Geocentric solution to 3-body prob - Galileo Was Wrong...the theory of geocentricity should yield a different solution than what is observed. In anticipation of that challenge Bob

A Geocentric Solution to the 3-body Problem 1

A GEOCENTRIC SOLUTION TO THETHREE-BODY PROBLEM

Prepared byGerard 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. Inanticipation of that challenge Bob Sungenis invited me to derive a geocentric solutionwithin the framework of the theory of geocentricity. Our purpose was to derive a viablegeocentric framework for the 3-body problem. We were successful and this paper is thefruit of our labors.

The Approach

The version of the 3-body problem we consider here is, as the name suggests, of threebodies in mutual orbit about each other. Classically, the 3-body problem assumes thattwo of the bodies are massive and the third body’s mass can be regarded as zero. Unlikemost modern dynamic derivations posted on the web which know the solutions ahead oftime, the approach in this paper is completely general and so could be used to locate anyadditional critical points besides the five Lagrangian points. Geocentrically, the 3 bodiesare: the sun, the earth, and a third body, of negligible mass, that appear at rest relative tothose two bodies. Fundamentally, we are looking to answer this question: “At whatlocations is the orbital speed of the negligible-mass body zero with respect to both thesun 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 Eand the sun is located at point S. The angular velocity vector, D represents the diurnalrotation of the firmament about the axis that runs from the center of the earth through itspoles. The angular velocity vector, A, is inclined at 23.5° to D, which is the angularvelocity of the earth-sun line about the earth in the course of the year. The geometry issuch that the sun is located at its southernmost point, namely the first day of winter, thatis, roughly 21 December. Thus the line from E to S goes clockwise around the point E asseen from A, with a period of one year. Likewise, as seen from the tip of D, the entirepage 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 shownspecifically for the diurnal (daily) rotation case that the system holds together; that thefirmament’s gravitational field, which we call inertia, holds the stellar universe togethereven though some may think that the orbital speeds would be immense. But rotation isnot the same as revolution, which is to say that if the entire universe rotates as one inertial

2 A Geocentric Solution to the 3-body Problem

field, the presence of that rotation cannot be felt by the atomic matter of the universe, orby photons traveling through it.

We shall shortly demonstrate that neither the daily rotation of the firmament nor theyearly orbital motion of the sun’s entire about the earth can disrupt the motions weobserve. When we are finished with that section, we can simplify our problem to thatpresented 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 ofchoice, used by physicists to describe motion. Usually physicists are well behaved intheir use of math, but at times, they introduce fudge factors to bridge what theorydemands and experiment lacks. Even then, the fudging is quite obvious from the namesgiven the fudge factors such as “guillotine factor,” for instance. But there are times whenreputations and careers are at stake and at those times, the fudging becomes quite subtle,even mean-spirited at times.


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 ofevidence for the geocentric universe, physicists resort to sleight of hand to keep the earthin orbit about the sun when all fundamental experimental results reveal earth to bestanding still in the firmament. In this case, they multiply one side of the generalizedequation of motion by the number one. Before multiplying by one, the equation is said tobe kinematic, describing the accelerations and velocities of the bodies but not taking themasses of the bodies into consideration. For instance, consider this equation thatdescribes the velocity, v, of a body in circular motion with a rotational speed of at adistance R from the center of the circle:

v= R. (1)This equation is said to be kinematic and even though it perfectly describes the velocityand 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 themass, m, divided by itself, i.e., m/m. This is equivalent to multiplying both sides of theequation 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 obviouslycancel out the m’s and simplify equation (2) back to equation (1). To hide this sleight ofhand, 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 cancelits 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 abuilt-in set of reaction rules. These rules dictate the behaviors of accelerating bodies andthe 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 onearth are specified by longitude and latitude, so a star’s coordinates are given by its rightascension and declination. A star’s longitude is specified by its right ascension and itslatitude 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 thefirmament—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 theargument at hand.


Figure 2: The Geocentric View of the Daily Rotation

The following is a derivation of the dynamical equations for the universe rotating aboutthe 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 throughthe north and south poles of earth: imagine it measured in degrees per secondalthough technically we use radians per second.2

The usual objection against geocentricity is that the earth is not massive enough to havethe universe “orbit” it once a day. In reality, neither the mass of the earth nor the mass ofthe 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:

2

2

dt

Rda (4)

2 There are 2 radians in the circumference of a circle, so a radian is roughly 57 degrees.


Here, R is the distance to a moving object and t is time. This can be rewritten moreexplicitly as:

dt

dR

dt

da (5)

where dR/dt is the velocity, v, of the moving object, the star in our case. This equationsays, “Acceleration is the rate of change in velocity.”

But we’re not trying to model the speed and acceleration of an automobile here but thatof a distant star rotating about the earth once every 23 hours and 56 minutes. We mustthus add the rotational velocity (Equation (1)) into the mix. This requires us to rewriteequation (5) as:

R

dt

dR

dt

da (6)

where is the angular velocity (measured in degrees per second, for instance) and R isthe distance of the star from the axis of rotation.

Distributing the derivative (d/dt) through the terms in parentheses of Equation (6) givesus equation (7):

Rdt

dRR

dt

d

dt

Rda

2

2

2

. (7)

Here the first term on the right-hand side of equation (7) is any acceleration that may be

imparted to the earth (the central point). The second term,

R

dt

d, is the Euler force,

which is not of interest here since it only kicks in if the length of the day changessignificantly over the course of a day. The third term (starting with the 2), is the Coriolisforce and the last term [R)] is the centrifugal force.

The Coriolis and centrifugal forces dominate the motion of the sun, planets, and stars in ageocentric system. We shall thus ignore the Euler and local acceleration terms ofequation (7) and work only with:

Rva 2 (8)where v is the orbital speed of the star. Since the firmament rotates and not the earth, thesign of v is in the opposite direction to the heliocentric system, and is thus negative. Thev in equation (8) is thus replaced by R .

After expanding v, equation (8) is now:

RRa 2 ;or

Ra . (9)

Distributing the cross-product through the term in parentheses gives us:


RRa . (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:

sinˆ2 DRa . (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 circlein Figure 1. This keeps the acceleration experienced by the star confined to the star’slatitude, 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 wherethey are shown as acceleration vectors. To make them dynamic, multiply each by thestar’s mass. The acceleration pictured by the sine term is aligned along the rotationvector, , and serves to keep the star’s rotationalplane from “falling” up or down the rotational axis.The second component is the cosine term. Thatacceleration pulls the star towards the axis ofrotation. If multiplied by the star’s mass it becomesa centripetal (non-fictitious) force, meaning that itpulls the star towards the axis of rotation. The netresult of these two accelerations is to keep the starin its place in the inertial field of the universe whichis the gravitational field of the firmament.Of course, equation (11) is kinematic, not dynamicand we have to show the geocentric model isdynamically correct. To do that, all we have to dois to multiply both sides by the star’s mass, m:

sinˆ2 DRmmaF (12)

Although we assumed the body was a star inderiving (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 thefirmament that rotates on the polar axis once a day and so photons, which are transmittedby the firmament, also participate in the daily rotation. Light, will also obey the above

Figure 3: Accelerations (Forces)


equations superimposed on its own motion. To object that it still exceeds the speed oflight we answer that the speed of light speed limit does not apply for rotation. In thiscase it is equivalent to claiming that when a supersonic jet flies faster than the speed ofsound, you could not talk to the person in front of you because you were flying fasterthan the speed of sound. But the air in the plane, too, was “flying” faster than the speedof sound, so you can talk to the person in the seat in front of you because the sound-bearing medium was carried with you, even as the light-bearing medium is carried alongwith 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 whatwe see and measure of the daily rotation whether that rotation is of the earth within theuniverse or the universe around the earth. In the final analysis, proofs based ondynamical equations are not proofs of anything; nor are they proofs against the geocentricuniverse.

By the same approach, we could show that the yearly orbital motion of the sun andplanets can be represented in the geocentric framework of the firmament. But for thispaper we shall only point out that every object in the universe obeys equation (12). SinceS, m, and even the c.m. in Figure 1 will all obey equation (12) for the rotational case, wecan 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.


There remains for us to derive the generalized equations of motion describing the path ofthe test mass, m. Then we need to solve the generalized equations for locations wherem’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 ofdiurnal forces acting on distant masses, we can assume that the two finite masses revolvein circles around their common center of mass. Since the test mass is assumedinfinitesimally 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 the3 masses = 1. This allows us to set the mass of one body equal to 1 - and the otherequal to. In our case, is the mass of the earth and 1 - is the mass of the sun. Herewe 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

Fksm 2 .

Set the origin of our coordinate system at the center of mass and let the -plane bedetermined by their mutual rotation (i.e., the ecliptic plane). Set the coordinates of thebodies so that 1 - and and the test mass be (and (respectively, and

221211 r

222222 r

Then the differential equations of motion for m are:

.1

'1

'1

32

31

2

2

32

2

31

1

2

2

32

2

31

1

2

2

rrdt

d

rrdt

d

rrdt

d

(13)

The choice of units makes the mean angular motion of the finite bodies be:


2

3

1

a

k

= 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 rotationis still in the -plane in the direction that the bodies move with uniform angular velocityunity see (14). The coordinates of the new system are defined by the equations:

z

tytx

tytx

,cossin

,sincos

(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:

.1

,cos1sin1

cos2sin2

,sin1cos1

sin2cos2

32

31

2

2

32

2

31

1

32

2

31

1

2

2

2

2

32

2

31

1

32

2

31

1

2

2

2

2

r

z

r

z

dt

zd

tr

yy

r

yyt

r

xx

r

xx

tydt

dx

dt

ydtx

dt

dy

dt

xd

tr

yy

r

yyt

r

xx

r

xx

tydt

dx

dt

ydtx

dt

dy

dt

xd

(16)

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:


.1

,12

,12

32

31

2

2

32

2

31

1

2

2

32

2

31

1

2

2

r

z

r

z

dt

zd

r

yy

r

yyy

dt

dx

dt

yd

r

xx

r

xxx

dt

dy

dt

xd

Assuming the x-axis as fixed to the centers of the finite bodies (earth and sun, forinstance), then y1=0 and y2=0 and the equations become:

.1

,12

,12

32

31

2

2

32

31

2

2

32

2

31

1

2

2

r

z

r

z

dt

zd

r

y

r

yy

dt

dx

dt

yd

r

xx

r

xxx

dt

dy

dt

xd

(17)

We now have the differential equations describing the motion of the test mass, m withrespect 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 explicitlyinclude time. When we started our analysis in equations (13); werefunctions of time.

Equations 17 can be integrated by Jacobi’s Integral if we let

21

22 1

2

1

rryxU

; (18)

so that equations 17 can be rewritten as:

.

,2

,2

2

2

2

2

2

2

z

U

dt

zd

y

U

dt

dx

dt

yd

x

U

dt

dy

dt

xd

(19)

Multiplying these equations by 2(dx/dt), 2(dy/dt) and 2(dz/dt) respectively, and thensumming, the resulting equation can be integrated because U is a function of x, y, and zalone. Doing so gives:


CUvdt

dz

dt

dy

dt

dx

22

222

where C is the constant of integration.

By definition this gives

C

rryx

dt

dz

dt

dy

dt

dx

21

22

222

21

2

(20)

Since our solution to the problem is of 6th order instead of the usual approach which is of5th order, we need five more equations to solve the problem. If we confine the motion ofthe 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 onemore equation.

Equation (20) is a relation between the square of the velocity and the coordinates of mreferenced 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; andconversely, given a velocity, (20) gives the locus of points where only m can be. Inparticular, if the velocity is set to zero in (20), it will define the surfaces at which thevelocity is zero. On the one side of these surfaces, the velocity will be real and on theother side, imaginary. We can regard this as saying that it is possible for the body tomove on one side of the surface and impossible to move on the other side.

The equation of the zero surfaces of relative velocity is:

.

,

,21

2

222

22

222

11

21

22

zyxxr

zyxxr

Crr

yx

(21)

Since only the squares of y and z occur, the surfaces defined by (21) are symmetricalwith respect to the xy and xz-planes, and, for the case that = 0.5, with respect to the yz-plane 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 intwo 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.


The equation of the curves of intersection of the surfaces with the xy-plane is obtainableby setting z = 0 in equation (21):

.

212

22

222

1

22 Cyxxyxx

yx

(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 canbe written as:

Cyx 22

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 greaterthe values of x and y which satisfy the equation. The smaller , the more circular thecurve 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 relativelyunimportant and the equation may be rewritten as:

Crr 21

1

These curves plot the locus of points of equal potential energy for the two centers offorce, 1- and . For large values of C they consist of closed ovals around each ofthe bodies E and S. For small values of C these ovals unite between the bodiesforming a dumbbell shaped figure in which the ends are of different size exceptwhen = 0.5. And for still smaller values of C, the handle of the dumbbell enlargesuntil the figure becomes an oval enclosing both of the bodies.

It thus follows that the approximate forms of the curves in which the surfaces intersectthe xy-plane are given in Figure 5. The curves C1, C2, C3, C4, and C5 are in order ofdecreasing values of the constant C. (Figures 5, 6, and 7 were not computed numericallybut are intended to show qualitatively the characteristics of the curves.)


Figure 5: xy-Plane Contours


Figure 6: xz-Plane Contours


Figure 7: yz-Plane Contours


The equation of the curves of intersection of the surfaces and the xz-plane is obtained bysetting y = 0 in equation (21):

C

zxxzxxx

22

222

1

2 212 (23)

Again we have two cases:

Case 1: large values of x and zFor 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 closerthe 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 theequation can be written:

2

1

21

C

rr

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).

Czyxzyx

y

2222

2221

2 212 (24)

Case 1: y, z largeFollowing the same reasoning as done for the earlier cases, for large y and z, we canwrite

y2 = C - ,which is near the asymptotic cylinder.

Case 2: small y and zFor small values of y and z, (24) may be written as

C

r1

12


which is the equation of a circle which becomes larger as C decreases. Hence theforms 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 theintegration constant. They may be roughly described as consisting of, for large values ofC, a closed fold approximately spherical in form around each of the finite bodies, and ofcurtains hanging from the asymptotic cylinder symmetrically with respect to the xy-plane. For smaller values of C, the folds expand and coalesce (Figure 5, curve C3); forstill smaller values if C the united folds coalesce with the curtains, the first points ofcntact being in every case in the xy-plane; and for sufficiently small values of C thesurfaces consist of two parts symmetrical with respect to the xy-plane but not intersectingit. (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 isreal and where it is imaginary.

The square of the velocity is:

C

rryxv

21

222 212

Assume C is so large that the ovals and curtains all are separate. The motion will be realin those portions of relative space for which the right member of the velocity equation ispositive. If it is positive in one point in a closed fold, it is positive in every other pointwithin 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 willbe positive, regardless of how great C may be. Therefore, the motion is real outside thecurtains. It is also clear that a point can be chosen so near to either 1- or (earth orsun), that is, either r1 or r2 may be taken so small that the right expression will be positivehowever great C may be. Therefore, the motion is real within the folds around the finitebodies.

If the value of C is so large that the folds around the finite bodies were closed, and if theinfinitesimal body should be within one of these folds at the origin of time, it wouldalways 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 mooninfinitesimal, 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 thenfinding by approximations the values of z which satisfy equation 20. Specifically, thecurves in the xy-plane are of interest because the first points of contact, as the various


folds coalesce, occur in this plane, and, indeed, on the x-axis as are evident from thesymmetries of the surfaces.

The equation of the curves in the xy-plane is:

C

yxxyxxyx

22

222

1

22 212

If this equation is rationalized and cleared of fractions the result is a polynomial of the16th degree in x and y. When the value of one of the variables is taken arbitrarily thecorresponding values of the other can be found.

This problem presents great practical difficulties because of the high degree of theequation (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 fixedpoints on the x-axis. We could not use this method if the curves were not symmetricalwith 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 polesare r1 and r2 respectively. To complete the transformation it is necessary to express x

2+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 , 1O , and 1O .

It follows that:

222122

12 2 xxrxry

222222

22 1121 xxrxry .

On eliminating the first power of x from these equations and solving for x2 + y2, we findthat:


.11 222

122 rryx

As a consequence of this equation, equation (22) becomes:

CCr

rr

r

1

221

2

22

1

21 (25)

If an arbitrary value of r2 is assumed, r1 can be computed from this equation: the points ofintersection of the circles around 1- and as centers, with the computed and assumedvalues respectively of r1 and r2 as radii will be points on the curves. As a result, we maylet equation (25) be written in the form:

.2

,2

11

,0

2

22

13

1

b

rr

Ca

barr

(26)

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 thisproblem, is real and positive; hence, the condition for real positive roots must beconsidered.

It follows from (25) that C is always greater than

2

22

2

rr for all real, positive values

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

0427 32 ab ; or, since b=2, if

a + 3 < 0. (27)

Given this inequality, we can find the cubic roots as:


,27

2sin

3a

b

2

,

.3

60sin3

2

,3

60sin3

2

,3

sin3

2

13

12

11

ar

ar

ar

(28)

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,

.2

,13

023

2

b

Ca

brar

(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. Thevalues of r2 to be substituted in (26) should be chosen at convenient intervals betweenthese roots.

Equation (29) will not have real, positive roots for all values of a ,́ the condition for real,positive roots being:

03 a ;

the limiting value of which is, in the original quantities,

,3

13

C

whence C´ = 3.

Therefore, C´ must be equal to, or greater than, 3 in order that the curves shall have realpoints in the xy-plane. For C´=3, the curves are just vanishing from the xy-plane and itfollows that equation (25) is satisfied by r1 = 1, and r2 = 1; i.e., the surfaces vanish fromthe 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 decreasesare all in the xy-plane. Therefore, it is sufficient here to consider the equation of thecurves in the xy-plane.


There are three pairs of points on the x-axis which appear when the ovals around thefinite bodies touch each other and when they touch the exterior curve enclosing the bothof them. Two more appear as the surfaces vanish form the xy-plane, at the two pointsmaking equilateral triangles with the finite bodies. These points are critical points oftheir respective contours and they are connected with important dynamical properties ofthe system.

Let the equation of a contour be written as:

0212,21

22

Crr

yxyxF

. (30)

Differentiating we get the conditions for the twin pairs:

012

1

012

1

32

31

32

2

31

1

r

y

r

yy

y

F

r

xx

r

xxx

x

F

(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 directioncosines of the normal at all ordinary points of the curves; and, since dx/dt and dy/dt arezero at the surfaces of zero velocity, it follows from (17) that the directions ofacceleration, i.e., the lines of effective force are orthogonal to he surfaces of zerovelocity.

Thus, if the infinitesimal body is placed on a surface of zero relative velocity it will startmoving in the direction of the normal. But at the twin points, the sense (direction andamount) of the normal becomes ambiguous. Hence, it might be conclude that if theinfinitesimal body were placed at one of these points it would remain relatively at rest.

The conditions imposed by (30) and (31) require that2

2

dt

dxand

2

2

dt

dy, which are the

accelerations in the x and y direction respectably, must vanish as per (17). The result ofthe latter constraint on accelerations is that if the infinitesimal body, m, is placed at a twinpoint with zero velocity, its coordinates will identically fulfill the differential equations ofmotion and it will remain forever at rest with respect to sun and earth unless an externalforce is brought to bear upon it.

Consider constraints (31); the second of which is fulfilled if we set y = 0. The twinpoints on the x-axis and the linear solutions of the problem statement are given by theconditions:


.0

,0

,012

32

2

2

2

32

1

1

z

y

xx

xx

xx

xxx

(32)

The first term of the first equation in (32), when taken as a function of x, is positive in thelimit as x goes to +. It is negative for x = x2 - where is a very small positivequantity. It is positive for x = x2 - ; and it is negative for x = x1 + It is positive forx=x1-; and it is negative for x=-. Therefore there are three positions along the linethrough 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 berepresented by . Then:

x-x2=x-x1=r1=1+

x=1-

Therefore, after clearing fractions, the first equation of (32) then becomes:

02233 2345 (33)

This 5th order equation has one change of signs in its coefficients, and thus has onlyone 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, theequation becomes:

.03323

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

expressed as a power series in 31

. One root will be real, the others complex.Therefore the real root has the form:

...33

33

2

23

1

1 aaa

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:


,3

332

1 a9

331

2 a , ,27

13 a …

And so:

.1

...39

1

33

1

3

1

3

3

3

2

3

1

2

r

r (34)

The corresponding value of C´ is found by substituting these values of r1 and r2 inequation (25).

Case 2:Let the distance from to the twin point on the x-axis between x2 and x1 berepresented by . Then x - x2 = -, x - x1 = r1 = 1 - , x = (1 - ) -. Therefore, thefirst of equation (32) becomes:

02233 2345

On solving as in Case 1, the values of r2 and r1 are

.1

...39

1

33

1

3

1

3

3

3

2

3

1

2

r

r (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 - berepresented by 1-. Then:

1

1

2

1

2

x

xx

xx

and the first equation of (32) becomes

07141213246197 2345 (36)

If =0, which is to say that both m and the lesser of the two bodies’ masses isnegligible, we obtain

01224197 2345


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 itwhen = 0. This root has the form:

...443

32

21 cccc

Substituting the expression for into (36) and equating to zero the coefficients of thevarious powers of , we find that:

,12

71 c 02 c , 4

2

312

723c , …

Hence

21

,1

...12

723

12

7

12

1

3

4

2

rr

r (37)

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 anytwo independent functions of them, define the twin points. Since y is distinct from zero,we may divide it into the second equation, which yields:

.01

13

23

1

rr

Multiplying this equation by x - x2 and x - x1, and subtracting the products separatelyfrom the first equation of (31) gives:

0

0

01

32

211

31

122

z

r

xxx

r

xxx

But, since 12x , 1x , and 112 xx ; we conclude:


0

01

1

01

1

32

31

z

r

r

The only real solutions are 121 rr so these points form equilateral triangles with the

two finite bodies, whatever their masses may be. Since z = 0, these points—calledTrojans—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 pointshave to exist. Such a potential candidate would look like the C6 contours on the y-axis ofFigure 7, which coincide with the L4 and L5 points.

In any case, we have taken the long way around to demonstrate that a geocentricderivation follows from the dynamic explanation of the geocentric system. Indeed, theso-called “fictitious forces” are brought into play because a geocentric system isconsidered “fictional.” Nevertheless, in a geocentric coordinate system they come intoplay because they are real, gravitational forces. Dynamic derivations lament the need forinvoking the fictitious forces in order to represent their Lagrange points derivations.


THE LAGRANGE POINTS: A MODERN APPROACH3

The derivation we have applied in the previous section is a rigorous one derived from thedefinition of force and geometry. In this section, we use another approach which is basedon Newtonian gravitational force. We again start with the same initial conditions as wedid 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. Theyare called Lagrange Points in honor of the French-Italian mathematician JosephLagrange, who discovered them while studying the restricted three-body problem. Theterm “restricted” refers to the condition that two of the masses are very much heavierthan the third. Today we know that the full three-body problem is chaotic, and so cannotbe solved in closed form. Therefore, Lagrange had good reason to make someapproximations. Moreover, there are many examples in our solar system that can beaccurately described by the restricted three-body problem.

The procedure for finding the Lagrange points is straightforward: We seek solutions tothe equations of motion which maintain a constant separation between the three bodies.If M1 and M2 are two masses, and 1r

and 2r

are their respective position, then the total

force exerted on a third mass m, at a position r

, will be

232

213

1

1 )( rrrr

mGMrr

rr

mGMF

(38)

3 Source: Neil J. Cornish, & Jeremy Goodman,http://www.physics.montana.edu/faculty/comish/lagrange.pdf


The catch is that both 1r

and 2r

are functions of time since M1 and M2 are orbiting each

other. Undaunted, one may proceed and insert the orbital solution for 1r

(t) and 2r

(t)

(obtained by solving the two-body problem for M1 and M2) and look for solutions to theequation of motion

2

2

dt

trdmtF

, (39)

that keep the relative positions of the three bodies fixed. It is these stationary4 solutionsthat are known as Lagrange points.

The easiest way to find the stationary solutions is to adopt a co-rotating frame ofreference in which the two large masses hold fixed positions. The new frame ofreference has its origin at the center of mass, and an angular frequency given byKepler’s law:

2132 MMGR (40)

Here R is the distance between the two masses [earth and sun —GB]. The only drawbackof using a non-inertial frame of reference is that we have to append various pseudo-forces 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

rmdt

rdmFF

2 . (41)

The first “correction” is the Coriolis force and the second is the centrifugal force. Theeffective force can be derived from the generalized potential

rrrvUU

2

1, (42)

as the general gradient

. Udt

dUF vr

(43)

The velocity dependent terms in the effective potential do not influence the positions ofthe 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 ofGeocentricity.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 fictitioussince they are necessary to obtain the correct equations.


motion about the equilibrium points. A plot of U with v

=0, M1 = 10, M2=1 and R = 10is shown in Figure 11. The extrema of the generalized potential are labeled L1 throughL5.

Figure 11: A contour plot of the generalized potential

Choosing a set of Cartesian coordinates originating from the center of the masses with thez-axis aligned with the angular velocity, we have

iRr

iRr

jtyitxr

k

ˆ

ˆ

ˆ)(ˆ

2

1

(44)

where

,21

2

MM

M

21

1

MM

M

. (45)


To find the static equilibrium points we set the velocity dtrdv /

to zero and seek

solutions to the equation 0

F , where

j

yRx

yR

yRx

yRy

iyRx

RRx

yRx

RRxxF

ˆ

ˆ

2/322

3

2/322

32

2/322

3

2/322

32

(46)

Here the mass m has been set equal to unity without loss of generality. The brute-forceapproach for finding the equilibrium points would be to set the magnitude of each forcecomponent to zero, and solve the resulting set of coupled, fourteenth order equations for xand y. A more promising approach is to think about the problem physically, and use thesymmetries of the system to guide us to the answer [which is what we did in the firstsection —GB].

Since the system is reflection-symmetric about the x-axis, the y component of the forcemust vanish along this line. Setting y = 0 and writing x=R(u+) (so that u measures thedistance from M2 in units of R), the condition for the force to vanish along the x-axisreduces to finding solutions to the three fifth-order equations.

,212331 43210003212 uuussussuuusu (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 thereis one real root to the quintic equation, giving us the positions of the first three Lagrangepoints. We are unable to find closed-form solutions to equation (47) for general values of


Identifying the remaining two Lagrange points requires a little more thought. We need tobalance the centrifugal force, which acts in a direction radially outward from the center ofmass, with the gravitational force exerted by the two masses. Clearly, force balance inthe 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 jyix ˆˆ and jxiy ˆˆ . The perpendicular

projection yields

.

113/2223/222

32

yRxyRxRyF

(49)

Setting 0

F 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||

2/3223

222 11

yRxRR

yx

(50)

Demanding that the parallel component of the force vanish leads to the condition that theequilibrium points are at a distance R from each mass. In other words, L4 is situated atthe vertex of an equilateral triangle, with the two masses forming the other vertices. L5 isobtained by mirror reflection of L4 about the x-axis. Explicitly, the fourth and fifthLagrange points have coordinates

RMM

MMRL

RMM

MMRL

2

3,

2:

,2

3,

2:

21

215

21

214

. (51)

Interestingly, the last two Lagrange points are stable because any deflecting of a particleinvokes 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 tosimplify 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 anygeocentrically-based coordinate system. The rule that it is “six-to-one, half-dozen of theother” when it comes to mathematical or physical proofs or disproof of heliocentric andgeocentric physics remains steadfastly true.

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