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Two particles of equal mass move in a circle of radius \(R\) under the action of their mutual gravitational attraction. The speed of each particle is

1. \(v=\sqrt{\frac{Gm}{2R}}\)
2. \(v=\sqrt{\frac{Gm}{R}}\)
3. \(v=\sqrt{\frac{2Gm}{R}}\)
4. \(v=\sqrt{\frac{4Gm}{R}}\)

Correct answer: \(v=\sqrt{\frac{Gm}{2R}}\)

Solution

The distance between the particles is \(2R\), so the gravitational force on each is \(F=\frac{Gm^2}{(2R)^2}=\frac{Gm^2}{4R^2}\). This provides centripetal force \(\frac{mv^2}{R}\), giving \(v^2=\frac{Gm}{4R}\), so \(v=\sqrt{\frac{Gm}{4R}}\).

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