Question

Is Gravity's weakness due to 2D dimensional reduction?

In dimensional balancing, the exponential dimensional profile on the right-hand side of the equation is always equal to the exponential dimensional profile on the left-hand side of the equation. Force and charge share an exponential dimensional profile of D² while mass and energy have an exponential dimensional profile of D³. Some examples of dimensional balancing are;

\[{Force}=m\times \frac{l}{t^2} = D_m^3 \times {D_l^1}\times {D_t^{-2}}=D_{Force}^2\]

\[Charge=m^{1/2}\times\;l^{3/2}\times t^{-1}=D_m^{3/2}\times D_l^{3/2}\times D_t^{-1}=D_{Charge}^2\]

\[{Energy}=F\times l= D_F^2 \times {D_l^1}=D_{E}^3\]

Dimensionally, a D² charge can transform directly into D² electrostatic force. In Coulomb's electrostatic force equation, for two charges, the Coulomb constant has a dimensional profile of D⁰.

\[{Coulomb\;Constant}\; k_e=\frac{m\;l^3}{C^2\;t^2} =\frac{D^3_m\;D^3_l}{D^4_C\;D^2_t}=D_{k_e}^{0}\]

 

For a pair of electrons the electrostatic force is;

\[F_{{e\;electrostatic}}=k_e\;\frac{q_e^2}{r^2}=D^{0}_{k_e}\;\frac{D^4_{q_e}}{D^2_r}=D^2_{F_e\;electrostatic}\]

Dimensionally, a D³ mass can be transform into D² gravitational force. In Newton's law of universal gravitation, for two masses, the Newton gravitational constant has a dimensional profile of D^-2.

\[{Newtons\;Gravitational\;Constant}\; G=\frac{l^3}{m}{\left(\frac{1}{t^2} \right)}=\frac{D^3_l}{D^3_m}\;{\left(\frac{1}{\color{red}{D^2_t}}\right)}=D_G^{-2}\]

For a pair of electron the gravitational force

A closer look at the Newton gravitational constant, reveals a frequency squared dimensional reduction of D^-2 is required due to the two electron mass. 

We  can hypothesize that the relative weakness of the gravitational to the electrostatic force is due to the difference in the above exponential dimensional profiles for charge D²  and mass D³?