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Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

$\begin{array}{1 1}(A)\;\sqrt{\frac{GM}{R} (1+2 \sqrt 2)} \\ (B)\;\frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt 2)} \\(C)\;\sqrt{\frac{GM}{R} } \\(D)\;\sqrt{2 \sqrt 2 \frac{GM}{R} }\end{array} $

Can you answer this question?
 
 

1 Answer

+1 vote
The speed of each particle is $\frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt 2)} $
Hence B is the correct answer.
answered Apr 7, 2014 by meena.p
 

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