archived as http://www.stealthskater.com/Documents/Astro_09.doc [pdf]

more related articles at http://www.stealthskater.com/Science.htm#Grant

note: because important websites are frequently "here today but gone tomorrow" the following was archived from http://www.grantchronicles.com/astro09.htm on December 27, 2005. This is NOT an attempt to divert readers from the aforementioned website. Indeed, the reader should only read this back-up copy if it cannot be found at the original author's site.

The Grant Chronicles:the Universal Laws of Gravity and Repulsion

Gravity -- a polarized particle that applies a force to a mass in the same direction during pass-through -- is only half the equation. The counter-force repulsion completes it.

Short Summary -- Gravity and Repulsion

● Gravity is still an unknown phenomena in Science.● Gravity is a polarized particle.● Gravity affects organic and inorganic matter differently.● Gravity flows evenly towards the center of a mass.● For every particle, there is an anti-particle.● The Repulsion Particle (new) is gravity's counter particle.● The Repulsion Particle flows out of the center of a mass in concentrated streams.● Concentrated repulsion streams of particles rip through mass without affecting it.● Grav. Form.= {[Grav. Cont/radius2] - [Repulsion Cont* e(k*radius)]}* [M1*M2*T]● Abbreviations: Grav. = Gravitational; Form. = Formula; Cont = Constant

Questions to Ask?

● Why is 90% of the mass of the Universe unaccounted for or has mankind under estimated?● If it is Dark matter, why doesn't it block visibility like in a nebula?● Why hasn't a space program put an orbiting laboratory around the Moon?● Maybe if they did, they could not explain it falling out of orbit around a mass with no atmosphere.● How is the Sun holding together if the gravitational force created to counter expansion of a hot

Hydrogen-Helium mass is insufficient to the calculated cubic density of the Sun?

Original VersionMore Articles 1More Articles 2Rebuttal 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 they would not work in other parts of the Universe.

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Gravity is only has half the equation. I will complete the equation with a force unknown except to a few -- the repulsion force.

Gravity

Gravity -- long rumored to be a wave or some cross wave-particle -- is truly a polarized particle. But it goes further than that. The gravity particle is composed of many sets of subatomic particles -- each assigned to a different task or assisting another but all operating as one. It has a very minute set charge when in equilibrium (or its original state) that moves from what seems to be the void, which contains a low density of Dark Matter, eventually to the center of the dominant mass with in the area where most gravity particles tend to gather. The flow rate of gravity particles in close proximity and on the surface of a cosmic object is determined by the size and average radii of the mass, the effects of any nearby mass, and the density [SS: doc pdf URL-doc URL-pdf ] in which the mass resides in.

Density is a new factor when dealing with particle movement [SS: doc pdf URL-doc URL-pdf ]at this level in the Universe. It determines the vibrational rate of all particles and their relative resonant velocities (or natural frequency of mass). It has been discussed [SS: doc pdf URL-doc URL-pdf ]that particles like gravity and light moves along the X-axis and oscillate back-and-forth along the Y-axis. The wave rotates as a function of the Z-axis and opposing particle pressures. The amplitude of the wave of the particle is determined by particle crowding a point where it is easier the reverse motion then plow ahead. This is where dark matter comes into play.

In our Universe, dark matter thinly spread throughout the void of space is in the greatest concentrations and lowest energy level of all the parallel universes. In the next parallel universe [SS: doc pdf URL-doc URL-pdf ], our laws of gravity and light would not work due faster velocities for the gravity and light particles. So a missing piece of the puzzle is that the concentration of dark matter not only determines the speed-of-light in our part of the Universe, but also the behavior of returning gravity particles.

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 it does not account for complex interactions for all motion. Here we 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. 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

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

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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 [SS: doc pdf URL-doc URL-

pdf ]). 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).

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.

The intensity of the gravitational force or flow rate is dependent on the amount of particles passing through a defined cubic area of the mass during a time period (represented by time as T approaches 0) and the subject mass elemental composition (inorganic or organic), because the bonds between the electron subatomic particle cloud and the nucleus are different and its density. T (a unit of time) is a variable in different parts of the Universe. This is why it is a component of the gravitational equation. Organic material with its complex molecular bonds and the density of a electron subparticle field structure tends to impede the gravitational particle flow rates. Constriction of available pass through space causes backpressures, which reflect back towards the incoming gravity particles. Hence a larger

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force applied to the pass thru mass due to a longer particle contact with the organic material. The weight of an object would be the cumulative total of all gravity particles applied pressure pushes during an instantaneous snapshot of specified time period, which accomplished as T approaches 0 revealing a more accurate the cumulative total.

Details on how a gravity particle interacts with an atom thru a series of steps

Behavior of a Gravity Particle During a Pass Thru of an Atom(for more details on the atomic structure and its electron particle field, consult astro33.htm [SS: doc pdf URL-doc URL-pdf ]

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.

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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.

Once the gravity particles reach the center of the host mass, a new phenomena to Earth science occurs. The repulsion force emanates from a particle that has a polarity opposite to gravity.

Scientists have observed the effects of a gravitational field but have arrived at the wrong conclusions. Gravitational fields do not bend the fabric of space and time. Are the stars near suspected black holes blurry or oblong? Many have observed the bending of light when it passes close to a star or large gravitational source during a total solar eclipse. Gravity deflects incoming photon particles due to pressure from a crowded field of gravity particles. The higher the concentration of gravity particles, the greater the deflection of photon particles.

This is the reason why British scientists while trying to prove Einstein’s theories observed light from distant not seen behind stars. Gravity affects mass as a whole or individual particles. Not dimensions of length, width, and breadth distorting the fabric of space. Is there a "lens effect" every sunrise and sunset? Is this due to light waves passing close to a gravity giant? No. If so, it should affect the view of the Sun through out its daytime arc across the daytime sky.

Repulsion

The repulsion particle and its different forms are new to Earth astrophysics. In essence, it is a transformed gravity particle due to acquiring a new charge while being subjected to forces within the core of the dominant mass where they are attracted to and tend to clump. This clumping from attraction overcrowds the core as gravity particles continue to pour in, thus pushing those particles that arrived earlier closer to the center of the core. The attractive charge emanating from the mass that has brought the particles to a point of equilibrium or neutrality within the center of the mass now faces pressure from the continuing influx of particles that are following from the rear.

Gravity particles -- now subjected compression from new arrivals -- are pushed further into the center of the core where their original static charge is overwhelmed by the opposite polarity charge at the center that attracted them. They begin to lose their neutral charge and align with the polarity of the dominant ambient attractive charge. As the number of particles build with like charges in this area of the core, segregation of like particles in cells begin stacking at specific points as repel each other, evenly dispersed randomly in a 360-degree spherical direction but shifting due particle pressures of relief and crowding. Ejection occurs in 2 primary ways: random intermittent dispersal about the mass, and a more gentle bleed-off about the equator of a rotating mass.

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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 subatomic particles remain over time. 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.

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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 particles 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 engaging a slipstream, 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 through 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 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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I will now apply mathematical gravitational equations to back the theory. Mankind assumes gravity or centripetal force exerted inward on planets in orbit is balanced by centrifugal force (NASA reference site) due to orbital motion. Otherwise the planets would be pulled into the Sun by gravity. All measurements below were acquired from standard reference books.

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.

Equations

Fg (Gravitational Constant) (Mass1) (Mass2)/ Radius2

Fc (Mass1) Velocity2 / Radius

Balancing the equations of gravity or centripetal force vs. centrifugal force, we have:

(Gravitational Constant) (Mass1) (Mass2) / Radius2 = (Mass1) (Velocity2) / Radius

Solving for Mass2:Mass2 = (Radius) (Velocity2) / Gravitational Constant

Since the gravitational constant does not vary, the mass at the center is dependent on the radius of the orbit and the speed of the object (Mass1 which has dropped out of the equation).

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Mass1: the EarthMass2: the SunGravitational Constant: 6.67x10-11

Radius (distance from the Sun to the Earth): 1.4993x1011 metersVelocity = (Distance) (2 x 3.1416 x Radius) / Time(seconds) = 29,811 meters-per-secondTime conversion to seconds: 365.25 x 24 x 60 x 60

The previous calculations yields the velocity for the centrifugal portion of the equation. Now we will solve for Mass2:

2.003x1030 kg = [1.4993x1011m x (29,851m/s)2] / 6.67x10-11

Using the same equations on the Earth and the Moon, we can see how the mass for the Earth was approximated at 5.97x1024 kilograms. Our answer below is an approximation due to the rounding of distances involving the radius.

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Mass1: the MoonMass2: the EarthGravitational Constant: 6.67x10-11

Radius (distance from the Earth to the Moon): 3.85x108 metersVelocity = (Distance) (2 x 3.1416 x Radius) / Time(seconds) = 1024.75 m/sec.Time conversion to seconds: 27 days 7 hours 43 minutes = 2360580

Using the same base formula to find the mass of the Sun, we can apply it to find the mass of the Earth.

6.06 *1024 kg = 3.85x108 m * (1024.75 m/s)2 / 6.67x10-11

An approximation on the mass of the Earth.

Noting that the mass of the Sun and Earth were solved using this gravitational-centrifugal force equation, where the equation falls can now be noted. To solve this reverse proof, 2 additional equations will be introduced.

Density = Mass / VolumeVolume of a Sphere = 4 π Radius3 / 3

We will use the formulas Density = Mass/Volume, and 4/3 π Radius3 for the volume of spheres to arrive at the average mass of a cubic centimeter of the Sun and the Earth. The radii of the respective objects (the Sun and the Earth) are 6.9552x108 meters and 6.378x106 meters.

Reference Radiithe Sun: 6.9552x108 metersthe Earth: 6.378x106 meters

The masses of the objects are given above in the text boxes. Substituting for mass and radius, the resultant density for the Sun and Earth respectively is 1,416 kg/m3 and 5,493 kg/m3.

Density of the Sun = [3* 2.003x1030] / [4 * 3.1416 * 6.9552x108] = 1.421 g/cc

Density of the Earth = 3* 5.97x1024] / [4 * 3.1416 * 6.378x106] = 5.49 g/cc

Converting to grams/cubic-centimeters {kg:g :: 1: 1,000}, {cubic meters: cubic centimeters :: 1:106}. The equations does breakdown to a conversion ratio rate of 1 to 1,000. The density of the Sun and Earth comes to respectively: 1.421 grams/cc and 5.49 grams/cc. The first sign of conflict in mankind’s theories should now begin to surface.

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 oC 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

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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 75% Hydrogen and 25% Helium. 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.

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 much higher than normal in favor of current theories to validate my point. Hydrogen and Helium at 293oK under Earth’s gravitational force has densities of 0.00008988 g/cc and 0.0001787 g/cc respectively.

Step 2: Establishing the elemental composition and weight of a cubic centimeter of the mass of the Sun on Earth.

Element Density (g/cm3) at 293oK

% of element in the Sun

Contribution in mass (g/cm3)

Hydrogen 0.00008988 71% 0.000063815Helium 0.0001787 27.1% 0.000048428Oxygen 0.001429 0.97% 0.000013861Carbon 2.62 0.40% 0.01048Nitrogen 0.00125 0.096% 0.0000012Silicon 2.33 0.099% 0.0023067Manganese 1.738 0.076% 0.00132088Neon 0.0008999 0.058% 0.000000522Iron 7.86 0.14% 0.011Sulfur 2.07 0.04% 0.000828

Totals for 1 gram ----- approx. 100% 0.026071326

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"

P = kρT/μmp

“2 forces keep the Sun in hydrostatic equilibrium in its current stage of evolution: (a) the gravitational force directed inward and (b) 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 the Sun was 276.16m/sec2 vs. the Earth was 9.8m/sec2 vs. -- a ratio of 28-to-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

G M / R2

G = 6.67x10-11

M = 2.0029x1030

R - 6.9552x108

Again using mankind’s equations, let's see what the gravitational acceleration would be if we maintain the mass of the Sun but reduced its radius.

% of the Sun's radius Gravitational acceleration (m/sec2)

(times) Earth gravity

100% 276.16 28.190% 340.94 34.880% 431.5 44.0370% 563.6 57.5160% 767 78.2750% 1,104 112.7140% 1,726 176.130% 3,068 313.1125% 4,418 450.820% 6,904 70410% 27,616 2,817

5% 110,465 11,2712.5% 441,861 45,087

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.

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Mankind used orbital velocity to estimate the mass of the Sun when you check against density and pressures the same equation falls far short or fails. 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.

Now moving on, we calculate the gravitational acceleration of a mass

Force= (Gravitational Constant)*(Mass)/Radius2

First solving for the gravitational force on Earth, substitute 6.67x10-11 for the gravitational constant, 5.97x1024 kg. for the mass of the Earth, and 6.378x106 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 .

The equation for the gravitational acceleration on the Sun is

Force= (6.67x10-11) (2.0029x1030) / (6.9552x108 meters)2 = 276.16 m/sec2 .

Let's 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/cm3 was compressed to 1.4 g/cm3 ?

1.4 / 0.026071326 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 the gravitational force needed, it is 525.64/m2 . So what reduction in the Sun’s radius with the same mass need to be to produced an acceleration of 525.64/ m2?

Plugging in the numbers, R2 = G x M/F from the original formula F = G M/R2 .

The square root of (6.67x10-11) (2.0029*1030) / 525.64 yields 5.0377x108 for the 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 293oK 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

P = Pressure V = Volume N = number of moles (Mass)R = Universal Gas Constant (8.3145 J/mole-oK)T = Temperature

Applying the Equation,

P = nRT/V

n is a constant, R is a constant, and V is now a constant with a radius 72.43%. We find that temperature in oK 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 300oK doubles the pressure and so on.

Mankind best estimates states that at .25 radii of the Sun, the temperature of the plasma is 7,500,000oK 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 300oK to 7,500,000oK is a factor of 25,000. So if we raise the temperature and go the original components to just the surface temperature of the Sun 6000oK, what does that do the pressure that gravity counters?

6000oK / 293oK 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 x 20.47 = 10,759.85

Now find the radius that the total mass of the Sun must be reduced to contain the pressures. The square root of (6.67x10-11) (2.0029x1030) / 10,759.85 yields 1.11426x108 = the new Radius or 16.02% of the present Sun (and well within the core of the present Sun).

What you have to realize is that mass does not 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. If it underestimated, then the Earth can maintain its orbit at the present velocity or it will fall into the Sun.

So we have G x Greater M * m/R2 = (m V2/R) +X . The 'X' is equal to the missing repulsion force.

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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)

[note: the effects of gravitation and repulsion of the other planets and the Dark One are omitted for simplicity sake although are minor.]

T (which is equal to time) is not a factor here for the basic equation when equal to 1, so it is omitted.Where M is the true mass of the Sun

Gravitational Equation valid only in the Resident Universe

e = exponential functionk = rate of decayM = Mass1

m = mass2

V = velocityR = radius

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 plays only 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 of gravitational particles on light particles are greatest when the angle of interaction and intersection approaches 90 degrees. Red wavelengths 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. 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 incoming 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.

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The 5 Primary Forces

Gravity exhibits all 5 primary forces in the Universe. First, magnetic in that it carries varied charges within its subatomic particles. Second, the nuclear force is responsible for the binding of all subatomic 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 without 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.

Mankind's Explanation for Gravity

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