Finding equilibrium distance of an orbiting particle.

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SUMMARY

The discussion focuses on deriving the equilibrium distance \( r \) of a mass \( m \) in circular orbit around a larger mass \( M_s \) with the same orbital period as another mass \( M_e \). Key equations include the eccentricity formula \( e = \sqrt{1 + 2mEh^2k^{-2}} \) and the expressions for \( r \) as \( r = \sqrt{\frac{-G^2M_s^2m}{2Ev^2}} \) and \( r = \frac{GM_s}{v^2} \). The participant realizes the need to calculate the orbital period of mass \( m \) and equate it with that of mass \( M_e \), indicating a deeper understanding of orbital mechanics and Lagrange points.

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Homework Statement



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

http://min.us/i/lprtU83D9cGR

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

Homework 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

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!
 
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I would think that you need to consider the net force on mass m.
 
Sounds like you're finding Lagrange points. Are you told to assume m << Me << Ms?
 

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