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

a. \(2\pi = \sqrt{\frac{d^{3}}{3Gm}}\)

b. \(\frac{1}{2\pi} = \sqrt{\frac{3Gm}{d^{3}}}\)

c. \(\frac{1}{2\pi} = \sqrt{\frac{d^{3}}{3Gm}}\)

d. \(2\pi = \sqrt{\frac{3Gm}{d^{3}}}\)

Answer: (a)

JEE Main 2021 24 Feb Physics Shift 1 Question 7 solution

⇒ G(m)(2m) / d2 = mω2 × 2d / 3

⇒ 2Gm / d2 = ω2 × 2d / 3

⇒ ω2 = 3Gm / d3

⇒ \(\omega = \sqrt{\frac{3Gm}{d^{^{3}}}}\)

We know that, ω = 2π / T so T = 2π / ω

⇒ \(T = \frac{2\pi }{\sqrt{\frac{3Gm}{d^{3}}}} \Rightarrow T = 2\pi\sqrt{\frac{d^{3}}{3Gm}}\)

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