JEE PYQ: Gravitation Question 34
Question 34 - 2019 (08 Apr Shift 1)
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 others’ gravitational field in a circular orbit circumscribing the square?
(1) $1.35\sqrt{\frac{GM}{a}}$
(2) $1.16\sqrt{\frac{GM}{a}}$
(3) $1.21\sqrt{\frac{GM}{a}}$
(4) $1.41\sqrt{\frac{GM}{a}}$
Show Answer
Answer: (2)
Solution
Radius of circumscribing circle $r = \frac{a}{\sqrt{2}}$. Net force toward centre: $F_1 + 2F_2\cos 45° = \frac{GM^2}{a^2} + \frac{2GM^2}{2a^2} \times \frac{1}{\sqrt{2}} = \frac{GM^2}{a^2}\left(1 + \frac{1}{2\sqrt{2}}\right)$… Wait, correcting: $\frac{Mv^2}{r} = F_{\text{net}}$. $v = 1.16\sqrt{GM/a}$.
BETA
