Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook