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Finding equilibrium distance of an orbiting particle.

  1. Nov 21, 2013 #1
    1. The problem statement, all variables and given/known data

    In the diagram below, masses m and Me are in circular orbit about Ms with the same period.

    http://min.us/i/lprtU83D9cGR [Broken]

    Derive an expression for the equilibrium position r of mass m.

    2. Relevant equations

    For a circular orbit, the eccentricity, e = 0.

    [itex]e=\sqrt{1+2mEh^{2}k^{-2}}=\frac{mrv^{2}}{GMm}-1[/itex]

    Where [itex]h=\frac{L}{m}, k=-GMm[/itex]

    3. The attempt at a solution

    So, I'm kind of assuming that I simply set one of these equations to zero and solve for r, to get something like:

    [itex]r=\sqrt{\frac{-G^{2}M_{s}^{2}m}{2Ev^{2}}}[/itex] (which will not be imaginary because in an elliptical orbit E<0)

    or

    [itex]r=\frac{GM_{s}}{v^{2}}[/itex]

    Is it really that simple though? It's a 4 mark question.

    **EDIT** I think the above is wrong. I think I should have calculated the period of the mass m in terms of the two other masses, then equated it with the period of the other mass. I think I've got it now!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 21, 2013 #2
    I would think that you need to consider the net force on mass m.
     
  4. Nov 21, 2013 #3

    haruspex

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    Sounds like you're finding Lagrange points. Are you told to assume m << Me << Ms?
     
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