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An isolated triple star system consists of two identical stars, each of mass m and a fixed star of mass M. They revolve around the central star in the same circular orbit of radius r. The two orbiting stars are always at opposite ends of a diameter of the orbit. The time period of revolution of each star around the fixed star is equal to:

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$\frac{4 {πr}^{3 / 2}}{\sqrt{G (4 M + m)}}$

$\frac{4 {πr}^{3 / 2}}{\sqrt{G (M + m)}}$

$\frac{2 {πr}^{3 / 2}}{\sqrt{GM}}$

$\frac{2 {πr}^{3 / 2}}{\sqrt{G (M + m)}}$

answer is A.

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Detailed Solution

Net gravitational force on any orbiting star provides necessary centripetal force.

$\begin{array}{l} So \frac{GMm}{r^{2}} + \frac{{Gm}^{2}}{4 r^{2}} = \frac{{mV}^{2}}{r} \\ \Rightarrow V = \sqrt{\frac{G (4 M + m)}{4 r}} \\ T = \frac{2 πr}{V} = \frac{4 {πr}^{3 / 2}}{\sqrt{G (4 M + m)}} \end{array}$

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