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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. [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/s^{2}.
We just need to find numbers relative to the Sun.
But we
haven't included the falloff 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
falloff 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 23^{2}
= 16,399. So the
mainstream had the right numbers, it just had the math
upsidedown, 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 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. 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=v^{2}/r
equation, which is a=32r/t^{2})
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.
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