal becomes ambiguous. Hence, it might be conclude that if the
infinitesimal body were placed at one of these points it would remain relatively at rest.
dx 2
dy 2
and
, which are the
dt 2
dt 2
accelerations in the x and y direction respectably, must vanish as per (17). The result of
the latter constraint on accelerations is that if the infinitesimal body, m, is placed at a twin
point with zero velocity, its coordinates will identically fulfill the differential equations of
motion and it will remain forever at rest with respect to sun and earth unless an external
force is brought to bear upon it.
The conditions imposed by (30) and (31) require that
Consider constraints (31); the second of which is fulfilled if we set y = 0. The twin
points on the x-axis and the linear solutions of the problem statement are given by the
conditions:
22
A Geocentric Solution to the 3-body Problem
x  1   
x  x1 
x  x  
1
3
2 2

x  x2 
x  x  
2
3
2 2
 0,
y  0,
(32)
z  0.
The first term of the first equation in (32), when taken as a function of x, is positive in the
limit as x goes to +. It is negative for x = x2 -  where  is a very small positive
quantity. It is positive for x = x2 - ; and it is negative for x = x1 +  It is positive for
x=x1-; and it is negative for x=-. Therefore there are three positions along the line
through the finite bodies at which the infinitesimal body can remain when placed there.
We now have three cases:
Case 1:
Let the distance from  to the double point on the x-axis between + and x2 be
represented by . Then:
x-x2=
x-x1=r1=1+
x=1-
Therefore, after clearing fractions, the first equation of (32) then becomes:
 5  3    4  3  2  3   2  2     0
(33)
This 5th order equation has one change of signs in its coefficients, and thus has only
one real positive root. The value of this root depends upon , the mass of the earth.
Consider the left member of the equation as a function of and For =0, the
equation becomes:
 3  2  3  3  0.


This has 3 roots: one of them is zero, and the other two are imaginary.
It follows that for  0 but sufficiently small, 3 roots of equation (33) can be
1
3
expressed as a power series in  . One root will be real, the others complex.
Therefore the real root has the form:
1
2
3
  a1  3  a 2  3  a3  3  ...
Substituting this into equation (33) and setting the result to zero, the coefficients (a1,
a2, a3, …) of corresponding powers of 1/3, we find that:
A Geocentric Solution to the 3-body Problem
a1 
And so:
23
2
3
1
3
1
3
3
, a3   , …
, a2 
3
9
27
1
2
3
  3 1   3 1   3
r2             ...
3 3 
9 3 
3
r1  1   .
(34)
The corresponding value of C´ is found by substituting these values of r1 and r2 in
equation (25).
Case 2:
Let the distance from  to the twin point on the x-axis between x2 and x1 be
represented by . Then x - x2 = -, x - x1 = r1 = 1 - , x = (1 - ) -. Therefore, the
first of equation (32) becomes:
 5  3    4  3  2   3   2  2     0
On solving as in Case 1, the values of r2 and r1 are
1
2
3
  3 1   3 1   3
r2             ...
3 3 
9 3 
3
r1  1   .
(35)
C´ is found by substituting r1 and r2 into (25).
Case 3:
Let the distance form 1- to the double point on the x-axis between x1 and - be
represented by 1-. Then:
x  x 2  2  
x  x1  1  
x    1  
and the first equation of (32) becomes
 5  7    4  19  6   3  24  13  2  12  14   7   0
(36)
If =0, which is to say that both m and the lesser of the two bodies’ masses is
negligible, we obtain
 5  7  4  19 3  24 2  12  0
24
A Geocentric Solution to the 3-body Problem
which has only one root, =0. Therefore,  can be expressed as a power series in 
which converges for sufficiently small values of this parameter and vanishes with it
when  = 0. This root has the form:
  c1   c 2  2  c3  3  c 4  4  ...
Substituting the expression for  into (36) and equating to zero the coefficients of the
various powers of , we find that:
c1 
7
23 7 2
,…
, c 2  0 , c3 
12
12 4
Hence
7
23  7 2 3

  ...
12
12 4
r1  1   ,

(37)
r2  1  r1  2  
It’s C´ can be found by substituting into equation (25).
To find the twin points not on the x-axis, we again turn to equation (31). They, or any
two independent functions of them, define the twin points. Since y is distinct from zero,
we may divide it into the second equation, which yields:
1
1  
 3  0.
r13
r2
Multiplying this equation by x - x2 and x - x1, and subtracting the products separately
from the first equation of (31) gives:
x 2  1   
x1  
x2  x1 
r13
x1  x2 
r23
0
0
z 0
But, since x 2  1   , x1    , and x 2  x1  1 ; we conclude:
A Geocentric Solution to the 3-body Problem
25
1
1
0
r13
1
1
0
r23
z0
The only real solutions are r1  r2  1 so these points form equilateral triangles with the
two finite bodies, whatever their masses may be. Since z = 0, these points—called
Trojans—are located where the surfaces vanish from the xy-plane.
Figure 9 plots the solutions on the xy-plane.
Figure 9: The Five Lagrange Points Plotted on the xy-plane
It is clear from the presence of the asymptotic cylinders that many other critical points
have to exist. Such a potential candidate would look like the C6 contours on the y-axis of
Figure 7, which coincide with the L4 and L5 points.
In any case, we have taken the long way around to demonstrate that a geocentric
derivation follows from the dynamic explanation of the geocentric system. Indeed, the
so-called “fictitious forces” are brought into play because a geocentric system is
considered “fictional.” Nevertheless, in a geocentric coordinate system they come into
play because they are real, gravitational forces. Dynamic derivations lament the need for
invoking the fictitious forces in order to represent their Lagrange points derivations.
26
A Geocentric Solution to the 3-body Problem
THE LAGRANGE POINTS: A MODERN APPROACH3
The derivation we have applied in the previous section is a rigorous one derived from the
definition of force and geometry. In this section, we use another approach which is based
on Newtonian gravitational force. We again start with the same initial conditions as we
did with Figure 4, here presented as Figure 10.
Figure 10: The restricted 3-body problem
There are five equilibrium points to be found in the vicinity of two orbiting masses. They
are called Lagrange Points in honor of the French-Italian mathematician Joseph
Lagrange, who discovered them while studying the restricted three-body problem. The
term “restricted” refers to the condition that two of the masses are very much heavier
than the third. Today we know that the full three-body problem is chaotic, and so cannot
be solved in closed form. Therefore, Lagrange had good reason to make some
approximations. Moreover, there are many examples in our solar system that can be
accurately described by the restricted three-body problem.
The procedure for finding the Lagrange points is straightforward: We seek solutions to
the equations of motion which maintain a constant separation between the three bodies.


If M1 and M2 are two masses, and r1 and r2 are their respective position, then the total

force exerted on a third mass m, at a position r , will be
 GM 1 m   GM 2 m  
F  
(r  r1 )   3 r  r2 
3
r  r1
r  r2
3
Source: Neil J. Cornish, & Jeremy Goodman,
http://www.physics.montana.edu/faculty/comish/lagrange.pdf
(38)
A Geocentric Solution to the 3-body Problem
27


The catch is that both r1 and r2 are functions of time since M1 and M2 are orbiting each


other. Undaunted, one may proceed and insert the orbital solution for r1 (t) and r2 (t)
(obtained by solving the two-body problem for M1 and M2) and look for solutions to the
equation of motion


d 2 r t 
,
F t   m
dt 2
(39)
that keep the relative positions of the three bodies fixed. It is these stationary4 solutions
that are known as Lagrange points.
The easiest way to find the stationary solutions is to adopt a co-rotating frame of
reference in which the two large masses hold fixed positions. The new frame of
reference has its origin at the center of mass, and an angular frequency  given by
Kepler’s law:
 2 R 3  GM 1  M 2 
(40)
Here R is the distance between the two masses [earth and sun —GB]. The only drawback
of using a non-inertial frame of reference is that we have to append various pseudoforces to the equation of motion.5 [Emphasis & footnote added —GB.] The effective

force in a frame rotating with angular velocity  is related to the inertial force

F according to the transformation
    dr
 

F  F  2m     m    r .
dt 



(41)
The first “correction” is the Coriolis force and the second is the centrifugal force. The
effective force can be derived from the generalized potential
  1    
U  U  v   r   r   r ,
2
(42)

d
F   rU    vU  .
dt
(43)

as the general gradient
 


The velocity dependent terms in the effective potential do not influence the positions of
the equilibrium points, but they are crucial in determining the dynamical stability of
4
I.e., stationary relative to both the earth and the sun, this is clearly geostatic, as per the theory of
Geocentricity.
5
This refers to the so-called fictitious forces which in a geocentric framework are real gravitational forces.
In short, this statement says that in a geostatic framework these forces cannot be dismissed as fictitious
since they are necessary to obtain the correct equations.
28
A Geocentric Solution to the 3-body Problem

motion about the equilibrium points. A plot of U with v =0, M1 = 10, M2=1 and R = 10
is shown in Figure 11. The extrema of the generalized potential are labeled L1 through
L5.
Figure 11: A contour plot of the generalized potential
Choosing a set of Cartesian coordinates originating from the center of the masses with the
z-axis aligned with the angular velocity, we have
 
  k

r  xt iˆ  y (t ) ˆj

r1  Riˆ

r2  Riˆ
(44)
where

M2
M1
, 
.
M1  M 2
M1  M 2
(45)
A Geocentric Solution to the 3-body Problem
29
 
To find the static equilibrium points we set the velocity v  dr / dt to zero and seek
 
solutions to the equation F  0 , where


  x  R R 3
  x   R R 3  ˆ
F   2  x 

i
2
2
2 3/ 2
2 3/ 2 

 x  R   y
x  R   y





 yR 3
2
 y

x  R 2  y 2



3/ 2



yR
x  R 
2
3
 y2

3/ 2
(46)

 ˆj


Here the mass m has been set equal to unity without loss of generality. The brute-force
approach for finding the equilibrium points would be to set the magnitude of each force
component to zero, and solve the resulting set of coupled, fourteenth order equations for x
and y. A more promising approach is to think about the problem physically, and use the
symmetries of the system to guide us to the answer [which is what we did in the first
section —GB].
Since the system is reflection-symmetric about the x-axis, the y component of the force
must vanish along this line. Setting y = 0 and writing x=R(u+ ) (so that u measures the
distance from M2 in units of R), the condition for the force to vanish along the x-axis
reduces to finding solutions to the three fifth-order equations.




u 2 1  s1   3u  3u 2  u 3   s0  2 s0 u  1  s0  s1 u 2  2u 3  u 4 ,
(47)
where s0 is the sign(u) and s1 is the sign(u+1). The three cases we need to solve have
(s0,s1) equal to (-1,1), (1,1), and (-1,-1). The case (1,-1) cannot occur. In each case there
is one real root to the quintic equation, giving us the positions of the first three Lagrange
points. We are unable to find closed-form solutions to equation (47) for general values of
<<1. To lowest order in , we find the first three Lagrange points to be positioned at
1
 
 
   3  
L1 :  R 1    ,0 ,
  3  
 
 
1
 
 
   3  
L2 :  R 1    ,0 ,
  3  
 
 
(48)

5  

L3 :   R 1   ,0 .
 12  

For the earth-sun system   3  10-6, R  1.5  108 km, and the first and second Lagrange
points are located approximately 1.5 million kilometers from earth. The third Lagrange
point orbits the sun just a bit further out than one astronomical unit [on the far side of the
sun from the earth].
30
A Geocentric Solution to the 3-body Problem
Identifying the remaining two Lagrange points requires a little more thought. We need to
balance the centrifugal force, which acts in a direction radially outward from the center of
mass, with the gravitational force exerted by the two masses. Clearly, force balance in
the direction perpendicular to centrifugal force will only involve gravitational forces.
This suggests that we should resolve the force into directions parallel and perpendicular

to r . The appropriate projection vectors are xiˆ  yˆj and yiˆ  xˆj . The perpendicular
projection yields

1

F   y 2 R 3 
  x  R 2  y 2



2/3

1
x  R 
2

.
2 2/ 3 
y


(49)

Setting F  0 and y  0 tells us that the equilibrium points must be equidistant from
the two masses. Using this fact, the parallel projection simplifies to read
F||    2
x2  y2
R
 1
1


3
R
 x  R  2  y 2



 
3/ 2



(50)
Demanding that the parallel component of the force vanish leads to the condition that the
equilibrium points are at a distance R from each mass. In other words, L4 is situated at
the vertex of an equilateral triangle, with the two masses forming the other vertices. L5 is
obtained by mirror reflection of L4 about the x-axis. Explicitly, the fourth and fifth
Lagrange points have coordinates
 R  M  M2  3 
,
L4 :   1
R ,
2
M

M
2
2 
  1

.
 R  M1  M 2 

3
,
L5 :  
R 
 2  M1  M 2  2 
(51)
Interestingly, the last two Lagrange points are stable because any deflecting of a particle
invokes a Coriolis force that brings it back to the point.
Since the only reason why the center of mass of the earth-sun balance was used only to
simplify the derivations, the earth could just as well have been taken as the center:
therefore, we expect no difference will be discovered for these points by any
geocentrically-based coordinate system. The rule that it is “six-to-one, half-dozen of the
other” when it comes to mathematical or physical proofs or disproof of heliocentric and
geocentric physics remains steadfastly true.