### Electric Potential Replaces Gravity

Posted:

**Tue Aug 19, 2014 9:22 pm**Anyone who has been following my recent posts should have given up all ambitions of understanding gravity....

Right, forget gravity, instead it appears that bodies move according to their surface potential, or more correctly stated, according the their difference in electric surface potential.

The Ground Potential law for relative velocity is:

\[\Delta v = c(\frac{\Delta V}{\Phi})\]

Where \(\Phi\) is the surface potential of the proton and \(\Delta V\) is the difference in surface potential between two bodies. The classical law for velocity under free fall is:

\[v = \sqrt{2gh}\]

Where g is accelleration due to gravity and h is the height in meters.

By combining these two equations we can find the potential difference at one meter above ground.

\[\sqrt{2gh} = c (\frac{\Delta V}{\Phi})\]

\[4.429 mps = c (\frac{\Delta V}{\Phi})\]

If we now solve this equation for height = 1 meter, we get \(\Delta V = 13.86\) volts, ie. the potential between ground and a body 1 meter above ground is about 13.9 volts.

Benjamin Franklin was way ahead of his time...

Steven

Right, forget gravity, instead it appears that bodies move according to their surface potential, or more correctly stated, according the their difference in electric surface potential.

*[ignore this:*

This electric potential decreases at the rate of 1/r^2 with height above ground. From this we can now work out the potential gradient above ground, with a two step process;]This electric potential decreases at the rate of 1/r^2 with height above ground. From this we can now work out the potential gradient above ground, with a two step process;]

The Ground Potential law for relative velocity is:

\[\Delta v = c(\frac{\Delta V}{\Phi})\]

Where \(\Phi\) is the surface potential of the proton and \(\Delta V\) is the difference in surface potential between two bodies. The classical law for velocity under free fall is:

\[v = \sqrt{2gh}\]

Where g is accelleration due to gravity and h is the height in meters.

By combining these two equations we can find the potential difference at one meter above ground.

\[\sqrt{2gh} = c (\frac{\Delta V}{\Phi})\]

\[4.429 mps = c (\frac{\Delta V}{\Phi})\]

If we now solve this equation for height = 1 meter, we get \(\Delta V = 13.86\) volts, ie. the potential between ground and a body 1 meter above ground is about 13.9 volts.

Benjamin Franklin was way ahead of his time...

Steven