Q 11-07-182JEE MainJEE Main 2019 (8 Apr, Shift 1)Medium
Four identical particles of mass $M$ are located at the corners of a square of side $a$. What should be their speed if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square?
Answer: (D) $1.16\sqrt{\dfrac{GM}{a}}$
The orbit radius is $r = a/\sqrt2$. Net force on one particle towards the centre: two neighbours at distance $a$ (each contributing $\frac{GM^2}{a^2}\cos45^\circ$) and the opposite corner at $\sqrt2a$:
$$F = \frac{GM^2}{a^2}\left(\sqrt2 + \frac12\right)$$
Setting $F = \dfrac{Mv^2}{r} = \dfrac{\sqrt2Mv^2}{a}$:
$$v^2 = \frac{GM}{a}\left(1 + \frac{1}{2\sqrt2}\right) = 1.354\frac{GM}{a} \Rightarrow v \approx 1.16\sqrt{\frac{GM}{a}}$$
Solution by Sreeraj P, M.Sc Physics