Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is :

1 \over 2\pi \sqrt d^3 \over 3Gm

1 \over 2\pi \sqrt 3Gm \over d^3

Gravitation PYQ — JEE Main 2021 · 24 Feb · Shift 1 · Q60 with Solution

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Correct answer: D

Set up the two-body system

Two stars of masses $m$ and $2m$ are separated by distance $d$ and revolve about their common centre of mass.

For two bodies interacting gravitationally, the angular speed of revolution can be found by equating gravitational force to the required centripetal force.

Locate the centre of mass

Let the distances of masses $m$ and $2m$ from the centre of mass be $r_1$ and $r_2$ respectively. Then,

r_1+r_2=d

and from centre of mass condition,

m r_1=(2m) r_2

r_1=2r_2

Using $r_1+r_2=d$ ,

2r_2+r_2=d \Rightarrow 3r_2=d \Rightarrow r_2=\frac d3

r_1=\frac{2d}{3}

So mass $m$ moves in a circle of radius $\frac{2d}{3}$ and mass $2m$ moves in a circle of radius $\frac d3$ .

Write the gravitational force

The mutual gravitational force is

F=\frac{G(m)(2m)}{d^2}=\frac{2Gm^2}{d^2}

Apply centripetal force condition

Consider the mass $m$ , which moves in a circle of radius $\frac{2d}{3}$ with angular speed $\omega$ .

Required centripetal force:

F_c=m\omega^2 r_1=m\omega^2\left(\frac{2d}{3}\right)

This is provided by gravity:

\frac{2Gm^2}{d^2}=m\omega^2\left(\frac{2d}{3}\right)

Cancel $2m$ :

\frac{Gm}{d^2}=\omega^2\left(\frac{d}{3}\right)

\omega^2=\frac{3Gm}{d^3}

Find the time period

We know

T=\frac{2\pi}{\omega}

So,

T=2\pi\sqrt{\frac{1}{\omega^2}}=2\pi\sqrt{\frac{d^3}{3Gm}}

Match with the options

Thus,

\boxed{T=2\pi\sqrt{\frac{d^3}{3Gm}}}

This matches Option D.

Verification with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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