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Gravitational Constant (Big G I think)

  1. Jul 17, 2013 #1
    G = (6.674215 ± 0.000092) x 10-11 m3/kg/s2 or there abouts is the constant that we multiply to the mass of m1 to m2 divided by the distance squared. This gives us the attractive force between bodies. In the equation F = G(m1m2/r2) it is obvious that the attractive force depends on the mass of each object and exponentially more the distance between them (inversely). This "Big G" though.. what is it? I get that it is the gravitational constant but where does it come from and what is its significance. Why do we need this very tiny number to accurately determine the attractive force of massive bodies? Also how does it represent the strength of gravity as a force? Is it because it is an independent variable?
     
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  3. Jul 17, 2013 #2

    pervect

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    We have a FAQ on why "c" has the value it does https://www.physicsforums.com/showthread.php?t=511385 [Broken], but not one for G.

    The rationale is pretty similar though.

    I'll mark-up the answer there for G. You might try some of the references there as well.

    Because [STRIKE]xxcxx[/STRIKE] G has units, its value is what it is only because of our choice of units, and there is no meaningful way to test whether it changes. These questions [STRIKE]are[/STRIKE] would be more meaningful [STRIKE]when[/STRIKE] if posed in terms of the unitless [STRIKE]fine structure constant.[/STRIKE] gravitational coupling constant http://en.wikipedia.org/wiki/Gravitational_coupling_constant.

    Except that unlike the case for the fine structure constant, I don't think we have any direct experimental data on the Gravitational coupling constant.

    Basically the numerical value of c, or G, has every bit as much physical significance as saying there are 2.54 centimeters per inch.

    I would particularly recommend http://math.ucr.edu/home/baez/constants.html in the references section of the FAQ.

    ...

    Thus, by a proper choice of units, one can (and often does) make G equal to 1 by using geometric units. These are similar to the plack units Baez discusses, except that we don't bother making planck's constant equal to 1.


     
    Last edited by a moderator: May 6, 2017
  4. Jul 17, 2013 #3
    I understand the number depends on the units you measure it in, that wasn't really what I meant with my questions though. I'll try to think of a better way to restate them I guess.
     
    Last edited by a moderator: May 6, 2017
  5. Jul 17, 2013 #4
    It is just dawning on me that multiplying by a really tiny number really reduces the overall value. In some respects I guess that makes it quite meaningful. It is that constant which makes it such a weak force.
     
  6. Jul 17, 2013 #5

    Nugatory

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    Or you could say that it is the weakness of the force that makes the constant so small. Physics is about using math to describe the world around us, so the world tells us what the math must be and not the other way around.
     
  7. Jul 17, 2013 #6
    Good point.
     
  8. Jul 17, 2013 #7

    pervect

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    Really good questions / comparisons probably won't involve constants with units. For instance you could take the ratio of the fine structure constant to the gravitational coupling constant. That would be more or less comparing the electrical repulsion between two electrons to their gravitational attraction.
     
  9. Jul 17, 2013 #8
    Having something to compare it too is probably a good start. I wasn't aware gravity was even noticeable at the molecular level.
     
  10. Jul 18, 2013 #9

    WannabeNewton

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    It's definitely hard to notice; for two protons the ratio will yield about 36 orders of magnitude in favor of the electrostatic force (i.e. ##F_{E}/F_{G}\approx 10^{36}## for the aforementioned system).
     
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