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

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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