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Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

1. √(GM/R)
2. √(2√2 GM/R)
3. √(GM/R (1+2√2))
4. 1/2 √(GM/R (1+2√2))

Correct answer: 1/2 √(GM/R (1+2√2))

Solution

For 4 masses on a circle, the net gravitational force on one (from two adjacent at sqrt(2)R apart and one diametrically opposite at 2R) provides centripetal force: GM^2/R^2 * (1/4)(1+2sqrt2) = Mv^2/R. So v = (1/2) sqrt(GM/R (1+2sqrt2)).

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