Verified Solution • 2025–26 Session moderate

QJEEM21Ph89ed: JEE Main 2021 Physics - Class 11 JEEM21Ph89ed Solution: Two stars of masses m and 2m at a distance d rotate about their …

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 :

Multiple Choice Options

  • A 12πd33Gm{1 \over {2\pi }}\sqrt {{{{d^3}} \over {3Gm}}}
  • B 2π3Gmd32\pi \sqrt {{{3Gm} \over {{d^3}}}}
  • C 12π3Gmd3{1 \over {2\pi }}\sqrt {{{3Gm} \over {{d^3}}}}
  • D 2πd33Gm2\pi \sqrt {{{{d^3}} \over {3Gm}}}

Step-by-Step Verified Solution

The given situation is shown below

JEE Main 2021 (Online) 24th February Morning Shift Physics - Gravitation Question 128 English Explanation
The gravitational force between these two stars provide the required centripetal force for rotation in a circle about their common centre.

Assuming 2 m at origin, the centre of mass of the system lies at

x=2m×0+m×d2m+m=d3x = {{2m \times 0 + m \times d} \over {2m + m}} = {d \over 3}

Hence, FG=FC{F_G} = {F_C}

where, FG is gravitational force between them and FC is centripetal force.

⇒Gm1m2r2=2mω2x\Rightarrow {{G{m_1}{m_2}} \over {{r^2}}} = 2m{\omega ^2}x

⇒G(2m)(m)d2=2mω2×d3⇒ω2=3Gmd3\Rightarrow {{G(2m)(m)} \over {{d^2}}} = 2m{\omega ^2} \times {d \over 3} \Rightarrow {\omega ^2} = {{3Gm} \over {{d^3}}}

⇒ω=3Gmd3\Rightarrow \omega = \sqrt {{{3Gm} \over {{d^3}}}}

We know that,

ω=2πT\omega = {{2\pi } \over T}

∴\therefore T=2πωT = {{2\pi } \over \omega }

3Gmd3\sqrt {{{3Gm} \over {{d^3}}}}

=2πd33Gm= 2\pi \sqrt {{{{d^3}} \over {3Gm}}} [using Eq. (i)]

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring