Finding equilibrium distance of an orbiting particle.

1. Nov 21, 2013

phosgene

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.

$e=\sqrt{1+2mEh^{2}k^{-2}}=\frac{mrv^{2}}{GMm}-1$

Where $h=\frac{L}{m}, k=-GMm$

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:

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

or

$r=\frac{GM_{s}}{v^{2}}$

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. Nov 21, 2013

voko

I would think that you need to consider the net force on mass m.

3. Nov 21, 2013

haruspex

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