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A
Mathematical Explanation of the Orbital Distance of
Mercury
by
Miles Mathis
Abstract:
I will show precisely why Mercury is orbiting at the distance it
is, rather than at any other distance.
I
said in my Third Wave
Series that I would have more to say about the mechanics of
the Solar System, and this paper will be the first in a series
explaining the Celestial Mechanics of nearspace in a new and
more rigorous manner. In
other papers I have shown that many “coincidences”
are not coincidences but the outcome of simple interactions, and
here I will continue to do that, with very simple math.
Historically,
the orbital distance of Mercury from the Sun has been considered
an accident, like all orbital distances. Currently, the standard
model cannot tell us why the orbital distance of Mercury is what
it is. Using my new Unified Field, I can. The standard model
cannot explain the orbital distance using simple mathematics,
since it does not have the correct fields. The standard model
explains the orbit using only two vectors, the centripetal
acceleration caused by gravity and the orbital velocity of the
planet. But the gravity vector in the standard model has no
mechanical life of its own. The gravity vector is only an
outcome of the orbital velocity. The gravity vector is not
measured independently, nor is it theorized independently. Both
historically and mathematically, the gravity vector was arrived
at by solving down from the orbital velocity, using the equation
a = v^{2}/r.* The orbital velocity is the data (from the
raw orbital period). It is what is seen and measured. The
gravity vector only balances the orbital velocity. It is an
outcome, not a postulate. This was true for Newton, it was true
for Einstein, and it is still true for the standard model. Up to
now, there has been no real theory of gravity and no real
mechanics underlying the vector.
But
I have shown in many
other papers that the gravitational field is actually a
compound field made up of two separate vectors. One of these
vectors I continue to call gravity, since it is an apparent
attraction. In the Unified Field, this vector points in. I call
it solo
gravity, since it is the gravitational field without the second
field. The second field is the foundational
E/M field or charge field. This is the field that underlies
both electricity and magnetism, and mechanically it causes the
charge between proton and electron. I have shown that although
the electrical field may appear to be either negative or positive
in interactions, the foundational E/M field is always positive or
repulsive. It is caused by simple bombardment, via a particle I
have dubbed the Bphoton.
This Bphoton
replaces the virtual photon or messenger photon of the standard
model. I also call it the charge photon in later papers. All
real photons in the spectrum can act as charge photons, and most
do. Although the charge field peaks or averages in the infrared,
any photon that passes through the nucleus (during charge
recycling) is part of the charge field.
Since
the foundational E/M field is always repulsive, it acts in vector
opposition to the solo gravity field. Both fields are active at
all levels of size, cosmic and quantum.
I
have shown how these fields are both found in Newton’s
gravitational equation. In the first instance, I have not added
any terms to Newton’s equation: it still stands as it
always has. I have simply broken it down into its constituent
fields, showing how G
works in the equation as a transform between the solo gravity
field and the foundational E/M field.
Once
the two fields are separated, the solo gravity field turns out to
be determined by radius alone. Gravity is an acceleration and
nothing more. To be rigorous, it is the acceleration of a length
or differential. This means it has nothing to do with density.
Density is a part of Newton’s equation, since it is a part
of the mass variables, but it turns out that the density buried
in the equation is actually a density of the foundational E/M
field. It is a density of Bphotons. All you have to do
is write each mass in Newton’s equation as density times
volume, giving the density to the E/M field and the volume to the
gravity field. G then acts as the transform between the two
fields. I have shown that G is also the relative diameter of the
Bphoton. In other words, the Bphoton is 6.67 x
10^{11} times smaller than the hydrogen atom.
Those
who have found my theory and math in these other papers to be a
bit abstract may be interested to see the field applied to a
standing problem of the solar system. I will do that now. I
will show that the orbital distance of Mercury is the distance of
balance for the three vectors involved.
If,
as I have claimed, the solo gravity field is determined by radius
alone, we should be able to find an acceleration at the distance
of Mercury quite easily, and it should not be the same as the
current number. In
another paper I have shown that the solo gravity field of the
Earth is not 9.81, but 9.82, and that the foundational E/M field
of the Earth is .009545. If these figures are correct, I should
be able to use these figures along with the known parameters of
the Sun to find its
separate fields. If these two fields do balance at the distance
of Mercury, I think I will have gone a long way to convincing
skeptics, since only the correct mechanics could hope to solve
the problem in this way.
The
Sun’s radius is 109 times that of the Earth. If gravity is
dependent upon radius alone, then the Sun’s gravitational
acceleration at its surface must be 109 times that of the Earth,
not 28 times as the standard model tells us. If so, then its
surface gravity must be 1070 m/s^{2}, not 274 m/s^{2}.
If that is true, then at the distance of Mercury, this
acceleration would be:
(1070
m/s^{2})(696,000/5.8 x 10^{7})^{2 }= .154
m/s^{2}
The
standard model currently believes the acceleration at the
distance of Mercury is:
Orbital
velocity of Mercury = 48,000 m/s
a
= v^{2}/r
a
= (48,000 m/s)^{2}/(5.8 x 10^{10}m)
a
= .04 m/s^{2}
The
dirty little secret here is that the standard model believes the
surface gravity of the Sun is 274 m/s^{2 }simply because
they did this math backwards. NASA has never measured the surface
gravity directly, of course. That number 274 is arrived at by
working backward from the orbital velocity of the planets. It is
not a measured number, it is a calculated number. And it
is calculated by assuming that gravity balances the
orbital velocity directly. The standard model has never proved
that gravity does this, it simply calculates the number required
to balance the orbital velocity and calls that number gravity.
There is no mechanics involved.
But
in my theory, all the numbers are grounded by field mechanics.
If gravity is dependent on radius alone, we have a logical field
from the beginning. Gravity is an acceleration in the equations,
and now in my equations it is an acceleration in the field as
well. It is an acceleration and nothing else. It does
not include density, so it has no mystery to it anymore. Where
Newton and the standard model have much to explain, I have
nothing to explain. My solo gravity field has nothing to do with
mass or density, so most of the historical questions immediately
evaporate.
Can
my mechanics explain the balance better than the standard model?
You be the judge. I have shown that the gravitational strength at
the distance of Mercury should be .154 m/s^{2 }and that
Mercury’s orbital velocity tends to give it an escaping
force of .04 m/s^{2}. Therefore, the E/M field must make
up the difference. That is, I must find that
.154
m/s^{2 }+ E = .04 m/s^{2}
E
= .114 m/s^{2}
I
must show that, at the distance of Mercury, the Sun is repulsing
at that strength. We can do that in two ways. One, use that
number to show what the Sun must be repulsing on its surface:
E_{S}
(696,000/5.8 x 10^{7})^{2 }= .114 m/s^{2}
E_{S}
= 792 m/s^{2}
Or,
go to the surface directly, and figure from the surface numbers:
1070
m/s^{2} + E_{S} = 274 m/s^{2}
E_{S
}= 796 m/s^{2}
To
balance the orbit of Mercury we must find that value for E_{S}.
Is it possible that this value confirms my value for the Earth
of .009545 m/s^{2}?
I found that value for the Earth simply by looking at the
parameters of the Earth and Moon, as you can see by returning to
that paper. That
paper was written several years ago, and when I wrote it I had no
idea it would be useful when looking at Mercury or the Sun. Many
will have thought I pulled that number out of a hat, by some kind
of magic (although the math is quite simple); but if I can use it
to balance the orbit of Mercury, many people will be eating their
own hats.
Watch
this:
I
showed that the foundational E/M field of the Earth has a
strength at the surface of the Earth of .009545 m/s^{2}.
The Sun has a mass of 333,000 Earths. It has a density ¼
that of the Earth.
(.009545
m/s^{2})(333,000)(¼) = 795 m/s^{2}
No
magic there, just transparent math and fully defined field
mechanics. By these calculations, the Sun is “attracting”
Mercury with a straight acceleration of .154 m/s^{2}. It
is repulsing Mercury by bombarding it with Bphotons at a
rate of .114 m/s^{2}. Mercury is fleeing the Sun at a
rate of .04 m/s^{2}. These three vectors balance at the
orbital distance of 58 million kilometers. That is the
mechanical reason Mercury is at that distance and at no other
distance.
All
this was made possible by mechanical postulates. I did not start
with orbital velocities and solve down from them, like Newton and
the standard model. I started with the theoretical postulate,
based on logic, that gravity, taken alone, must be proportional
to radius and radius only. If gravity is an acceleration in our
math, we should attempt to make it an acceleration in our
mechanics and fields, unpolluted by other factors like density.
If Einstein was correct in his equivalence principle—and I
believe that he was—then gravity should be expressed by
acceleration alone. If we follow Einstein’s lead and
reverse the gravity vector, then the acceleration must apply to a
length, not to a mass. If we reverse the vector g, then
it must be the acceleration of the radius. The equivalence
principle is equivalent only if the vector applies to the same
parameter in both directions, both before and after the reversal.
The acceleration and gravity vectors are equivalent and
reversible only if they both apply to lengths. We must have the
acceleration of a length in both directions. In other words,
when we treat gravity as a pull down, we do not have to consider
the mass of the object being pulled. All objects,
regardless of mass or density, are pulled down at a rate of 9.81
on the surface of the Earth. Therefore, if we reverse the vector,
we should not have to take the mass or density of the Earth into
consideration. If we reverse the vector, a la Einstein,
the Earth is then expanding. But the acceleration is given to
the radius and the radius only. Logically then, gravity should
vary with radius.
This
postulate, applied to Earth and Moon, allowed me to generate
numbers for both fields in my Unified Field. The only other
postulate necessary was that the E/M field filled the gap, and
did so by being always repulsive, in vector opposition to
gravity. This postulate was not pulled from thin air, either. It
was a logical requirement of the charge field. Up to now, the
charge field has been mechanically undefined. It has been
mediated by virtual particles, which created a force and a field
with no mass and no energy. Impossible. I simply gave this
field the energy required to make it work mechanically, and that
energy had to be inserted into the universe. It could not exist
at the quantum level and fail to exist at the macrolevel. I
inserted it into Newton’s equation, and it happened to fit
quite well. Since I had already changed the strength of solo
gravity, there was plenty of room in Newton’s equation. The
foundational E/M field fit the empty space like a hand in a
glove. In this way I was able to keep Newton’s compound
field exactly like it was. I made no correction to Newton, I
simply reexpanded his equation, showing its constituent parts.
One
final clarification. Some will see my equation above and ask why
it cannot be applied to the Moon. To find the number for the
Sun, I multiplied the Earth’s E/M field by the mass
differential and the density differential. If we do that for the
Moon, we get
(.009545
m/s^{2})(1/81)(.6) = .00007 m/s^{2}
But
I have shown in another paper that the Moon’s E/M field is
actually greater than the Earth’s, with a number of 1.05
m/s^{2}. How do I explain this? I explain it because,
as I have said before, you can’t just start plugging
numbers into equations. You have to look at how the field
mechanics are working beneath the equations. These differentials
are relative numbers, and in one case we are looking at an object
smaller than the Earth, the Moon; in the other case we are
looking at an object larger, the Sun. And I have shown that the
Unified Field unifies differently at different sizes. This is
because we aren’t concerned directly with the macrodensity
of the macroobject, we are concerned directly with the density
of the bombarding field of Bphotons. In a smaller
spherical object, this field is automatically compressed on the
surface of the object, due to the surface area only. Our
equations must therefore express this. In other words, if the
Sun and Moon had the same density, they still would not have the
same E/M field density. The E/M field density is not dependent
upon object density, it is dependent on mass and surface area.
The entire mass emits the field, and it emits it into a certain
space. The Moon is emitting into a very small space compared to
the Sun, you see. Therefore, all other things being the same, the
Moon must have a denser field.
Of
course all other things are not the same, since the Sun has a
mass 27 million times that of the Moon. It has a huge amount of
mass summing behind the surface, and this takes the strength of
the Sun’s E/M field back above that of the Moon, by a large
margin (757x). It may seem strange that both the Sun and Moon
have stronger E/M fields at their surfaces than the Earth, but if
you do the math and follow the fields, it makes perfect sense.
Let’s
look at the surface area differential in both problems, to show
this. To find the Sun’s E/M field from the Earth’s,
you can multiply by the mass differential and the density
differential, as I showed. But, although this is the most
efficient method, it does not show the mechanics as clearly.
Using the surface area differential, we would expand the
transform for the Sun like this:
(.009545
m/s^{2})(SA)(D) x (M)/(SA) = 795 m/s^{2}
The
first term, SAD, gives us the density of the macrofield on the
surface; and the second term, M/SA, gives us the strength of the
Bphoton field backing up or feeding that surface area.
The two together give us the E/M field strength at the surface.
But
if we do the same thing with the Moon, we cannot use the same
equation. Instead we must use this equation:
(.009545
m/s^{2}) (SA)M/D x (SA)/M = 1.05 m/s^{2}
The
second term is the inverse, as you see, since we must find the
field feeding the surface. And the first term is also slightly
different, since this term expresses the size difference. Objects
larger than the Earth will be emitting into a larger space and
objects smaller than the Earth will be emitting into a smaller
space. Therefore, the transform must work differently in each
direction. The Sun must have a less dense field on the surface,
and the Moon must have a more dense field on the surface, and the
equation must express this. We divide by the density instead of
multiplying, as you see, and this expresses the move to smaller
instead of larger. But we need one final change, and that is to
multiply by the mass differential once more. This is because if
we divide by the density differential we lose the mass
differential; so we need to reinsert it. Another way to look at
it is that we really multiply by the volume differential when we
go smaller.
(.009545
m/s^{2}) (SA)V x (SA)/M = 1.05 m/s^{2}
That
keeps the field mechanics in order. I know this last part is a
bit dense, but if you study the mechanics—as I have
described it above and in several other papers—you will see
why transforming smaller and transforming larger, using the field
differentials, must be done with different equations. Remind
yourself occasionally that these variables do not stand for the
actual densities or volumes or masses. In these equations they
stand for the relative numbers: one object’s mass relative
to the other object’s mass, for instance.
*Some will be
surprised to find me using the equation a = v^{2}/r, when
I have shown that it is false. I must use it in this paper,
simply because I am using numbers derived from it, like the
orbital velocity of Mercury. The number 48,000 isn’t really
the orbital velocity of Mercury, it is 2πr/t for Mercury.
Although the term 2πr/t isn’t the orbital velocity, it
is the term that works with a = v^{2}/r, so if I am going
to use one I must use the other.
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