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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.

kanchana Rao , 14 Years ago
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vikas askiitian expert

Last Activity: 14 Years ago

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]

sai raghava rajeev penagamuri

Last Activity: 14 Years ago

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

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