JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Gravitation

Each mass moves on the circle that circumscribes the square, of radius $r=\frac{a}{\sqrt{2}}$. The net gravitational pull on any one mass points to the centre and comes from its two neighbours a distance $a$ away and the diagonally opposite mass a distance $a\sqrt{2}$ away. The two neighbours contribute only their components toward the centre, $2\cdot \frac{G{M}^2}{a^2}\cos 45^{\circ }=\sqrt{2}\frac{G{M}^2}{a^2}$, while the diagonal mass adds $\frac{G{M}^2}{2a^2}$. Setting the total equal to the centripetal need $\frac{M{v}^2}{r}=\frac{\sqrt{2}M{v}^2}{a}$ gives $v^2=\frac{GM}{a}\left(1+\frac{1}{2\sqrt{2}}\right)\approx 1.35\frac{GM}{a}$, so $v\approx 1.16\sqrt{\frac{GM}{a}}$.