
Gravitational force on each mass 'M'
=2a2GM2+(2a)2GM2=(2+21)a2GM2
This force will provide required centripetal force
(2+21)a2GM2=(a/2)Mv2
v=(1+221)aGM=1.16aGM
Four identical particles of mass M are located at the corners of a square of side ‘a’ . What should be their speed if each of them revolves under the influence of other’s gravitational field in a circular orbit circumscribing the square?

Held on 8 Apr 2019 · Verified 6 Jul 2026.
1.35aGM
1.21aGM
1.41aGM
1.16aGM
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Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. 
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