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Grade:

                        

Four particles of equal masses M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

9 years ago

Answers : (2)

vikas askiitian expert
509 Points
							

net force on any of particle due to remaining three = GM2 (81/2+1)/4R2

centripital force is provided by mutual gravitational force

GM2 (81/2+1)/4R2 = MV2/R

  V = [GM(81/2+1)/4R]

9 years ago
sai raghava rajeev penagamuri
14 Points
							

SQUARE ROOT OF GM/R(2(1.414)+1)/4

9 years ago
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