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Q.
Four particles each of mass $M$, move along a circle of radius $R$ under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is:
$F _{\text {net }}=\frac{ MV ^{2}}{ R }$
$\sqrt{2} F + F _{1}=\frac{ MV ^{2}}{ R }$
$\sqrt{2} \frac{ GMM }{(\sqrt{2} R )^{2}}+\frac{ GMM }{(2 R )^{2}}=\frac{ MV ^{2}}{ R }$
$\frac{ GM }{ R }\left(\frac{1}{\sqrt{2}}+\frac{1}{4}\right)= V ^{2}$
$\frac{ GM }{ R }\left(\frac{4+\sqrt{2}}{4 \sqrt{2}}\right)= V ^{2}$
$V =\sqrt{\frac{ GM (4+\sqrt{2})}{ R 4 \sqrt{2}}}$
$V =\frac{1}{2} \sqrt{\frac{ GM (2 \sqrt{2}+1)}{ R }}$
