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Two particles of equal mass 'm' go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is

1. √(Gm/4R)
2. √(Gm/3R)
3. √(Gm/2R)
4. √(Gm/R)

Correct answer: √(Gm/4R)

Solution

Mutual force = Gm^2/(2R)^2 provides centripetal force on each mass orbiting at radius R: Gm^2/(4R^2) = m v^2/R -> v = sqrt(Gm/(4R)).

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