Home Pharos Fiction Site Map Updates Search


                                                                                                                        Back Next

Halexandria Foundation
Sacred Mathematics
Connective Physics
Chronicles of Earth
Justice, Order, and Law
Extraterrestrial Life
Creating Reality
Tree of Life

Nibiru ala Kepler

New -- December 15, 2003


© 1994, 2003 Dan Sewell Ward

Interlude I -- Nibiru ala Kepler

Keplerís First Law states that all planets revolve about the Sun in elliptical orbits with the Sun located at one focal point (in the figure, the Sun is located at F1).

The equation for an ellipse is given by:

x2/a2  + y2/b2   =  1

where 2a is the length of the major axis, 2b the length of the minor axis, and 2c the distance between the focal points, F1 and F2.  In this case,

a2  =  b2  +  c2

Solving the ellipse formula for y yields:

y  =   (b/a)  [a2  -  x2  ]1/2

The area under a curve from point x1 to point x2 (the limits on the definite integral) in this case the ellipse or a portion of the ellipse, is given by:

A  =   Ú  y dx  =  (b/a)  Ú  [a2  -  x2  ]1/2 dx

A  =  (b/2a) { x [a2  -  x2 ]1/2 + a2 [arc sin (x/a)] }   (from x1 to x2)

The area, AT, of the ellipse from perihelion (closest point to F1) to aphelion (closest point to F2) is given by setting x1 = - a and x2 = + a.  Thus,

AT  =  (b/2a) { [ a (a2 - a2 )1/2 + a2 arc sin (a/a) ] - [ -a (a2 - a2 )1/2 + a2 arc sin (- a/a) ] }

AT  =  (b/2a) { [ 0 + a2 w ] - [ 0 + a2 (-w) ] }  =  a b w

where w = arc sin z  is the solution for sin w = z.  If z = 1, sin w = 1 and therefore w = p / 2.  For z = - 1, sin w = -1 and w = - p / 2.  Therefore,

AT  =  a b p / 2

Consider now, the shaded area where the distance from F1 to an arbitrary point, f, on the ellipse is sf is also at a distance j along the x-axis and k along the y-axis from the origin.  The area under the curve from perihelion (x = -a) to the point f is given by:

Af  =   (b/2a) { [ - j (a2 - j2 )1/2 + a2 arc sin (- j/a) ] + [ a2 p/2 ] }

The shaded area, Af, is given by:

As  =  Af  -  k ( c - j ) / 2

where k2  =  s2  -  ( c - j )2

It follows from Keplerís Second Law, that the time, t, for a planet to move from perihelion to point f, will be related to the time, T, for a planet to move from perihelion to aphelion, by the relation:  t / T  =  As / AT.   In other words, the ratio of the area swept out in time t, As, to the area swept out in time T, AT, equals the ratio of time t to that of time T.

IN THE CASE OF NIBIRU, let us consider the situation where the orbital period of Nibiru is assumed to be 3600 Earth years.  The question is then:  For how long or for what portion of its orbit is Nibiru within the confines of the known solar system?  If we know this answer, we can then have some idea as to when it would be visible, and also for what period of time ďtrafficĒ between Earth and Nibiru might be realistic.

According to Keplerís Third Law, the mean distance from the Sun, D, is related to the time of revolution about the Sun, T, by the following:

D3 / T2   =   K

where K is a constant.  If D is in Astronomical Units (1 A.U. = 93 million miles; the distance from the Sun to the Earth), and T is in years (Earth), then K = 1.

From Keplerís Third Law, we can calculate the mean distance from the Sun to Nibiru.  The result is:

D  =  234.89  (about 235 A.U.)

This can be compared to the mean distances of other solar system planets.  For example, Saturn is 9.54 A.U. from the Sun, Uranus is 19.17 A.U., Neptune is 30.03 A.U., and Pluto is 39.42 A. U.  Note, however, that these distances are the mean distances, i.e. an average of the perihelion and the aphelion.  At the same time, the aphelion is simply the length of the major axis of the ellipse less the perihelion distance ( 2 a - p ), i.e.

D  =  ( p + h ) / 2  =  [ p + (2 a - p)] / 2 =  ( 2 a ) / 2  =  a

In the case of Pluto, its perihelion is just slightly within the orbit of Neptune (about 29.74 A.U.), and its aphelion is 49.10 A.U.  In the case of Nibiru, the orbit is much more eccentric, with the distance to aphelion the order of 467 A.U.

In order to calculate the length of the minor axis of Nibiruís elliptical orbit, we must make a second assumption.  For this case, let us assume that Nibiruís perihelion is within the asteroid belt, at approximately the mean distance of Ceres (2.75 A.U.).  Then,

b  =  [a2 -  c2 ]1/2   =  [a2  -  (a - p)2  ]1/2  =  [2 a p -  p2 ]1/2   =  [5.5 a - 7.56]1/2 A.U.

b   =  35.8 A. U.

We can now calculate the value of AT ( = a b p / 2 ).  We obtain:

A  =  13,224 A.U.2

Let us now assume several values of the point f, corresponding to distances within the known solar system.  For simplicity, we will consider the five cases where Nibiru is located at a distance within the outer reaches of the asteroid belt, a second distance where the position of Nibiru is on a line perpendicular to the major axis, a third distance at the equivalent to the mean distance of Saturn, a fourth distance at the equivalent of Uranus, and a fifth situation where f is located at the edge of our known solar system, just beyond the furthermost reach of Pluto. 

For simplicity, we will assume:


a)  s = 3.20 A.U. -- Asteroid Belt


b) s = 5.47 A.U. -- Nibiru perpendicular to major axis


c) 9.53 A.U., -- mean distance to Saturn


d)  s = 19.17 A.U., -- mean distance to Uranus


e)  s = 50 A.U. -- just beyond the furthermost reach of Pluto

Using the equation for the ellipse, with x = j and y = k, along with the relationships s2  =  k2  +  (c - j)2 and c = a - p; we obtain for values of j:

a) 234.29 A.U., b) 232.14, c) 227.98 A.U., d) 218.26 A.U., and e) 187.08 A.U.

This yields for the values of Af

a)  4.41 A.U.2, b) 9.68 A.U.2, c) 39.18 A.U.2, d) 147.68 A.U.2, e) 706.23  A.U.2

We can then calculate values for As, in which case we obtain:

a) 6.96 A.U.2, b) 9.68 A.U.2, c) 21.32 A.U.2, d) 55.92 A.U.2, e) 218.00  A.U.2

We can now solve for t ( = T As / AT = 1800 As / AT years), the time that Nibiru would be within 3.20, 5.47, 9.53, 19.17, and 50.00 A.U. as it approached or when it departed from its perihelion with the Sun:


a) 0.95 yrs, (within the asteroid belt)


b) 1.32 yrs,


c) 2.90 yrs, (within the orbit of Saturn)


d) 7.61 yrs, (within the orbit of Uranus)


e) 29.67 yrs (within the outermost boundary of Pluto)

The total time that Nibiru would be within these distances is just 2 t, or:


a) 1.90 yrs, (within the asteroid belt)


b) 2.64 yrs,


c) 5.80 yrs, (within the orbit of Saturn)


d) 15.22 yrs, (within the orbit of Uranus)


e) 59.35 yrs (within the outermost boundary of Pluto)

In essence, Nibiru would be within the 50 A.U. limit for approximately 60 years, and within the orbits of Uranus and Saturn for only roughly 15 and 12 years.  If Nibiru is approximately 4 times the size of the Earth, it should be visible to a powerful telescope within a reasonably wide time span.  At the same time, however, its orbital speed would increase dramatically, averaging some 15,000 mph between the outer limit of Pluto and Uranus, 22,000 mph between Uranus and Saturn, 27,000 mph between Saturn and Jupiter, 55,000 mph in the vicinity of Jupiter, and 65,000 mph between Jupiter and the asteroid belt.

Episode II -- The Beginning (Revisited)

Forward to:

Episode III -- The Evolution and/or Creation of Life


                                                                                      The Library of ialexandriah       

2003© Copyright Dan Sewell Ward, All Rights Reserved                     [Feedback]    

                                                                                                            Back Next