Satellites - Orbital Velocity & Period question from MHT CET 2026 (12 May, Shift 1)

Answer: (d)

1. Adjacent masses are $\sqrt{2}R$ apart, the opposite one is $2R$ away.
2. Net force towards centre: $F = 2\cdot\frac{GM^2}{2R^2}\cos 45^\circ + \frac{GM^2}{4R^2} = \frac{GM^2}{4R^2}(2\sqrt{2}+1)$.
3. $\frac{Mv^2}{R} = \frac{GM^2}{4R^2}(1+2\sqrt{2}) \Rightarrow v^2 = \frac{GM}{4R}(1+2\sqrt{2})$.
4. So the answer is (d) $v = \frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}$.