Kepler Ratio

Kepler Ratio Explained by Volume Theory

The ratio was provided as a constant of proportionality, W.

d^3 / p^2 = W

This essay derives the formula for Kepler's ratio.

d^3 / p^2 = NV / (tau second * 8 pi^3)

where d is distance from Sun to planet, p is year for planet period,
N is number of baryons in Sun, V is one baryon volume,
tau is 5.13ns, S is speed of space.

The period of Earth's orbit is p ~= 31 million seconds.

Gravity Volume Theory provides S2, the Speed of Space

S2 = NV / (2 pi tau d^2)

S2 = 6.074*10^-3 meters per second

S2 is about 30mm per second as the Earth falls toward the Sun for a
1 second test. A circular orbit during 1 second must have an arc
that falls a certain distance determined by the geometry of a circle.
The angle change made during part of an orbit is theta, when t equals
1 second

theta = 2 pi t/p

The distance traveled in the orbit in one second is: x. t equals 1 second.
The sine of theta is theta for small values of theta. Jan. 2, 2017 Figure added:

x = d sin(theta) ~= d * 2 pi * 1 second/p

The z distance fallen towards the Sun is

z = x sin(theta) ~=  (d * 2 pi * t/p)* 2 pi t/p = 4 pi^2 d t^2 / p^2
z3 = 5.874*10^-3

S3 = 5.874*10^-3 meters per second

S3 =  4 pi^2 d t^2 / (1 second * p^2) = speed falling in 1 second

Expecting the S2 speed of space to equal S3, but off by 2% in
this draft, the speed the circle turns for an orbit,

S2 = S3

N(V/tau)/(2 pi d^2) = 4 pi^2 d t^2 / (1 second * p^2)

Solve for p^2

p^2 = tau 2 pi d^2 4 pi^2 d t^2 / (VN second)


d^3 / p^2 = NV / (tau second *8 pi^3)


Inverse Square Time of Kepler

A Law from Kepler was

d^3 / p^2 = W

where p is the period of time for an orbit.
People sometimes ask why is the denominator
a time squared. What does a square second have
to do with gravity?

This has been answered today in a 1 second
calculation. For the Earth

p = 31,000,000 seconds

d = 148,000,000 km from Earth to Sun

As the planet orbits the Sun in a circle,
in one second its distance traveled is x,
and it turns an angle of theta

sin(theta) = 2 pi t / p

x = d sin(theta)

As the planet orbits for one second , it falls
a distance z

z = x sin(theta)

z = d sin(theta) sin(theta)

z = d (2 pi t/p) (2 pi t/p)

z = d pi^2 t^2 / p^2

That is where the p^2 comes from, time squared. It occurs because an orbiting planet
travels as far as the period allows and it falls as far as the period allows.

December 16, 2016

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