JEE Main · 2014 · 06 Apr 2014 · Offline · Gravitation

The four equal masses sit at the corners of a square whose diagonal is $2R$ and whose side is $\sqrt{2}R$. Look at one mass: the one directly opposite, a distance $2R$ away, pulls it towards the centre with $\frac{G{M}^2}{4R^2}$, while the two neighbours, each $\sqrt{2}R$ away, contribute their components towards the centre, together $2\times \frac{G{M}^2}{2R^2}\times \frac{1}{\sqrt{2}}=\frac{G{M}^2}{\sqrt{2}{R}^2}$. The total pull towards the centre is $\frac{G{M}^2}{R^2}\left(\frac{1}{4}+\frac{1}{\sqrt{2}}\right)=\frac{G{M}^2}{R^2}\cdot \frac{1+2\sqrt{2}}{4}$. Setting this equal to the required $\frac{M{v}^2}{R}$ and solving, $v=\frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}$.