:<math>dE/dx=2mcdc/dx</math> ...... (3)
The above equation implies that (if it is going to be the real ''"gravitational force"'' <math>F=ma</math>)then
:<math>dc/dx=-a/c/2</math> ...... (4),
:<math>F=-dE/dx=2mcdc/dx=-2mca/c/2=ma</math> ...... (5).
And so the ''"gravitational force"'' comes out as <math>F=ma</math> and , more accurately than the Newtonian force given by the Newtonian equation
:<math>F_N=GMm/r^2</math> ...... (6),