# The GF (The gravitational force)

1. Nov 28, 2003

### QuantumNet

|---------------------|
s = ct

The amount of motion p of the smallest particles moving in circles might or might not be mc/(2(pi)) = mv
and thereby v = c/(2(pi)) (the average value anyway, the acceleration is constant.)

since:

v0/(1 - (v/c)2)½
= v1/(1 - (v/c)2)½

we know that:

a0/(1 - (v/c)2)½ + v0/c2/(1 - (v/c)2)½
= a1/(1 - (v/c)2)½ + v1/c2/(1 - (v/c)2)½ = the acceleration.

and thereby:

a0/(1 - (v/c)2)½ + c/(2(pi))0/c2/(1 - (c/(2(pi))/c)2)½
= a1/(1 - (c/(2(pi))/c)2)½ + v1/c2/(1 - (c/(2(pi))/c)2)½ = the acceleration.

the second term (c/(2(pi)) should be the gravityconstant.

But the force of gravity decreases sphearically. So the gravityconstant we use is actually 1/(8(pi)2c).

At large range the speed of light can be aproximated to 299792458 m/s.

If anyone can confirm this, please do...

(It's easy to prove that it stands if you prove that the speed is
c/(2(pi))).

Last edited by a moderator: Dec 1, 2003
2. Dec 4, 2003

### Sariaht

e = electron charge, c = lightspeed, h = plancks constant, pi =b

Delta = Y, y = 1/(1-v2/c2)½

A particle has Yv = e22(pi)c2/(107*h)

(also Yv >= h/(4(pi))mYx))

yvaverage for a particle inside the atom is ve + vp*mp/me / (1 + mp/me) = 2ve/(1 + mp/me)= 2e22(pi)c2/(107*h)/(1 + mp/me) = 2381.612376 m/s (taken from bohr's atommodel)

we know that Yv0*y0 = Yv1*y1

We derivate both sides and get:

Yay0 + YaYv/c2y0 = Yay1 + y12YaYv/c2

but Ya0*y0 is said to be Ya1*y1, so we guess that the second term is an attractionconstant.

this attractionconstant divided with 4(pi) might be G. If a is v then c for platinum-iridium is around 60.

althought;

( (ec/r)^2/10^7 = a , r taken from bohrs atommodel. )

G = n2c643b5me2e8/(1028(1 + mp/me)h3)

I have not controlled this yet.

Last edited: Dec 10, 2003