JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Gravitation

Two identical particles each of mass 'm' go round a circle of radius

a

under the action of their mutual gravitational attraction. The angular speed of each particle will be :

(A) $\frac{G m}{2 a ^{3}}$
(B) $\frac{G m}{8 a ^{3}}$
(C) $\frac{G m}{4 a ^{3}}$ ✓ Correct
(D) $\frac{G m}{a ^{3}}$

Answer: (C)

Solution

The two particles sit at opposite ends of a diameter, separated by

2 a

. Their mutual gravitation supplies the centripetal force for each:

\frac{G m ^{2}}{( 2 a ) ^{2}} = m ω^{2} a

. This gives

ω^{2} = \frac{G m}{4 a ^{3}}

, so the angular speed of each particle is

\frac{G m}{4 a ^{3}}

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option

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