Angular Velocity for Constant Separation Distance

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Dustinsfl
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Determine the required angular velocity $\omega$ of the relative position vector in order to maintain a constant separation distance $r_0$ for two equal masses.

What equations are needed for this?
 
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To me it seems as if you have to balance the potential energy due to gravity against the rotational kinetic energy. That is, it seems as if you're in outer space with two equal masses, and they are rotating around each other like a binary star system. If the distance between them is constant, how fast are they rotating around each other?
 
Ackbach said:
To me it seems as if you have to balance the potential energy due to gravity against the rotational kinetic energy. That is, it seems as if you're in outer space with two equal masses, and they are rotating around each other like a binary star system. If the distance between them is constant, how fast are they rotating around each other?

Fast enough to not be pulled in by the others gravity and a slow enough that they don't fly off.
 
dwsmith said:
Fast enough to not be pulled in by the others gravity and a slow enough that they don't fly off.

Exactly. So I think you'd have the force due to gravity being

$$|F_{g}|=\frac{m^{2}G}{r_{0}^{2}},$$

and the centripetal acceleration being

$$a_{c}=\frac{v_{\tan}^{2}}{r_{0}}=\frac{F_{c}}{m}.$$

Also note that $v_{\tan}=\omega \,r_{0}$.
 
Is this question talking about the position vector for the center of gravity of the two bodies?

If we set the origin to be the center of gravity, the bodies orbit it in a uniform circle with radius 1/2r_0.
I can just find the angular velocity on this circle?
 
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dwsmith said:
Is this question talking about the position vector for the center of gravity of the two bodies?

If we set the origin to be the center of gravity, the bodies orbit it in a uniform circle with radius 1/2r_0.
I can just find the angular velocity on this circle?

It shouldn't matter. You could choose your frame of reference to be at the center of one of the bodies just as well.