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
Properties
& Origin of the Gravity Sub-Atomic Particle
More
Articles 1
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Articles 2
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.
Gravity
Gravity, long rumored to be a wave
or some cross wave-particle, is a polarized particle. It has a very minute set
charge when in equilibrium or its original state that moves from the void of
space eventually to the center of a mass. The flow rate of gravity particles in
close proximity and on the surface of a cosmic object is determined by the size
and volume of the mass. This dictates the magnitude of the attracting charge to
the gravity particles and the available space, which has gathered at the center
of the mass. A similarity to this description is water pressure, where the size
of the pipe (mass) and the power of the motor propelling the water (attracting
charge) yield rate of water moved (gravitational particle flow rate). 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).
During this particle-mass encounter, the independent mass is subjected to a
gravity particle pass through, a pressure force precedes the particle which is
dependent upon the complexity of the molecular bonds and the density of the
molecular structure with slight variations accounted for by anomalies in the
mass’s structure. This cumulative force of all gravity particles is applied or
pushes on the mass in the direction of the flow of gravity particles or towards
the center of the primary mass. The intensity of the gravitational force is
dependent on the amount of particles passing through a defined cubic area of the
mass during a time period represented by T and the subject mass elemental
composition, inorganic or organic 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 molecular
structure blocks the gravitational particle flow rates. Constriction of
available pass through space causes backpressures to reflect back towards the
incoming gravity particles, hence a larger force applied to the 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
a specified time period T. Once the gravity particles reach the center of mass,
a new phenomena, the repulsion force, emanates from a particle that has a
polarity opposite to gravity occurs. 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. Many have observed the bending
of light when it passes close to a star or large gravitational source. 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. The distortion in the fabric of space and time is not true.
Gravity affects mass not dimensions of length, width, and breadth.
Repulsion
The repulsion particle new to earth science is a transformed gravity
particle due to acquiring a new charge while being subjected to forces in the
core of the dominant mass. This takes place from overcrowding as gravity
particles continue to pour in. 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 original
charge and align with the polarity of the dominant 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, evenly
dispersed randomly in a 360-degree spherical direction. A two dimensional
example can be shown in the following thread where we only take into account net
X and Y distances when the balls occupy the same the plane at different times
and ignore the Z distance of up and down. {The effect of dispersion can be seen
when two balls are randomly set free in a slowly sloping cone. Their circular
path is finally separated by 180 degrees. This is not by chance. The balls, as
they travel, create pressure zones that precede their path in conjunction with
producing low-pressure zones in their wake. The equilibrium point between the
high and low pressure zones created between the balls sets them on opposite
sides of the circular path.} 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. This particle exhibits
some of the following new characteristics not yet seen by scientists. Any mass
at close range encountered 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. This is
accomplished as a stream of repulsion particles follow in each other wakes
allowing minimal push to affect the mass. {Similar to a Nascar racer who follows
a leader, the car’s drag is reduced while in close proximity to the rear wake
air stream in front.} Repulsion particles move in the form of a needle being
expelled at a high speed. Powerful, but passes through any mass without really
applying any force due to concentration. 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 dispersed apply an
outward force to the mass as they pass through. The point where this force
neutralizes gravity is the repulsion point or the mean distance of an orbit 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. This is why gravity is the same at all
parts of the globe except where surface anomalies are present.
I will now apply mathematical
gravitational equations to back the theory. Mankind assumes the force of gravity
exerted on planets in orbit is balanced by centrifugal force due to orbital
motion. Otherwise the planets would be pulled into the Sun by gravity. All
measurements below were acquired from standard reference books.
Setting
up the equations gravity equals centrifugal force, we have:
Gravitational
Constant * Mass 1 * Mass 2
=
Mass
1 * Velocity2
Radius 2
Radius
Solving
for Mass 2:
Mass 2
= 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 (Mass 1 which has dropped
out of the equation).
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) = 29,811 meters per second
Time (seconds)
Time conversion to seconds: 365.25 * 24 * 60
* 60
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 (Mass 1 which has dropped
out of the equation).
Mass2 =
Radius * Velocity2
1.995 *1030 kg = 1.4993
*1011 m * (29,811 m/s)2
6.67 * 10-11
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 is an
approximation due to the rounding of distances involving the radius.
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)
Time conversion to seconds: 27 days 7 hours 43 minutes
= 2360580
Mass2 =
Radius * Velocity2
Gravitational Constant
6.03 *1024 kg = 3.85
*108 m * (1024.75 m/s)2
6.67 * 10-11
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
Volume
Volume of
a Sphere = 4
* Pi * Radius3
3
We will use the formulas Density = Mass / Volume, and 4/3
* Pi * 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 6.9552 *108 meters, 6.378 * 106
meters.
Radius:
The Sun: 6.9552 * 108 meters
The Earth: 6.378 * 106 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 Earth = 3*
5.97 * 1024
4 * 3.1416 * 6.378 * 106
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: 1.416 grams / cc. and 5.49
grams / cc. The first sign of conflict in mankind’s theories should now begin
to surface.
Density of the mass of the Earth: 5.49
grams/cc.
Density of the mass of the Sun: 1.416
grams/cc.
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 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.
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 20
degrees C. under Earth’s gravitational force has densities of .00008375 grams
/ cc. and .0001664 grams / cc. respectively.
Densities at 200
centigrade
Hydrogen =
.00008375 grams/ cc.
Helium = .0001664
grams/ cc.
Under Earth’s
gravitational field
These density measurements of both hydrogen and helium gases would be much
significantly lower if the temperature of the gases were brought up to the
average internal temperatures of the Sun. Using the density figures from above
at the H - He distribution of the Sun, the combined density of the two elements
at Earth’s gravity is .0001044125 grams / cc. I will now introduce another
equation Pressure = Force / Area. I will substitute [Mass * Acceleration] for
Force, and Gravitational Force for Pressure. Now I will replace acceleration
with the gravitational constant. This yields Gravitational Force = [Mass *
Gravitational Constant] / Area. Since I am using a cc. for the standard area.
Both the area and the gravitational constant are fixed. Resulting in the mass
and the gravitational force being proportional to each other.
Force = Mass* Acceleration
Substituting: Gravitational force = Pressure
Gravitational constant = Acceleration
Yields: Gravitational Force = [Mass * Gravitational
Constant]/Area
Now according to mankind, the Sun’s gravity at it’s surface is approximately
28.1 times that of Earth’s, using the standard formula of Gravitational Force
or weight = Gravitational Constant * Mass / Radius2. It would take a
gravitational force approximately 13,562 times that of Earth to compress an H -
He mixture of the Sun to a density of 1.416 grams / cc. at 20 degrees C. This
would be significantly less than the force needed to support the internal
temperatures of the Sun. The gravitational force to compress H-He mixture to a
density of 1.4616 grams / cc. would not occur until the radius of the Sun was
reduced to approximately 4.5% of its original radial distance with the same
mass. To arrive at that conclusion, we will use the same gravitational formula
or weight formula above and solve for the radius as a final result. 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. Since
the gravitational force on the surface of the Sun according to physics is 28.1
times Earth, the force needed to achieve the desired density of hydrogen and
helium within its mass is 482 times the Earth’s gravitational force of
9.788837/ meters /sec2. Calculating the mass of the Sun and the
gravitational force needed to compress a hydrogen-helium mixture to the density
established by scientists supporting today’s theories could only be achieved
near the core, which would have to a maximum radius of 3.167 * 107 m
could yield sufficient force. If you compare the volume of the Sun’s present
sphere with the radius to achieve the hydrogen-helium compression level
equivalent to density of the Sun, the volume of the Sun would be only .094 % of
its original size. We can now see that the mass of the Sun is grossly under
estimated. Conclusion, the velocity of the planets and other objects in motion
around the Sun, does not have a fraction of the velocity to maintain their
positions.
This concludes that there is another force present to
the balance the velocity of the orbit and gravitational attraction, the
Repulsion Force. Now I will complete the new universal law of gravity and
repulsion. Flow Rate of Gravity Particles, which varies according to the density
level of the occupying universe or for this part of the universe, replaces the
accepted gravitational constant. So [Flow Rate of Gravity Particles / Radius2]
– [Flow Rate of Repulsion Particles, which varies according to the density
level of the occupying universe or for this part of the universe, becomes the
new Repulsion Constant * e(k *
Radius)]. The repulsion force, which is emitted out of a mass in
concentrated streams cannot use Radius2 as a rate of change, so the
exponential function e will be used, which that would more accurately describe
the rate of change. Where k is the rate of exponential change for is unit of
radius. So the exponential function becomes a factor of the repulsion constant
and the divisor radius2 drops out. The ratio between Mass1
and Mass2 must be large enough for the repulsion force to be invoked.
Otherwise the smaller mass will be captured and crash to the surface of the
larger object, due to insufficient repulsion particles flowing to over take
gravity. Mass1, Mass2 and T (Time) a new component to the
gravitational formula is a variable in the universe, which is still unknown to
most of Earth science become factors to both parts of the gravity and repulsion
equations, * [Mass1 * Mass2 * Time] = Total Force (the new
term replacing Gravitational force, which now only represents a positive
attracting force, Repulsion force the negative side.
T F = [Gravitational
Constant - Repulsion Constant * e(k
* Radius)] * [M1*M2*T]
·
T F = Total Force (Positive value is a gravitational force, a
negative value is a repulsive force.)
·
e
= exponential function
·
k = exponential rate at which the repulsion particle flow increases
or decreases to distance
·
T = Time (in our case is equal to 1)
To
test this theory we would have to look towards the Moon. The Moon’s absence of
an atmosphere will eliminate all of mankind’s excuses for backing Newton. 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 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.
I
want to thank God for bestowing knowledge
All
Rights Reserved: Copyright 2000
Mankind's
Explanation for Gravity
