return to homepage 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 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 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 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 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 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 (1070 m/s The standard model currently believes the acceleration at the distance of Mercury is: Orbital velocity of Mercury = 48,000 m/s a = v a = (48,000 m/s) a = .04 m/s The dirty little secret here is that the standard model believes the surface gravity of the Sun is 274 m/s 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/s .154 m/s E = -.114 m/s 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 E Or, go to the surface directly, and figure from the surface numbers: 1070 m/s E To balance the orbit of Mercury we must find that value for E 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 (-.009545 m/s No magic there, just transparent math and fully defined field mechanics. The Sun is “attracting” Mercury with a straight acceleration of .154 m/s 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 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/s 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 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 write the transform for the Sun like this: (-.009545 m/s The first term, SAD, gives us the density of the macro-field on the surface; and the second term, SA/M, gives us the strength of the 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 The second term is the same, as you see, since we still must find the field feeding the surface. But the first term is 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 (-.009545 m/s 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 If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many |