JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Gravitation

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Two identical particles of mass 1 kg each go round a circle of radius R , under the action of their mutual gravitational attraction. The angular speed of each particle is :

(A) $\frac{1}{2} \frac{G}{R ^{3}}$ ✓ Correct
(B) $\frac{G}{2 R ^{3}}$
(C) $\frac{1}{2 R} \frac{1}{G}$
(D) $\frac{2 G}{R ^{3}}$

Answer: (A)

Solution

Each particle travels on a circle of radius

R

about their common centre, so the separation between them is

2 R

. The mutual gravitational attraction provides the centripetal force:

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

With

m = 1

kg,

ω^{2} = \frac{G}{4 R ^{3}}

The angular speed of each particle is

\frac{1}{2} \frac{G}{R ^{3}}

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option

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