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Q.

Four particles of equal mass M move along a circle of radius .R under the action of their mutual gravitational attraction [Fig.]. The speed of each particle is

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a

GMR

b

(22GM/R)

c

[(22+1)GM/R)]

d

[(22+1)GM/4R]

answer is D.

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Detailed Solution

The gravitational force on each particle due to other three particles provides the necessary centripetal force.GM2(2R)2cos⁡45∘+GM2(2R)2cos⁡45∘+GM2(2R)2=Mv2R2GM22R2×12+GM24R2=Mv2R2GM2R+GM4R=v2 ∴ v=GMR22+14

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