Mechanics> gravitation...

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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2 Answers

vikas askiitian expert

net force on any of particle due to remaining three = GM² (8^1/2+1)/4R²

centripital force is provided by mutual gravitational force

GM² (8^1/2+1)/4R² = MV²/R

V = [GM(8^1/2+1)/4R]

Last Activity: 14 Years ago

sai raghava rajeev penagamuri

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

Last Activity: 14 Years ago

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Mechanics> gravitation...

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

vikas askiitian expert

sai raghava rajeev penagamuri

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