JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Gravitation

Two stars of masses

m

and

2 m

at a distance

d

rotate about their common centre of mass in free space. The period of revolution is:

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

Answer: (C)

Solution

The center of mass divides the distance in the ratio

2 : 1

, so the star of mass

m

is at distance

\frac{2 d}{3}

from the center.

The gravitational force provides the centripetal force:

m ω^{2} \cdot \frac{2 d}{3} = \frac{G \cdot m \cdot 2 m}{d ^{2}}

Solving for

ω

ω = \frac{3 G m}{d ^{3}}

The period is:

T = \frac{2 π}{ω} = 2 π \frac{d ^{3}}{3 G m}

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option

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