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Position as a function of time throught a varying gravitational field

  1. Jul 23, 2010 #1
    I've been told this is a differential equation, but I have no knowledge of the subject, so I'm hoping it is and that an answer can be found in this forum. The two equations x=1/2at2 and a=gm/r2, when r is replaced with (r0-x), with r0 being the initial radius from the mass from which the gravitational field originates, and x being the distance traveled since t=0, due to the acceleration of the field. Combining these equations gives

    x=1/2(gm/(r0-x)2)t2
    I don't know how to solve that equation for x, and I'm hoping this is the place to look for help. Thanks.
     
  2. jcsd
  3. Jul 24, 2010 #2

    Filip Larsen

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    Gold Member

    The equation x = 1/2at2 is the position assuming constant acceleration a, so it does not make much sense to insert a variable acceleration, like from gravity, into it.

    The solution you are looking for is a Kepler orbit [1], and if you have the degenerate case of purely radial motion then the solution is a so-called rectilinear Kepler orbit where the eccentricity is 1. For the special case where the initial speed is zero you can also find a relationship between fall time and distance in [2].

    [1] http://en.wikipedia.org/wiki/Kepler_orbit
    [2] http://en.wikipedia.org/wiki/Free_fall#Inverse-square_law_gravitational_field
     
  4. Jul 24, 2010 #3
    awesome, exactly what I was looking for. Thanks a lot.
     
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