# The Gravity Theory (TGT)

1. Nov 20, 2003

### QuantumNet

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s = ct

I claim that ptot = mc for all particles, and thereby:

p = mc = mv/(1 - (v/c)2)½

and thereby v = c/2½

If the Loretz-transformation is true for x, y, z and tcoord (and also for v) inside the referencesystems:
((1 - (v0/c)2)½) = ((1 - (v1/c)2)½)

The derivata is also true for dx, dy, dz and dtcoord (and also for dv) even if v is constant:

-v0/c2(1 - (v0/c)2)½ = -v1/c2(1 - (v1/c)2)½

(This is seen from a third referencesystem, the equation is alternatively 1 = -v1/c2(1 - (v1/c)2)½ seen from a second)

so:

dv0v/c2(1 - (v/c)2)½ = dv0v/c2(1 - (v/c)2)½

and (thereby)

a0v/c2(1 - (v/c)2)½ = a1v/c2(1 - (v/c)2)½

so m0a0v/c2(1 - (v/c)2)½ = m1a1v/c2(1 - (v/c)2)½ = Ftot

This means that the force m0 attracts m1 with is the force m1 attracts m0 with. It also means that if m0 x-doubles, a1 do to. if m0 thereafter y-doubles, the force has xy doubled.

We understand that we probably should look at this from the second referencesystem as in the case of de Broglie's instead of the third:

m0a0 = m1a1v/c2(1 - (v/c)2)½ = Ftot

and get that:

m0m1[/sub]v/c2(1 - (v/c)2)½ = F since v/c2(1 - (v/c)2)½ is what causes the acceleration.

But the force decrease sphearically and thereby we get that the total force between two bodies is m1m2v/c2(1 - (v1/c)2)½/(4$$\pi\$$r2)

But v/c2(1 - (v/c)2)½ = 1/(2c).

Thereby F = m1m2/(16$$\pi\$$r2c)

But c variates from medium to medium, therefor c in this case is the average lightspeed, ca, in the mediums between the masses.

G = 1/($$8$$$$\pi\$$$$c_a$$)

I must add that the force variates with the form of the object, The force between two cylindershaped platinum weight at the distance one meter, both with the weight 1 kg is hard to calculate but seems to be more or less 1/($$8$$$$\pi\$$$$c_a$$)

This means that nasa's problem with their satellite should not exist.

And that the force of gravity variates with length and temperature.

Last edited by a moderator: Nov 25, 2003