### Kepler's third law

September 21, 2020
Charles deAnne

To prove Kepler's third law, set gravitational force equal to centrifugal force:

$\frac{GM}{R^{2}}m=m\frac{V^{2}}{R}$

Divide both sides by mR:

$\frac{GM}{R^{3}}=\frac{V^{2}}{R^{2}}$

Now the geometry:
Replace V with the circumference of a circular orbit, divided by the period:

$\frac{GM}{R^{3}}=\frac{(2\pi R/T)^{2}}{R^{2}}$

$\therefore \frac{GM}{(2\pi)^{2}}=\frac{R^{3} }{T^{2}}$

This is the explicit expression of Kepler's third law.

### Physics creates Geometry

At its root physics uses integer arithmetic to count discreet interactions between quantum particles. Time has no meaning other than a count of interactions compared to the count of interactions from a system we use as a clock. Distance has no meaning other than the integer clock count accumulated between the emission of light by one quantum particle and the absorption of light by another quantum particle.

This limits the geometric precision of all natural systems. For example, integer arithmetic can only estimate the value of 2PI:

$2\pi \approx \frac{861 }{137}\equiv 2\pi _{q}$

This is nature's best estimate of 2PI, and implies the existence of a quantum angle:

$\alpha =\frac{2\pi _{q}}{861}=\frac{1}{137}\;\;radians$

Alpha is known as the fine structure constant and defines the relationship between many fundamental lengths:

#### Classical electron radius and the Compton wavelength

$r_{e}=\alpha \frac{\lambda _{C}}{2\pi _{q}}=\frac{\lambda _{C}}{861}$

#### Compton wavelength and the Bohr radius

$\frac{\lambda _{C}}{2\pi _{q}}=\alpha \: r_{Bohr}$

$\therefore \: r_{e}=\alpha^{2} \: r_{Bohr}=\frac{r_{Bohr}}{137^{2}}$

#### Quantum angle of gravity

While electric and magnetic forces arise from interactions between electrons and protons,  gravitational and inertial forces arise from interactions between neutrons and protons. Both neutrons and protons are composite particles, each containing three quarks.

Emerging evidence suggests quarks might be the gravitational quantum particles. It appears that each nucleon always has three simultaneous gravitational interactions, causing the quantum angle of gravity to be three times the fine structure constant.

$\alpha _{g}=3\alpha =\frac{2\pi _{q}}{287}=\frac{3}{137}\;\;radians$