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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 near-space in a new and more rigorous manner. In other papers I have shown that many “coincidences” are 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 = v2/r.* The orbital velocity is the data. 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 B-photon. This B-photon replaces the virtual photon or messenger photon of the standard model.

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. 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 B-photons. 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 B-photon. In other words, the B-photon 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 figure. 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/s2, not 274 m/s2. If that is true, then at the distance of Mercury, this acceleration would be:

(1070 m/s2)(696,000/5.8 x 107)2 = .154 m/s2

The standard model currently believes the acceleration at the distance of Mercury is:

Orbital velocity of Mercury = 48,000 m/s

a = v2/r

a = (48,000 m/s)2/(5.8 x 1010m)

a = .04 m/s2

The dirty little secret here is that the standard model believes the surface gravity of the Sun is 274 m/s2 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 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/s2 and that Mercury’s orbital velocity tends to give it an escaping force of .04 m/s2. Therefore, the E/M field must make up the difference. That is, I must find that

.154 m/s2 + E = .04 m/s2

E = -.114 m/s2

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:

ES (696,000/5.8 x 107)2 = -.114 m/s2

ES = -792 m/s2

Or, go to the surface directly, and figure from the surface numbers:

1070 m/s2 + ES = 274 m/s2

ES = -796 m/s2

To balance the orbit of Mercury we must find that value for ES. Is it possible that this value confirms my value for the Earth of -.009545 m/s2? 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/s2. The Sun has a mass of 333,000 Earths. It has a density ¼ that of the Earth.

(-.009545 m/s2)(333,000)(¼) = -795 m/s2

No magic there, just transparent math and fully defined field mechanics. The Sun is “attracting” Mercury with a straight acceleration of .154 m/s2. It is repulsing Mercury by bombarding it with B-photons at a rate of -.114 m/s2. Mercury is fleeing the Sun at a rate of .04 m/s2. 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 macro-level. 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 re-expanded 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/s2)(1/81)(.6) = -.00007 m/s2

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/s2. 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 macro-density of the macro-object, we are concerned directly with the density of the bombarding field of B-photons. 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/s2)(SA)(D) x (M)/(SA) = -795 m/s2

The first term, SAD, gives us the density of the macro-field on the surface; and the second term, M/SA, gives us the strength of the B-photon 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/s2) (SA)M/D x (SA)/M = -1.05 m/s2

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/s2) (SA)V x (SA)/M = -1.05 m/s2

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 = v2/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 = v2/r, so if I am going to use one I must use the other.


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