## The gravitational force can be derived from the electric force between quantum charges.

Electric force between two groups of charges is defined without reference to charge, or invoking the nebulous property of space known as permittivity.
$F=\frac{r_{e}c^{2}}{R^{2}}m_{e}ZN$

The above equation for electric force is a prototype for gravitational force. General Relativity predicts mass dilation, time dilation, and length contraction for regions of space with large negative gravitational potential. After matching empirical measurements to the prototype, it appears that the ratio, electron mass/nucleon mass, is equal to the length contraction ratio, as well as the reciprocal of the time dilation ratio for our region of the universe. Also, the gravitational equation provides clear evidence that neutrons and protons are composite particles, each containing three quantum gravitational particles.

Applying length contraction to the classical electron radius, and time dilation to the frequency, c/R:

$F=\frac{r_{e}c^{2}}{R^{2}}\left ( \frac{m_{e}}{m_{\overline{n+p}}} \right )^{3}m_{e}ZN=3\frac{r_{e}c^{2}}{R^{2}}\left ( \frac{m_{e}}{m_{\overline{n+p}}} \right )^{4}M_{Z}N$

(Z is the total 'Z' mass divided by 1/3 the average nucleon mass.)

Dividing by N (the number of quantum gravitational masses in the 'N' mass):

$\frac{F}{N}=\frac{F}{N\left ( \frac{1}{3}m_{\overline{n+p}} \right )}=\frac{F}{M_{N}}\equiv \ddot{R}_{N}=3\frac{r_{e}c^{2}}{R^{2}}\left ( \frac{m_{e}}{m_{\overline{n+p}}} \right )^{4}M_{Z}$

The rhs of the above equation is the dimensional equivalent of a constant force for any given 'Z' mass. This, however, belies the fact that it actually represents the force acting on each and every 1/3 nucleon in the 'N' mass. While each increment of N increases the total tension between the two masses, it also increases the mass of the 'N' group by 1/3 nucleon, keeping the force/mass ratio constant. Force/mass ratio is, by definition, acceleration; i.e., acceleration if the mass is not restrained by an equal opposing force.

$\therefore F=3\frac{r_{e}c^{2}}{R^{2}}\left ( \frac{m_{e}}{m_{\overline{n+p}}} \right )^{4}M_{Z}M_{N}$

Conclusion: Gravitational force has dimensions of mass-squared times acceleration.

Parsing out the portion of the above equation referred to as the gravitational constant:

$G=3r_{e}c^{2}\left ( \frac{m_{e}}{m_{\overline{n+p}}}\right )^{4}\newline \newline\frac{m_{e}}{m_{\overline{n+p}}}=\frac{1}{1836.9}\newline\newline r_{e}=2.818\times 10^{-15}m\newline\newline c=2.998\times 10^{8}m/s\newline\newline \therefore G=6.674\times 10^{-11}m^{3}/(s^{2}kg)$

The mass ratio used in this evaluation represents a distribution of 72% protons and 28% neutrons.

The kg in the dimensional attribute comes from the fact that G x MZ / R2 is the gravitational acceleration at MN (and vice versa). Kilograms must be put in the dimensional denominator to make the gravitational constant's dimensions agree with its empirical dimensions. While local acceleration arises from the mere existence of a remote mass, the total tensile force is the product of local mass, remote mass, and G/R2. Despite the mysterious dimensional origin of total force, the relationship between total force, local acceleration, and local mass is consistent with our concept of F=ma.