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Linear motion in central field

  1. May 20, 2013 #1
    1. The problem statement, all variables and given/known data

    I have two masses of finite width, m_1 and m_2. The force is newton's gravity, so U = k/r. I want to work out their relevant equations of motions r_1(t) and r_2(t) as they start off from rest and collide. I don't want to consider any rotational motion.

    2. Relevant equations

    [itex]U = \frac{k}{r}[/itex]
    [itex]r = r_{1}(t) - r_{2}(t)[/itex]

    0.5μ[itex]\dot{r}^2 = E_{tot} - \frac{k}{r}[/itex]

    dt = [itex]\frac{μ}{2}[/itex]∫[itex]\frac{dr}{\sqrt{E_{tot} - \frac{k}{r}}}[/itex]

    3. The attempt at a solution

    I try to integrate the above equation it gives me something very complicated (I end up integrating cosec^3 after making the substitution [itex] \frac{1}{r} = \sin^{2}(\theta) [/itex]) which gives me t = t(r), but I can't invert this to give me r = r(t).

    Am I doing something wrong? Is there another way of doing it that doesn't involve lagrangian or hamiltonian dynamics (which I haven't studied).

    Another way would be to solve the the 2nd order differential equation directly:

    [itex]\frac{dr^{2}}{dt^{2}} = \frac{k}{r^{2}}[/itex]

    but I can't seem to do this (I fee I'm missing something very simple here). I know that if I let [itex] r = At^{\frac{2}{3}}[/itex] this is a solution but it doesn't have enough constants.

    Any pointers?


  2. jcsd
  3. May 20, 2013 #2
  4. May 20, 2013 #3


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    I think what you are doing is correct. Did you allow for the fact that both k and Etot are negative numbers? Anyway, as you say, you can get an expression for t(r). I agree that the result looks too complicated to invert for r(t).
  5. May 20, 2013 #4
  6. May 20, 2013 #5
    So strange that for such seemingly simple problem there is no closed form result for r(t). t(r) is actually all I wanted so that's ok!
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