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
Grade
2 Answers
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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