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The GF (The gravitational force)

  1. Nov 28, 2003 #1
    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.)


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


    ( (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
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