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Is body to body gravity scale independent?

  1. Jun 25, 2011 #1
    I'm working on a gravity simulator and I can't help but feel that I've done something wrong. Extremely large bodies take a longer time to fall into each other than smaller ones. Is this right? Like for example, if you had to bodies of solar mass at 1 AU apart, would it take a longer, shorter or the same amount of time for two other bodies of half solar mass at half an AU to fall into each other?
  2. jcsd
  3. Jun 25, 2011 #2


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    In the second case (two 0.5-solar mass bodies), the force is 1/4 as much, the acceleration is 1/2 as much, and the time is greater by a factor of [itex]\sqrt{2}[/itex].
  4. Jun 25, 2011 #3
    In both cases the force is the same, [tex]F = \frac{G M_1M_2}{r^2} = \frac{G (M_1/2) (M_2/2)}{(r/2)^2}[/tex]

    The acceleration is different. In the first case:

    [tex]a = \frac{GM_1}{r^2}[/tex]

    In the second case:

    [tex]a = \frac{G(M_1/2)}{(r/2)^2} = \frac{2GM_1}{r^2}[/tex]

    so the acceleration is twice that of the first case.

    The distance (x) to travel is also half that of the first case, so:

    In the first case [tex]t= \sqrt{\frac{2x}{a}} [/tex]

    for the time to fall a short distance x relative a large r.

    In the second case [tex]t = \sqrt{\frac{2(x/2)}{2a}} = \sqrt{\frac{x}{2a}} [/tex]

    which is less time to fall in the second case.

    You would of course have to integrate the equations over the total distance for them to come together and the analysis above is just for a short segment of the fall, but overall the smaller closer bodies fall together faster.

    If the initial distance apart (r) was the same for the smaller and larger bodies, then the falling time would for the lighter bodies would be longer because:

    For the larger bodies [tex] t = r\sqrt{\frac{8x}{GM_2}}[/tex]

    and for the smaller bodies [tex] t = r\sqrt{\frac{16x}{GM_2}}[/tex]
    Last edited: Jun 26, 2011
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