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- Thread starter dwsmith
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- Jan 26, 2012

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

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- Jan 26, 2012

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Exactly. So I think you'd have the force due to gravity beingFast enough to not be pulled in by the others gravity and a slow enough that they don't fly off.

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

and the centripetal acceleration being

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

Also note that $v_{\tan}=\omega \,r_{0}$.

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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?

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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- Jan 26, 2012

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