The Universal Laws of Gravity and Repulsion Particles
| Gravity, a polarized incoming particle applies a force to its host mass as it vibrates in a wave like forward fashion while rotating about 360 degrees its directional axis thus creating multiple wave tubes through the host matter and it is the interactions of this movement, which creates a force perceived as gravity towards the mass to which these particles are returning to. The surprise is that this is only half of the equation, the counter-force exiting the mass, the repulsion particle completes it. |
Properties & Origin of the Gravity Sub-Atomic Particle
Rebuttal to: Southwestern Oklahoma State University
The laws of gravity that mankind has accepted from Newton and Einstein provided a great foundation and work for simplistic applications, but would not work in other parts of the Universe. Gravity is only has half the equation, I will complete the equation with a force unknown except to a few, the repulsion force.
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In this simplified model, we are isolating the motion and behavior of a few particles in the same plane. It does represent the basic motion, but does not account for complex interactions for all motion. Here have a gravity particle entering a zone where pressure between the nuclei in the area was in a state of temporary equilibrium. Its introduction into the system crowds the right side of the nucleus and to relieve this new force the nucleus moves into an area of less pressure towards the other nucleus. |
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In this diagram the pressure coming from the approaching nucleus forces the nucleus on the left to back up until the counter force of nuclei behind it equalizes, thus the nucleus now holds position and the same particle crowding the nucleus on the right was escaping has built up in front of its path. Having left a pressure void in its wake, the area behind it is now the easiest path. The nucleus reverses direction to repeat the cycle all over again when its path is blocked with particle crowding |
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The Oscillating motion is a result of the nucleus as it moves from one crowded particle area towards a void overshoots into another crowded particle reverses and starts the cycle all over. The basic particles, which permeate matter and occupy the void of space provide consistency to this motion. Primarily this particle is responsible for the oscillating motion of all particles related to gravity in the universe. This oscillation motion of the nucleus and associated parts within in matter is known as its resonant vibrational rate, basis for density and parallel universes (details). The basic particle, known only in theory is Dark Matter. |
Gravity is an attractive force applied to an independent mass as gravity particles flow towards the center of the dominant mass (the destination due to a charge).
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Gravity Particles
1. As the gravity particles pass thru the nucleus of a random mass, a force (F) is applied to the object in the direction of the gravity particle stream. 2. Gravity particles also move the nuclei of a mass out of its path by crowding an area that was in equilibrium, thus the nuclei shifts away. 3. Gravity is nothing more than the cumulative total of all force applied the individual nuclei of a mass as T for time of the interaction approaches 0. |
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Behavior of a Gravity Particle During a Pass Thru of an Atom
Step 1: A random approaching gravity particle. Step 2: Certain subatomic particles within the gravity particle adapt to the electron field's charge of the atom to be penetrated and is now attracted to the nucleus temporarily supplanting the attraction of the primary mass within the region. Step 3: The gravity particle enters the nucleus and the polarity of those certain subatomic particles, which is part of the whole moves as it pushes ahead towards its core. The result is a force is released in the direction of the particle flow. Step 4: As gravity particles follow the same path, pressure builds from behind and the neutral gravity particle adapts a positive charge and is ejected from the nucleus while releasing another force again in the same direction. Step 5: As the ejected gravity particle enters the electron field in which attraction plays a part, the positive charge acquired in the nucleus erodes quickly. Step 6: The gravity particle induced passed thru charge returns to neutral and returns to its path towards the dominant mass. |
For more details on the atomic structure and its electron particle field, consult astro33.htm.
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Repulsion Particles The repulsion particle stream is different from gravity particles, because they are pulse released from the core of a mass in a laser-like sinusoidal stacked stream of particles, resulting in a pushing action 360 degrees tube-like zone perpendicular about the directional axis of the atoms in the mass away from the stream. Thus no force is invoked to the mass as the nuclei are parted by a concentrated pulsed stream of repulsion particles that rip through following the wake created by preceding repulsion particle. |
The repulsion particle is divided into 2 primary types; in the first type, the repulsion particle ejected randomly about the globe and regulates the distance between the masses. Once the incoming gravity particle has assimilated a like charge that is great enough to overcome the containment pressures from the stacked particles on top, it is ejected or repulsed from the center of the mass in a high speed, concentrated stream. The force propelling the particle is in relation to the particle charge vs. the total charge at the center of the mass. Hence, the name and result is a repulsion particle. The second type of repulsion particles are ejected from the core in a different manner carry a slightly different charge than the first type. As the core of the host mass rotates the repulsion particles crowd the equatorial area and are pushed to the edge of the core. At the pressure points the repulsion particles bleed off in a more subtle action small, more frequent bursts and carry a smaller charge, but still rip through all mass within the area. For the type 2 repulsion particle, expulsion within the equatorial plane allows for rotational torque to be applied to mass in the direction of the rotation of the host's mass core. The potential of the charge applied to certain sub atomic particles remain over time and after drifting over vast regions of space a return flow sets up. What we have is a particle flow which creates rotation around the dominant mass and the return particle flow keeping and mass caught in a thin ecliptic plane. This creates a particle flow just above the ecliptic plane with eddies and turbulence developing between the 2 flows. As the particles return to the Sun they encounter the exiting field, which takes on the form of a hurricane with the core representing the eye. So the particle enter back into the Sun through the path of least resistance the poles of the Sun located above and below the core. For more details read Orbital Plane and Motion. The alternate repulsion particle controls the orbital velocities of any mass around the host mass as long as the core has rotational movement.
This particle exhibits some of the following new characteristics not yet seen by scientists. Any mass encountered at close range (this range is dependant on the size of the host mass) while exiting the primary mass is not effected, since the repulsion particles tend to rip through a mass due to speed of the particle ejection and the concentration of the stream without applying a force to the pass-thru mass. This is accomplished as a stream of ejected repulsion particles follow in each other wakes allowing minimal push to affect the mass. Similar to a engaging a slip stream, any mass that follows a leader has its friction or particle interaction reduced while in close proximity to the rear wake of the pressure void, which precedes it. The key here is that the nuclei, which comprises the mass is pushed to the side, thus the repulsion particle rarely passes thru the nucleus of matter near the host mass. It is when the repulsion particles saturate the immediate area surrounding a mass, that their effect takes place. Pressure precedes the repulsion particles thus deflecting and crowding out the incoming gravity particles and thus not allowing an incoming push to be applied to the mass. Simultaneously the repulsion particles now dispersing apply an outward force to the object of sufficient mass as they pass through and into the core where if enough repulsion particles have gathered, their like charges interact and repulse. The influx of repulsion particles crowd the area in and around the subject mass thus crowding out incoming gravity particles, further reducing the attraction between the host and subject mass. The point where this force neutralizes gravity is the repulsion point or the mean distance of an orbit or separation in mankind’s eyes. As the repulsion particles return to space, an equalization process occurs. Mankind sees space as having zero gravity, but it is really repulsion and gravity particles applying an equal force on a mass with a net force of zero. The repulsion particle settles in space and its charge bleeds off slowly, dissipating back to its original static charge and once again becoming a gravity particle as the cycle repeats.
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Equations
Mankind assumes the force of the Sun's gravity or centripetal force exerted on a planet in orbit is balanced by centrifugal force due to orbital motion. |
Balancing the equations of gravity or centripetal force vs. centrifugal force, we have:

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The Constants
Mass 1:
The Earth Mass 2:
The Sun Gravitational
Constant: 6.67 * 10-11 Radius
(Distance from the Sun to the Earth): 1.4993 * 1011 meters Velocity = Distance * (2 * 3.1416 * Radius) = 29,851 meters / second Time (seconds)
Time
conversion to seconds: 365.25 * 24 * 60 * 60
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Using the same equations on the Earth and the Moon, we can see how the mass for the Earth was approximated at 5.97 * 1024 kilograms. Our answer below is an approximation due to the rounding of distances involving the radius.
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Mass 1: The Moon Mass 2: The Earth Gravitational Constant: 6.67 * 10-11 Radius (Distance from the Earth to the Moon): 3.85 * 108 meters Velocity = Distance (2* 3.1416 * Radius) = 1024.75 m/sec. Time (seconds) version to seconds: 27 days 7 hours 43 minutes = 2360580
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Using the same base formula to find the mass of the Sun, we can apply it to find the mass of the Earth.
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Base Equation
Density
= Mass Volume
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Reference Radii The Sun: 6.9552 * 108 meters The Earth: 6.378 * 106 meters
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Density Reference Density of the mass of the Earth: 5.49
grams/cc. Density of the mass of the Sun: 1.421 grams/cc.
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The mass of the Earth when expressed in a comparison with the Sun's equated
density per cubic centimeter; seems to be too low when compression of matter has
to be taken into account. Examining the density of iron, a common element
located in the middle of the periodic chart, is a major component of core of the
Earth. Iron has an established density of 7.874 grams/ cc. at 20 degrees C.
under normal atmospheric pressure at sea level, but Earth’s average density is
only 69.76% of iron. With most of the light elements located in the crust, which
is about 20 miles thick, how did Newton account for the mass under extreme
compression below the crust and in the core of the Earth? One only has to
contemplate the depths of the ocean to get an idea of molecular compression,
which would crush most reinforced metallic structures on Earth. This leads to a
conclusion that the mass of the Earth is under estimated, which then negates the
current approximations of the Sun’s mass. Finally, to show that centrifugal
force does not balance gravity, we will conduct a density comparison of equal
volumes of the Sun’s and the Earth’s masses. Mankind assumes the composition
of the Sun is approximately 71% hydrogen, 27% helium and 1.9% other. This is due to spectrum
analysis of the light emanating from the Sun by astronomers. Most scientists
have overlooked the effects of gravity, light absorption of the stellar mass,
and the surface light overwhelming internal sources as to be undetected. The
gravitational force of the solar mass impedes and collects most of the light
spectrum of the heavy elements emitted from the core. Only trace amounts of the
heavy element’s spectrum that mixes with the light hydrogen and helium elements
burning in the outer shell are detected giving mankind a false representation a
of mass which is non homogenous especially in its core.
For those who want a precise break down according to mankind’s theories go to http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/suncomp.html
Step
1: I will now set up a mathematical proof to shed doubt on the current
estimates of the mass of the Sun and Earth. Setting up an Earth-Sun gravity-mass
comparison, I will use standard measurements, but we use the reference that
favor current theories to validate my point. Hydrogen and helium at 293
degrees K. under Earth’s gravitational force has densities of .00008988 grams
/ cc. and .0001787 grams / cc. respectively.
Step 2: Establishing the elemental composition and weight of a cubic centimeter of the mass of the Sun on Earth.
| Element | Density per cm3 @293K in grams | Percentage of element in the Sun | Contribution in mass per cm3 in grams |
| H | 0.00008988 | 71% | .000063815 |
| He | 0.0001787 | 27.1% | .000048428 |
| O | 0.001429 | 0.97% | .000013861 |
| C | 2.62 | 0.40% | .01048 |
| N | 0.00125 | 0.096% | .0000012 |
| Si | 2.33 | 0.099% | .0023067 |
| Mg | 1.738 | 0.076% | .00132088 |
| Ne | 0.0008999 | 0.058% | .000000522 |
| Fe | 7.86 | 0.14% | .011 |
| S | 2.07 | 0.04% | .000828 |
| Totals for 1 gram | --- | Approximately 100% | .026071326 |
Columbia University:
The interior of the Sun is entirely gaseous and the great majority of atoms are stripped of their electrons. The solar gas behaves under these physical conditions nearly like a perfect gas, governed by the "equation of state"
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“Two forces keep the Sun in hydrostatic equilibrium in its current stage of evolution: the gravitational force directed inward and the total pressure force directed outward. The equation of hydrostatic equilibrium is "
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where
P is the pressure, r the radial distance from the center, Mr
the mass within a sphere of radius r,
the
matter density, and G the gravitational constant.
This equation is consistent with radius changes, but requires the kinetic energy involved in expansion or contraction of the solar body to be small compared to the gravitational potential of the Sun
Source: http://www.columbia.edu/~ah297/unesa/sun/sun-chapter1.html
The equations from Columbia University will be the basis to showing a short fall in gravitational pressure to balance the hydrostatic equation.
Density of elements per cm3 in grams. You can double check at http://www.vcs.ethz.ch/chemglobe/ptoe/_/1.html
Other
References: http://science.nasa.gov/ssl/pad/solar/interior.htm
We also found that
gravitational acceleration on Earth was 9.8m/sec2 vs. the Sun @
276.16m/sec2 a ratio of
28.1.
The same mass on Earth would exhibit a force
28.1 times greater on the surface of the Sun.
Mankind’s equation for determining the gravitational acceleration on any mass is
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G * M R2
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The Constants G = 6.67 *
10-11 M = 2.0029
* 1030 R = 6.9552
* 108 |
Again using mankind’s
equations lets see what the gravitational acceleration would be if we maintain
the mass of the Sun but reduced its radius?
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This table shows how gravitational acceleration would change on the sun if we reduced its radius, but maintained the original mass. The base @ 100% matches current estimates on how many times heavier would a mass on Earth be while on the surface of the Sun. |
Mankind used orbital
velocity to estimate the mass if the Sun when you check against density and
pressures the same equation falls far short or fails due to the Factor
10 Problem. The mass of the Sun is far
greater than mankind assumes and it is ionized iron in the core that is
responsible for the Sun’s magnetic field and larger mass.
Formula Force= (Gravitational Constant)*(Mass)/Radius2
First solving for the
gravitational force on Earth, substitute 6.67 * 10-11 for the
gravitational constant, 5.97 * 1024 kg. for the mass of the Earth, and 6.378 *
106 m. for the radius of the Earth.
This equation yields a gravitational acceleration force
equal to 9.788837/m2
Now we know that mankind has stated that the Sun maintains
its present physical form as force of gravity counters the expansion of the Sun’s gases. For
those who doubt go to: http://science.howstuffworks.com/sun.htm
Equation for the gravitational acceleration on the Sun
Force= (6.67*10-11) * (2.0029*1030)/ (6.9552 *
108 meters)2
Result 276.16m/sec2
Lets introduce Pressure = M/V
How would the gravitational pressure have increase if a
cm3
was maintained and the original components of the Sun measured
at .026071326 g/c3 was compressed to 1.4 g/cm3
1.4/.026071326 g/cm3 = 53.69 times that on Earth yet we
know the gravitational pull on the surface of the Sun is 28.1 times that of
Earth.
If you solve for gravitational force needed it is
525.64/m2
So what reduction in the Sun’s radius of the with the
same mass need to be to produced an acceleration of 525.64/m2?
Plugging in the numbers
R2 = G * M/F from the original formula F = G
* M/R2
Square Root of (6.67*10-11) * (2.0029*1030)/525.64 yields
5.0377 * 108 new Radius or 72.43% of the Present Sun
Now you are going to think that the combination of elements are in a plasma state on the Sun. This is true, but a made appoint to 293K and Earth’s gravity, which everybody can relate to and what you find is that the mass of the Sun does have enough gravitational force to contain the components of the Sun at Earth conditions
Now lets Introduce Another new Formula: The Ideal Gas Law
This is the basis for mankind's hydrostatic equation describing the balance of stellar interiors and gravity
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PV=nRT |
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The Constants & Variables
P= Pressure V= Volume n=number of moles (Mass) R= Universal Gas Constant (For the brains 8.3145 J/mol K) T= Temperature |
ANALYZING THE EQUATION
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P=nRT/V |
n is a constant
R is a constant
V is now a constant with a radius 72.43%
We find that temperature in K is directly proportional to pressure, this now complicates matters
So whether I use hydrogen,
helium or any combination if the volume and mass are constant a rise of 300K
doubles the pressure and so on.
Mankind best estimates states that at .25
radii of the Sun, the temperature of the plasma is 7,500,000K with a density
of 20 g/cm3.
The increase in pressure for
any gas to a plasma state according to the Ideal Gas law from 300K to 7,500,000 is a
factor of 25,000.
So is we raise the temperature go the original components to just the surface temperature of the Sun 6000K, what does that do the pressure that gravity counters?
6000 K/293K increases by a factor of 20.47
525.64/m2(the original force needed to compress components
to 1.4) with heat added this rises 20.47 fold
525.64 * 20.47=10759.85
Now find the radius the total mass of the Sun must be reduced to contain the pressures?
Square Root of (6.67 * 10-11) * (2.0029*1030)/10759.85 yields
1.11426 * 108 new Radius or 16.02% of the Present Sun or well with the core of
the present Sun.
What you have to realize is that mass does create enough
gravity to contain the fusion process, yet it exists. This leads to the
conclusion that the mass is grossly underestimated.
As questions whirl through your mind many will say ok so there is a shortfall of gravitational pressure at the reduced Radii, what about all the mass above the point where you are measuring?
When we calculated the gravitational force at the new radii points we used the total mass of the Sun within the new radii eliminating thought of thin shell of mass trying to contain the fusion process of the Sun.
A Numerical Simulation of a collapsing nebula cloud does examine the flaws in theoretical star formation giving rise to the Factor 10 deficit associated with matter compression and its associated gravitational acceleration. All of which gives insight to the hydrostatic equilibrium which occurs in the Sun.
If it underestimated than the Earth can maintain its orbit
at the present velocity or it will fall into the Sun.
So we have G * Greater M * m/R2 = (m * V2/R) +X
The X is equal to the missing repulsion force
Since we know that small masses fall into the Earth we can
deduce that the repulsion force gains at an exponential rate and a large enough
mass must be present to counter the initial force of gravity.
Thus the basic equation would look like this:
*(Gravitational Constant - Repulsion Constant * e(k * Radius)) Mm/R2= (mV2/R)
*The effects of gravitation and repulsion of the other planets and the Dark One are omitted for simplicity sake although are minor.
Gravitational Equation valid only in the Resident Universe
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Equation Key
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To test this theory we would only have to look towards the Moon. The Moon’s absence of an atmosphere will eliminate all of mankind’s excuses for backing Newton, although his theories do work and do explain elementary motion around planets. When satellites are placed in orbit around Earth, the reason given for the decaying orbit is friction due to contact with Earth’s thin upper atmosphere. This contact only plays a small part in the decay process. An object moving in a circular motion loses forward momentum as direction of force is towards the center.. The centrifugal force counteracting gravity or centripetal force decreases and results in the satellite falling out of orbit. If a satellite were placed into an orbit around the Moon at the precise orbital velocity and distance to match its gravitational force, it would eventually crash to the surface. Since the object would have insufficient repulsion generated because of its small mass to overcome incoming speed and the gravitational force of the Moon to prevent a collision. This experiment would leave mankind mystified, since there isn’t any atmosphere to account for the decay in orbital momentum.
How does gravity work with respect to bending light?
The effects gravitational particles on light particles are greatest when the angle of interaction and intersection approaches 90 degrees. red wave lengths bend more due longer interaction with gravity particles, the increase of red light gathered because of the bending of light particles give the illusion of a larger disk when viewed at dawn or dusk and this is the reason for increases the size of the disk. Does dust in the sky redden the sky no, it allows more red light to reflected towards earth, but does not change the frequency of in coming light waves. If you were to look into the sky when dust fills sky after a volcanic eruption or fire at noon it would just dim the Sun not deepen its color to red or increase its visual size, Mankind's explanations fall short here.
The 5 Primary Forces
Gravity exhibits all 5 primary forces in the universe first, magnetic in that it carries varied charges within its sub atomic particles, second, the nuclear force is responsible for the binding of all sub atomic particles into what seems one particle. Third and fourth, gravity and repulsion as it attracts and repulses mass. Finally the fifth force, vibrational as the gravity particle excites the atoms in matter to a particular frequency dependant upon the amount and energy of dark matter in the resident universe.
| With Knowledge Comes Responsibility | I want to thank God, for with out him in my life this paper and its upgrades would have not been possible. | The Meek Will Inherit |
I want to thank God for bestowing knowledge
All Rights Reserved: Copyright 2000, 2004
Mankind's Explanation for Gravity