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Gravitational Constant

  1. May 23, 2010 #1
    I'm just trying to find how to derive the gravitational constant G. In other words what is G in terms of pi, r, [tex]\epsilon[/tex], etc. And how does it relate to k, the permittivity of free space. Like in: f679cf3e13506e73340d176eb06c80b2.png Any help would be appreciated. Thanks.
  2. jcsd
  3. May 23, 2010 #2

    Vanadium 50

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    It doesn't. G is an independent, separate constant.
  4. May 23, 2010 #3
    ok, but in the equation F=G*(m1*m2)/(r^2) I know the r^2 isn't just there for no reason. Its a coulombic force so the r^2 comes from 4*pi*(r^2). There is a geometric meaning. You could easily write G as x/(4*pi*epsilon0) which is the same as x*k (where k= 1/4*pi*e0) or x*(Boltzmann constant) in this case G is related to e0 (permittivity of free space) like this: e0 = x/(4*pi*G). This is just algebraic manipulation, but the geometric meaning is what I'm after. The units of G are N*(m/kg)^2 so as long as the units match in the end you could find other fundamental constants that G is factor of.
  5. May 23, 2010 #4


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    What? You're not making much sense. Sure, the [itex]\frac{1}{r^2}[/itex] factor in Newton's law has a geometrical interpretation, but G is just a proportionality constant. It has to be measured, there's no way to calculate it from anything else.

    And sure, you could write G = x/4πε0 for some value of x, but then x would just be some random value that doesn't have any particular physical meaning.
  6. May 23, 2010 #5
    Why wouldn't x have any physical meaning?
  7. May 23, 2010 #6


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    The [itex] \frac 1 {r^2} [/itex] in both the gravitational constant and Coulomb's law are related by only by the fact that both systems have spherical symmetry. The fact that both the gravitational force and the Coulomb force are spherically symmetric may have a deeper significance but to the best of my knowledge mainstream physics does not recognize it.

    The similarity in the two forms is one of the factors that drives the search for a GUT.
  8. May 23, 2010 #7

    Just measure all values and solve for G.
  9. May 23, 2010 #8
    gth759k, I may see the source of the confusion here. Correct me if I'm wrong, but you're wondering why we can express the Coulomb force constant as,

    [tex]k_e = \dfrac{1}{4\pi\epsilon_0}[/tex]

    Whereas G has no equivalent expression. The reason is part history and part convenience. In electrodynamics, we deal with the phenomena of polarization and magnetization, which change the electric and magnetic fields inside materials. Mathematically this can be analyzed by considering adjusted quantities called the electric permittivity [itex]\epsilon[/itex] and magnetic permeability [itex]\mu[/itex] So we can express the electric and magnetic force constants in terms of the vacuum values of these quantities. In the case of gravity, there's only an attractive force, meaning that we don't have gravitational dipoles. And so we don't need to modify the force constant inside a medium. Gravity inside matter behaves just like gravity outside of matter. Technically you could make up some variable and define it in terms of G, but that variable would be physically meaningless.

    Now as to the part about how G relates to r in Newton's Gravity Law, I'm not sure what you're referring to. Maybe you could be more specific.
  10. May 23, 2010 #9
    Hi gth759k;
    I think you are interested in the vacuum values...

    In linearized General Relativity, where the non-linear terms are ignored (in the to weak field / low velocity limit) the gravitational permittivity of free space, epsilon(g), is sometimes used:
    ............ e(g) = 1/(4pi)(G)

    and also the "gravitoMagnetic" permeability, upsilon(g),:....
    ..............u(g) = 4(pi)(G)/c^2

    .....which is analogous to (not the same as) the electromagnetic permittivity and permeability in vacuum.

    So that the speed of gravitational waves becomes the analog ...
    ....... c = 1/ sq.rt.[e(g) x u(g)]

    ... Is that what you are getting at?

    Last edited: May 23, 2010
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