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HOW TO
CALCULATE THE ECCENTRICITY OF THE EARTH Using 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:
Δa_{Ven}
= GM/R^{2}
= 2 x 10^{7}
m/s^{2}
Δa_{Jup}
= GM/R^{2}
= 3.6 x 10^{7}
m/s^{2}
Δa_{Mars}
= GM/R^{2}
= 1.37 x 10^{8}
m/s^{2}
Δa_{Sun}
= GM/R^{2}
= 6 x 10^{3}
m/s^{2}
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/s^{2},
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 Tshirts 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 230. 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. 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.
But let us use my reader's math to rerun the equations,
perhaps making 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 planets have a charge .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. We just multiply by 23
again to include the field variation, to get .736. Since I have
shown in my
Lagrange point paper that the Sun's number at the Earth is
.163, we find the number for the big planets to be .122. That's a
lot of charge density, admittedly, but it is not more than the
Sun's, not even at maximum perturbation.
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 it takes a
variation in charge density almost 75% that of the ambient charge
to create that much eccentricity. In other words, the charge from
the big planets is only 25% less than the charge from the Sun, as
a matter of density. Not 16,000 times less, but 1/4 less. The
current math isn't even close. Currently, it is thought that
eccentricity is low because the perturbations are low. But, as we
have just seen, the eccentricity is low because the outside
influence is near the inside influence. It is the low percentage
change that causes the low eccentricity, not the low relative
value of the perturbation.
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 sologravity of the Sun is 1070 m/s^{2},
which drops off by radius. So at the Earth this number is
1070/23,456=.0456. As we saw above, the charge is .163. To
create a stable orbit in the unified field, those numbers must
balance, but the given velocity can't
balance those numbers, since (by
my correction to the a=v^{2}/r
equation, which is a=32r/t^{2})
it adds an acceleration of only .0048. 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 with a combined
charge acceleration of .163 + .0048  .0456 = .122. Please pause
to drink in that last number, since it is very important. Go back
two paragraphs, where you will find the same number, achieved by
different math. That number 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 brings up more
questions. 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.
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