The GF (The gravitational force)

[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))).

 

[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))

yv<sub>average 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.</sub>