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Clarification
of the equation a = v^{2}/r
by Miles Mathis
Abstract: I will show the arc and
tangent are equal only in one specific place on the circle, and
that place is not at zero or the limit. It is at 1/8th of the
circle. This being so, we must rework all of Newton's orbital
math. Once this is done, I show that we can easily calculate a
time for the centripetal acceleration. Yes, the acceleration is
not instantaneous, and I can now prove it, by showing you the
number for the time.
I was able to
continue my analysis of Newton's orbital math by discovering a
disclarity between two of my own papers. In
my first paper on a = v^{2}/r,
I show many problems with the historical proofs, and conclude,
among other things, that the orbital velocity cannot be equal to
the tangential velocity. But in a
more recent paper on π, I show that the tangent is actually
equal to the arc. In fact, they are equal not at the limit but at
a finite real length. How can these two findings be
commensurate?
Well, if we take the orbital velocity as v =
2πr/t, then the tangential velocity cannot equal the orbital
velocity. The tangential velocity cannot be expressed that way.
Not only is it in the wrong form, it is the wrong number.
However, as I made clear in that first paper, the orbital
velocity cannot be expressed that way either. So, NEITHER
velocity equals 2πr/t. That expression is simply a heuristic
ratio that we like, since it is easy to measure from visual data.
However, what I did not make clear in that first paper is
that once we find new correct expressions for both the orbital
velocity and the tangential velocity, they MAY equal eachother
under certain very specific conditions. The tangent DOES equal
the arc, provided the tangent is the same length as the radius.
In my "Extinction of π" paper, I let the tangent equal
the radius in length, and show that the tangent equals the arc.
This appears to confirm Newton and the standard analysis,
but it doesn't. The tangent and arc aren't equal at the limit,
they are equal only
when the tangent equals the radius, in which case the radius,
tangent and arc are all equal. Even in this situation, none
equals the chord, so my analysis of Newton's
lemmae was not wrong. At the limit, the arc approaches the
chord, but at the limit, the arc does not equal the tangent. So
at the limit, the orbital velocity and tangential velocity are
not equal. Newton's proof fails. The current proof fails. The
equality of the tangent and the arc can only be proved by making
the tangent equal to the radius. Therefore, the orbital velocity
is equal to the tangential velocity only when both velocities are
"equal" to the radius.
You will say, How can the
tangent equal the radius, when one is a velocity and one is a
length? The answer: the numbers must be equal. If the velocity is
30, the radius must be 30. But if we match meters to meters, can
we just use one second to create the equality? Good question.
Another way of asking it is to ask if the number we are given for
any radius can be used as a straight velocity in this way. If we
are told that a circle has a radius of 5m, does that mean that
our velocity along it would be 5m/s? How do the radius and the
tangential velocity really relate to one another? I have shown in
my π paper that they CAN be related kinematically, but are they
NECESSARILY related?
In the case of an orbit, they must
be related, since the equation tells us they are. We cannot have
an arbitrary velocity at a given radius, we must have a specific
velocity. A velocity that is too great will cause escape, and a
velocity that is too small will cause a crash. So the answers are
yes and yes. The velocity is necessarily related to the radius,
and if we get our equations and dimensions right, the velocity
should equal the radius. All we have to do is get the time right.
We don't just use 1 second to make the equality, however.
Using my new equations, we find that the velocity in orbit is
8r/t. Therefore, to make the velocity equal the radius, we just
let the time period equal 1/8 of the orbit. As I show in the
diagrams in the π paper, the radius is 1/8 of the circumference;
therefore, in 1/8 of the orbital period, the velocity must equal
the radius, numerically.
A critic will say, "That
doesn't work in real life, as we know from the Moon. Just look at
the numbers from the Moon. The orbital speed of the Moon is
1.022km/s, and the radius is 384,400km." Well, it doesn't
work because the orbital speed of the Moon is wrong. It is
developed from a faulty equation. Let us make all the
corrections. If the radius is 384,400km, then the distance
travelled by the Moon in one orbit must be 8 times that, or about
3 million km. The orbital period is 2,360,534 seconds. That is a
velocity of about 1.3km/s.
My critic will now say, "Even
if you are right, the velocity still doesn't equal the radius,
numerically or otherwise. The number value of the velocity is
1.3, and the number value of the radius is 384,400." No, I
meant to point out a number equality between the radius and the
distance travelled by the velocity in 1/8th of an orbit. You see,
they ARE the same. The distance travelled at 1.3km/s in 1/8 of an
orbit is 384,400km.
In fact, an onboard satellite
speedometer, calibrated to work for straightline distances, will
not work in orbit, and engineers know this. That is precisely
what caused the Explorer
Anomalies. The thrusts were set for velocities as calibrated
here on Earth, in straightline motions, then transformed into
orbital motions by the common equations. Since the common
equations were wrong, the orbits established from the thrusts
were wrong. The satellites flew too high and were thought to be
lost.
Not only is the velocity wrong, the acceleration of
gravity is wrong. Using the current equations and the value of π,
the acceleration of gravity is calculated from an equation that
reduces to a=39.5r/t^{2}.
But the correct equation is 32r/t^{2}.
Therefore, the acceleration at the distance of the Moon is not
.002725. It is .002208.
Some will look at those numbers
and say, "That means that using the given numbers for the
Moon, in these equations g
would not be 9.8, but 7.947!" Well, if you assume that g
balances the Moon's velocity by itself, yes. But that isn't the
way it works. The Moon is moving in a threebody problem, at its
simplest, not a twobody problem. If you calculate that number
7.947, you have left the Sun out of it. To see what I mean, you
need to visit my paper where I run the
full unified field math on the Moon, showing how the distance
and the velocity are caused by Earth and
Sun.
As you see, the current equations are hiding many
secrets. As another example of a secret being hidden, we can
continue to study the arc and tangent. Let us return to the
diagram from my π paper:
There
I proved that AD + DC = arcAC. This means that the tangent equals
the arc. This works only when the tangent is equal to the radius,
as I said above. The angle at O must be 45^{o},
so that DC = DB. If the angle is not 45, then the tangent cannot
equal the arc, because AD + DC is not equal to AB.
This
is important because if we assign the tangent to the tangential
velocity and the arc to the orbital velocity, as Newton did, we
find they are equal not at the limit, but only when the tangent
equals the radius. In fact, as
I have shown, the tangent and the arc are NOT equal at the
limit. At the limit, the tangent remains longer than the arc. And
this means that the tangential velocity and orbital velocity are
equal only when the length of the tangent is equal to the radius,
or when the time is equal to 1/8th of the orbital period. An
orbital velocity found by any other method will get the wrong
answer. This is why 2πr/t is wrong: 2πr/8
is not equal to r.
With all this
under our belts, we are now in a position to see that we may
assign the acceleration a to the line segment BC. Furthermore, if
a = v^{2}/2r, and
AB = r, then r = v, and a = r/2. Consulting the diagram above
again, that is also 2BC = CO. What this means is that we have a
new way to find a centripetal acceleration, currently called
gravity. The equation a = r/2 gives us a distance of acceleration
over 1/8th of the orbit, so the total distance of acceleration
over the entire orbit is 4r. Or, we can find the acceleration
over any subinterval. Say we want to find the acceleration of the
Moon over 1 second. We are given that the period for the Moon is
2,360,534s. Since the Moon is orbiting at 384,400,000m, the total
distance of acceleration over the orbit is 4 times that, or 1.538
x 10^{9}.
Dividing gives us a = 1.538 x 10^{9}/2,360,534^{2}
= .000276m/s^{2}
over 1 second.
For more fun, we can even show that the
centripetal acceleration is not found at an instant. If the
Moon's acceleration over 1 second is .000276, then an
acceleration of .002208 cannot happen over an instant. We can
even find the time, with simple math. With the same math, we find
that a = .002208 when t = 8s. That is not an instant or an
infinitesimal, since it is a calculable number. You cannot have a
real acceleration over an instant: we should have known simply
from logic that the centripetal acceleration we have always had
could not be an acceleration over an instant or infinitesimal. I
have just calculated the real time of orbit during the
"instantaneous" acceleration, so I have proved that
Newton did not go to a limit or approach zero. As I have said in
my paper on the
derivative, the calculus does not work by going to zero or a
limit, it works by going to a subinterval, and I have just shown
you the subinterval in a specific problem.
You will say,
"But if we can find an acceleration at that small time, we
should be able to find an acceleration closer to zero, over an
even smaller time." Yes, we can, but that acceleration would
not be the centripetal acceleration. In seeking a centripetal
acceleration, we are not seeking an acceleration at or near zero
time or length. We are seeking the derivative of the orbital
velocity, and the derivative of any motion is found by going to a
subinterval, as I show in great detail in my calculus paper. We
have just gone to that subinterval mathematically, since what I
am doing here is calculus without the calculus. I have found the
subinterval underneath the orbital motion where that motion
becomes uncurved, which is defined as the derivative. And that
gives us the acceleration we were seeking. Acceleration is not
defined as the change in velocity at or near zero, it is defined
as the change in the velocity, period. I showed in my calculus
paper that you don't have to go toward zero to find any
derivative, and I have shown here in a specific problem that the
subchange is always happening over a real interval.
Now
you may ask, "But that number, 8s, where is that coming
from? You just calculated it, but I still don't get it."
Normally, the derivative is found at a subinterval of 1, since
that is how it is (or should be) defined. The derivative is not
found by going to zero or to a limit, it is found by going to a
subchange or subinterval where we have a constant differential
of 1. So normally, we would be looking for a derivative at 1
second. But in the case of the circle, this logic changes
slightly. As you can see from the diagram above, I did my
"calculus", or my calculations, over 1/8th of the
circle. My solution is therefore over 1/8th of the circle.
Therefore, to find a solution for the entire orbit, I have to
multiply everything by 8. Even the time of the subchange, or what
we call the derivative, must be multiplied by 8. That is why the
time here is 8 seconds rather than 1 second.
You can see
that it took a complete reworking of Newton's postulates to crack
open the orbital math, but once I did it, everything began to
make sense. We have always been taught that the centripetal
acceleration is the acceleration at an instant, but that is
illogical. Acceleration, like velocity, is motion, and you cannot
have motion at a instant, by the definition of motion.
Acceleration is also defined as a change in velocity, but you
cannot have a change at an instant. Change requires an interval
of change. Motion can only take place over a differential. This
being so, we should have been able to find that differential. If
the acceleration is not taking place at an instant, it must be
taking place over some real time, and we should have been able to
find that real time. Problem is, Newton couldn't solve this one,
and no one else since then could either. They were looking in the
wrong place. They were looking near zero, and the answer was
hiding at 1/8th of the circle. The answer is found only when the
tangent equals the radius. Because physicists could not solve
this, they decided to hide it. Once again, they hid it in the
instant. They buried it in the zero and covered it over with
centuries of slippery math and slippery explanations. As you now
see, the solution is simple.
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