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four particles each of mass m placed at the four corner of square of edge a,move with velocity v,along a circle,which circumscribes the square,under the influence of their mutul gravitational force.find the speed v
The gravtiatation force on the a single particle is Gm2/a2 + Gm2/a2 + Gm2/2a2 = 5Gm2/2a2
= 5Gm2/2a2= root(2)mv2/a (Centripetal force)
= (5Gm/2a(root(2)))^1/2
Sorry for my mistake,
The gravtiatation force on the a single particle is Gm2root(2)/a2 + Gm2/2a2 = Gm2(2root(2)+1)/2a2
= Gm2(2root(2)+1)/2a2= root(2)mv2/a (Centripetal force)
= (Gm(1+1/2(root(2))/a)^1/2
Daer Aashish,
The gravtiatation force on the a single particle is Gm2root(2)/a2 + Gm2/2a2 = Gm2(2root(2)+1)/2a2 = Gm2(2root(2)+1)/2a2= root(2)mv2/a (Centripetal force) = (Gm(1+1/2(root(2))/a)^1/2
Best Of luck
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