JEE Main 2014 — Gravitation Question with Solution

All those particles are moving due to their mutual gravitational attraction.

The force between each masses are repulsive force.

On mass M at C, due to mass at D the repulsive force is F in the vertical direction.

On mass M at C, due to mass at B the repulsive force is F in the horizontal direction.

On mass M at C, due to mass at A the repulsive force is F'

Net force acting on particle at C,

= $$2F\cos 45^{\circ }+F^{\prime }$$

Where $$F=\frac{G{M}^2}{{\left(\sqrt{2}R\right)}^2}$$ and $$F^{\prime }=\frac{G{M}^2}{4R^2}$$

$$\Rightarrow \frac{2\times G{M}^2}{\sqrt{2}{\left(R\sqrt{2}\right)}^2}+\frac{G{M}^2}{4R^2}$$

$$\Rightarrow \frac{G{M}^2}{R^2}\left[\frac{1}{4}+\frac{1}{\sqrt{2}}\right]$$

This net force will balance by the centripetal force F_cp = $$\frac{M{v}^2}{R}$$

$$\therefore$$ $$\frac{M{v}^2}{R}=$$ $$\frac{G{M}^2}{R^2}\left[\frac{1}{4}+\frac{1}{\sqrt{2}}\right]$$

$$\Rightarrow$$ $$v=\sqrt{\frac{Gm}{R}\left(\frac{\sqrt{2}+4}{4\sqrt{2}}\right)}$$

$$=\frac{1}{2}\sqrt{\frac{Gm}{R}\left(1+2\sqrt{2}\right)}$$