HOW to CALCULATEthe ECCENTRICITY of the EARTHUsing the Charge Field by Miles Mathis First published February 2011 In a series of other papers, I have calculated the axial tilt of many planets, the eccentricity of the Moon, the Bode series, the magnetopause of both Earth and Venus, and many other numbers using my new unified field equations. Here, I will calculate the Earth's eccentricity. The Earth's eccentricity is said to be caused by gravitational perturbations from other bodies, but once again both the math and the theory fail. It is interesting to go to Wikipedia on this one, which normally has all kinds of math on things like this. This time we get nothing. Wiki doesn't even give us a section on the cause of orbital eccentricity. It doesn't even mention that eccentricity must have a cause. Eccentricity is used as a cause of climate, but it doesn't have a cause itself, I guess. The Earth has a low eccentricity of .0167, about 1/3 that of the Moon. If we try to explain that with gravity alone, then we must look mainly to Jupiter and Venus. All other perturbations will be small relative to those. According to current math and theory, Jupiter should be the main influence, with a maximum force almost twice that of Venus and 26 times that of Mars. These perturbations are at closest approach: ΔaVen = GM/R2       = 2 x 10-7 m/s2 ΔaJup = GM/R2       = 3.6 x 10-7 m/s2 ΔaMars = GM/R2       = 1.37 x 10-8 m/s2 ΔaSun = GM/R2      = 6 x 10-3 m/s2 But we can't even add the effects from Venus and Jupiter, since when the two are in line at closest approach, they subtract. This means the maximum perturbation is by Jupiter and Mars in line, at 3.7 x 10-7 m/s2, and the minimum perturbation is 0, when Jupiter, Venus, and Mars are stacked behind the Sun. The difference between maximum and minimum should give us the eccentricity, and that is still 16,000 times less than the effect from the Sun. There is no way that can cause an eccentricity of .0167. Put simply, the gravitational math is all magic. It is fudged from top to bottom. I have already shown you in other papers how the Moon's orbit is a mess of fudged and unsupported equations. If the Moon's orbit is fudged, why would anyone trust the math of perturbations between Jupiter and the Earth, or Venus and the Earth? If they can't solve the Moon's orbit, then they can't solve anything. The Moon is orbiting the Earth, so there is an apparent gravity field between them. But there is no gravity field between the Moon and Venus. Venus is not perturbing the Moon gravitationally, or the reverse. They only influence one another via charge. This is what Newton, Kepler, Laplace, Lagrange, and Einstein could not comprehend. They had equations that contained this information, as I have shown, but they could not unlock them. Rather than look at the math again, let's look at the field again. The top physicists brag on their T-shirts that they have gotten rid of force at a distance, although you wouldn't know it from perturbation theory. If there is no force at a distance, how is Jupiter perturbing the Earth? Don't tell me gravitons, since 1) they haven't been discovered, 2) they aren't logical to start with. You can't explain pulls with fields of particles. Beyond that, many of these perturbations we are given by the mainstream are pertubations at right angles or tangents. The historical gravity field is centripetal: it can't provide such forces or accelerations or perturbations, and both Newton and Einstein agreed on that. They said it out loud, in simple declarative sentences. I will be told that space is curved, and that perturbations are geometric, not dynamic. But that explains nothing, especially not orthogonal perturbations. Gravity can't curve enough at such distances to provide these perturbations. Einstein's space is curved around a body. As such, it can explain orbits (sort of), but it still can't explain straight line forces between bodies that aren't orbiting one another. A perturbation between Jupiter and the Earth is a straight line force, since the Earth isn't orbiting Jupiter, or the reverse. The Earth isn't in Jupiter's curve, or the reverse. The Earth is only in the Sun's curve. Remember the rubber mat and the ball bearing they like to show you, to help you visualize curved space? Well, that explains the curve of the orbiter only, and thereby the appearance of a force. What curved mat is between here and Jupiter? And if Jupiter perturbs the Earth and the Earth also perturbs Jupiter, which way is the mat curving? Can it curve both ways at once? In other words, can space be concave and convex at the same time? No, it must be one or the other. Einstein didn't really explain anything about orbits, either. He just gave physicists two theories to ignore and corrupt instead of one. Without the charge field, neither Newton's equations, Einstein's equations, Laplace's equations, nor Lagrange's equations can explain the motions we see. Yes, maybe Lagrange came closest, since he was the best cheater. He saw that we needed a differential equation, with gravity somehow resisting itself, and he cleverly pushed the math to create one. But the Lagrangian is just as smelly as all the rest of the historical field math, and it is way past time we threw it out and started over. I will do that now. I have already shown (in my paper on Axial Tilt) that the Earth's tilt is caused mainly by charge forces from the four big outer planets (Jupiter, Saturn, Uranus, Neptune). It turns out that the same thing causes the eccentricity of the Earth's orbit. You will say that if they are caused by the same thing, they should show the same percentage changes in the field, so that the amount of eccentricity should match the amount of tilt. They don't match, therefore I can't be right. Well, this is why no one has solved this, but it can be solved quite easily. The tilt and eccentricity do match, as I will now show. I will once again limit my math to the four big planets, simply estimating an answer. In my Axial Tilt paper I showed—using simple unified field equations—that the four planets were responsible for a tilt+inclination of the Earth of 33.66*, which, being 37.4% of 90, indicates a 37.4% difference between the planets' effect and the Sun's. Since the planets are an average of 23 times further away from the Earth than the Sun, we divide .374 by 23 to obtain .0163. That would be the eccentricity just from the four planets and the Sun. Since the actual eccentricity of the Earth is .0167, I am very close already. To understand how my simple math worked, we have to go back to the ellipse, as proposed by Kepler. Kepler proposed an ellipse with two foci, the Sun being at one focus. But that isn't the way real ellipses are made. There is no body at the other focus, for example, so the forces in a Kepler ellipse are ghost forces. They don't exist. The second focus in the Earth's ellipse should be outside the ellipse, as we recognize when we start doing the actual perturbations. As I just showed, the second “focus” is the average distance of the four big planets. It is sort of their orbiting center of mass. Since it is at 24AU, it can't be inside the Earth's ellipse. This would be true even in current theory, since whether you are calculating gravitational perturbations or charge perturbations, you can't propose that the source of your perturbations is coming from within the Earth's ellipse. Yes, Venus' perturbation is inside the ellipse, but all the other important ones aren't, and the average perturbation certainly isn't. Therefore the second center of force can't be at Kepler's second focus. I will be told that you can solve with barycenter forces, since the center of mass of the four orbits can be found, and it will be found to be at the other focus, inside the Earth's ellipse. But even if that were true, and even if the problem could be solved with gravity, that would be a mathematical solution only. Again, no real body exists at the barycenter, so no real force or effect emanates from there. It is better to attach the math to the real bodies, so that the real influences can be seen and studied. These problems I am solving persist to this day precisely because physicists have given us mathematical instead of mechanical solutions. We don't need to know how abstract or geometric ellipses are created, we need to know how physical ellipses are created. Geometric ellipses are created with two interior foci, but celestial ellipses clearly are not. Celestial ellipses are caused by forces outside the circle. Besides, the gravitational field can't solve this one, so the question is moot. I might be able to solve by finding the barycenters of charge, in a like way, but I find that pointless. It is much preferable to describe the real mechanics. You will say, “But didn't you just find a center of mass to solve?” Kind of, but that is not a mathematical trick, that is just an average distance. You take the four orbital radii of the big planets and divide by four. It is not a center of mass, like a barycenter, it is an average orbiting distance of mass. The force is still coming from outside the Earth's orbit, so the direction of the force is clear. I have not moved the second “focus” inside the circle to misdirect you. This means that the ellipse of the Earth is not caused by a second pull from within, it is caused by a second push from without. The big planets are pushing on the Earth with their charge fields, with their photons. [Addendum, July 2015: With some help from a conference attendee, I finally realized how this ties into the Cassini oval. In the late 1600's, Domenico Cassini was able to better estimate the orbits of many real bodies by taking the product from two foci, rather than the sum. This indicated the orbits were not actually ellipses, but only looked like ellipses. Well, the math that includes this exterior force from the Jovians mirrors Cassini's oval, not the ellipse. As we have seen from my expanded mechanics, summing distances to create an ellipse was always naïve. You don't simply sum distances, you integrate forces, and these forces cannot be found only from distances. You need to know both the distances and the charge field influences, which means the creation of the oval was always more complex than anyone understood. Not only must you know the distances and the charge field strengths, you must know that the charge field influences work in the opposite way depending on whether charge is moving toward the Sun or away from it. Since we have an integration of influences, it is no surprise Cassini's product was a better estimate in most cases than the sum that created the ellipse. This also goes a long way to explaining why they needed perturbation theory and then chaos theory to fill holes in old field equations. Since they were expressing orbits with incomplete and naïve theory and math, of course they were going to find “remaining inequalities.” {See my earlier paper on Laplace for more on that.} If you think your orbit is an ellipse caused by interior foci, of course you are going to get field equation failures. All this was caused by a failure to correct Kepler. Newton assumed Kepler's ellipses were correct, and everyone since then has been filling holes in the ellipse equations. But if we create the oval by a more complex field mechanics, the holes in the equations evaporate. Chaos theory is not necessary, since the chaos was not in Nature, but in our misunderstanding of the field.] You will say, “Above you showed that the eccentricity was just 1/23 times the tilt. But you already used the number 23 to find the tilt, in that other paper. How can you use the same number again? First you find the planets have their charge compressed by distance, which means you multiply by 23; then you divide by 23 to find the eccentricity. You seem to be going around in circles. I am lost!” Well, we have to look at the way eccentricity and tilt are measured. The quickest way to make you see this is to remind you that if the force outside the circle were constant, there would be no eccentricity. Remember that we are explaining the ellipse now by a perturbation from outside the circle, not a second focus inside the ellipse. If the force from outside (by the big planets upon the Earth) were constant in strength and direction, this would only change the radius of the orbit; it would not create an ellipse. To create an ellipse requires a varying force from outside. You can see this in the way I did the gravitational perturbations above. We calculate the variance from maximum to minimum. So the number .0167 is a measurement of this variation. In other words, it is not telling us the difference between the charge field of the Sun and the charge field of the four planets. It is telling us the variation in this difference. So the tilt is a measurement of the difference, and the eccentricity is a measurement of the variation in the difference. This is why they aren't the same number. One number is the change in the other number. The reason we can use the number 23 to go from tilt to eccentricity is that the number 23 tells us the variation in the number from the planets. In seeking the variation, you seek the difference between the maximum and the minimum, right? Well, we have found that 23AU is the average distance of the four planets from the Earth. How do you calculate an average? You add the four distances and divide by 4. How do you find a difference between maximum and minimum? Maximum is when the four planets align and minimum is zero (when the four planets are stacked behind the Sun). When the four planets align, you find the charge distance by adding the four distances and dividing by 4. The average distance and the variation are the same number! The average distance is 23, and the variation in charge difference is 23-0. So the first time I use the number 23, it stands for average distance. The second time I use the number 23 (as above), it stands for variation. The number is the same only because the math is very similar for average distance and variation. Doubters will say, “That math is beautiful in a strange and curious way, but it doesn't make any sense. If charge toward the Sun increased with distance, then things even further away would have even greater charge. The Oort cloud would run the Solar System.” But I have never said that all distant objects have their charge increased by this method. It only applies to objects in orbit around the Sun. It is a unified field effect, so the object in question has to be IN the Sun's unified field. Its charge must be captured by the vortex of the Sun, and this vortex has a limit. I haven't been able to calculate yet where this limit is, but it is probably somewhere just beyond Pluto. [The mainstream has now found this Kuiper Cliff to be at about 47.8 AU.] Farther than that, the charge is not channeled toward the Sun in an efficient manner. In other words, charge out from the Sun will have dissipated so much that it cannot channel charge in. Remember, we have charge moving in both directions. It is doubtful, for instance, that much charge from the Oort cloud makes it back to the Sun. But, yes, the charge that does make it back will be compressed and its density will be increased, increasing its effect. You will now say, “But in other papers you have gotten rid of positive and negative charge. For you, all charge is positive. How can you have charge going in both directions? How are the potentials created?” They are created by poolball mechanics, not strange charge mechanics. In other words, it is not mathematical potentials, created out of nothing with plusses and minuses, that cause the charge motion, it is simply the recycling of the field. As I have said in other papers, you have to think of the charge field as wind, not as mysterious potentials. Basically you have two winds: you have the incoming wind of charge from the galactic core to the Sun. The Sun takes in this wind at the poles, due to spin, and emits it at the equator. The emitted wind moves opposite to the incoming wind, but the two are interpenetrable. The photon wind is so fine, as a matter of particles, that it doesn't interfere with itself much. Photons do collide, but the collisions only affects the spins, not the linear velocity. Which means that only the magnetism is affected by these collisions; the linear charge (which causes electricity) isn't. Although photon fields are mostly interpenetrable with eachother, they are not interpenetrable with normal matter in planets. Since baryons are quite large compared to photons, we have many orders (more than 10; see the number G) more collisions, and these collisions create measurable forces or motions. The sum of all these collisions is what causes perturbations. Now that we see how the ellipse is really created, we see that the big planets are not only the cause of the inner ellipses, they are the reason these ellipses are so small. The fairly large charge forces from the outside keep the Earth's orbit constrained. Even if it were bumped into a greater eccentricity by an impact, the charge field would resist this eccentricity, and would tend to push it back to a lower eccentricity over time. This is why even large impacts are rarely fatal to an orbit. Planets and moons will break up before they will crash into a primary or eject from the system altogether, as we see from the debris around Jupiter and Saturn, as well as from the asteroid belt. ~~~~~ In closing, I will answer some more questions. A reader has commented on my tilt math in this way, “I have studied your math from that other paper on tilt, and it appears that you have found that the four big planets have a charge effect greater than that of the Sun. How is that logical? Look here, according to you, altogether the four big planets have a charge .0014 that of the Sun, and an average distance of 23 times more than the Sun from the Earth. You have calculated in your paper on Mercury that the Sun has a charge of 796. So the fourth root of that is 5.93. That is the charge at the Earth. The charge of the big planets is 796(.0014)23= 25.6. The planets would appear to have 4.3 times as much charge as the Sun.” Well, that math isn't correct, but this reader is correct that the planets have a large charge density and a large field effect, both on tilt and eccentricity. I have already explained this in part 2 of the tilt papers. In looking at perturbations, we are looking at charge density, not charge strength. Yes, the perturbations from the planets, especially Uranus and Neptune, have very high charge densities in the math, since they are compressed by the long distance. This gives them great power as perturbers, but it does not mean they are responsible for greater overall charge than the Sun. You will say that charge density should be a measure of charge strength, and in some cases that is so. But here it isn't. A body with greater charge density would have greater charge strength than a second body only if equal amounts of charge are arriving at the spot in question over the same time. But that isn't the case with the Sun and the planets. The same amount of charge is always arriving from the Sun to the Earth, but the amount of charge from the planets varies according to positions. Therefore, over the course of a year, say, the Earth gets much more charge from the Sun. But because that charge is steady, it doesn't perturb. It doesn't cause a change because it is the baseline, and a perturbation is a change from normal. The high density charge perturbations from the planets peak about four times a year, but that is plenty to cause a change in the tilt or the eccentricity. Since celestial bodies are very large, with large inertias, you do not have to have continuous forces to create perturbations. Four peaks a year is more than enough. [March 2019: these next few paragraphs have been slightly rewritten and extended in an attempt to clarify.] But let us correct my reader's math, rerunning and then expanding the equations to make them a bit clearer, especially for those who haven't read the tilt papers closely. He has tried to mirror my math, but hasn't quite got the feel for it yet. He has actually done a bit too much math, but we will use his numbers to get the right answer. If the four planets have a total raw charge density at their surfaces .0014 that of the Sun, then we only need to multiply that by 23 to get .032. Since the Sun has a charge of 1 in that math, the number .032 is already the number we need, without bringing in the real charge number of the Sun. You see, the number .032 is already a fraction of 1, so we don't need any other manipulation. That is the percentage of charge from the planets, as matter of charge density. As I have shown most famously in my Bode's Law and Axial Tilt papers, to do unified field equations, we don't need absolute numbers, like accelerations in m/s2. We just need to find numbers relative to the Sun. But we haven't included the fall-off in charge from the Sun. In this simplified math, you only feel a charge density of 1 from the Sun at the surface, but the field density dissipates with distance as it goes out. There are various ways we could include that fall-off at this point in the math, but the simplest is just to multiply by 23 again. Rather than decreasing the Sun's density, it increases the planet's density by the same amount. You see, because the Sun's number is one in the math, we can't manipulate it by going to a fourth root or something, so we have to do the math on the other number. So we get .736. The perturbing charge density from the big planets is now almost 74% that of the Sun. And so I have proven my point above. I said that gravity perturbations 16,000 times less than the main orbital forces could not create an eccentricity of .0167. We have just seen that the charge density that causes eccentricity is only 31 times less than the ambient (Solar) field [1/.032]. Currently, it is thought that eccentricity is low because the perturbations are low. But, as we have just seen, the perturbation is actually very large. Also notice that 31 x 232 = 16,399. So the mainstream had the right numbers, it just had the math upside-down, in a way. But neither .032 nor .736 get us to the eccentricity number .0167. So how do I finish the math from this direction? Well, let's return to the number .736. That would give us the right percentage relative to the Sun only if the Earth could receive charge from the planets from all sides equally. But it can't because the Sun is in the way on one side. In other words, if we sum over any period of time, half that time the planets will be opposite the Sun. In that position, the Sun captures their returning charge, and it doesn't make it to the Earth. Therefore we have to divide by two, getting the number .368. Which brings us back to my first abbreviated math above, where I imported the number .374 from my axial tilt paper. We see almost the same number. So again, we just divide by 23 to get .016. And why do we divide by 23 this time? Because this is where we include the variance from maximum to minimum. As I just told you, eccentricity isn't caused by the raw charge densities or perturbations, it is caused by the variance from max to min. Of course, I could have just divided .032 above by 2 to get .016, simplifying the math and skipping a couple of steps, but then you would have missed all the field mechanics. I started with the abbreviated math way above, so this time I wanted to expand it, showing where the numbers come from. Importing a number from a previous paper enabled me to do the math in two lines, but it forced you to read that paper to see how I got that number. Here, with math that is just as simple, I have again showed you how the real charge field works, and how it generates numbers. For kicks, let us simplify the math back down, using what we just discovered. Using mainstream numbers for mass and density, we find the big four planets have .0014 the amount of charge as the Sun. But their charge is compressed in the return, so we multiply by 23, getting .032. We then halve that to get the eccentricity number .016. Again, that math is compressed, since we don't include the variance anywhere there. It gets the right answer only because the number for variance and the number for distance are the same. We multiply and then divide by the same number. I hope that clarified things. ~~~~~ Let's see how this fits into the unified field at the Earth. We know that the Earth must be balancing charge from both directions, since the actual distance of the Earth from the Sun can't be explained by forces from the Sun alone. Let's return to the numbers. Using my unified field numbers (found by separating current numbers into solo gravity and charge)**, the solo-gravity of the Sun is 1070 m/s2, which drops off by radius. So at the Earth this number is 1070/23,456=.0456. The charge is -.163. To create a stable orbit in the unified field, those numbers must balance, but the given orbital velocity can't balance those numbers, since (by my correction to the a=v2/r equation, which is a=32r/t2) it adds a centripetal acceleration of only .0048. So the Earth isn't anywhere near where it would be with the Sun alone, and no other planets. It would be a good deal farther away, to get that charge number down. This is logical, because it is the large outer planets that are pushing it back into its current radius. According to this simple math, the big planets should be pushing back with a combined charge acceleration of .163 + .0048 - .0456 = .122. Please pause to drink in that last number, since it is very important. See above, where we found the number .736. If the Sun's real charge number at the Earth is .163, then the big planets' charge number is .163 x .736 = .12. That number match proves I am right about the unified field and its mechanics. If I weren't correct, these numbers wouldn't match. So, as you see, I have balanced the unifed field once again, showing all the factors. This might possibly bring up more questions, from those who are following this math closely. You will say, “If the Sun's gravity drops off by radius, you should use the Sun's radius to calculate the drop off, not the Earth's. And yet you use the number 23,456, which is the number of Earth radii in one AU. How does that make sense?” Well, the number 23 above was found using 1AU as the baseline distance. The number 23 means 23AU. It does not mean 23 solar radii. Since we are comparing forces from the planets to forces from the Sun, we must use the same distances for both. The numbers won't be comparable, unless the distances under the numbers are the same. The number .122 comes from using the number 23, which comes from using 1AU, you see. Since I am comparing .122 to .163, I have to take that into consideration. And since .163 was found using both 1AU and 1 Earth radius, I have to use 1 Earth radius in the Solar gravity number as well. You may need to consult the math in my Lagrange point paper to comprehend my point here. I am combining numbers in three different fields here (Solar gravity, Solar charge, planetary charge), and combining them at the Earth, so I have to scale them all to the same baseline field. *The Earth's actual tilt to the Sun's equator is 30.55, but we have to include Mars and Venus to find that number.**See my papers on the unified field, starting with this one. For my simple method of finding the numbers for the Sun, you may go here.