The distance between two stars of masses $3 M_{S}$ and $6 M_{S}$ is $9 R$ . Here $R$ is the mean distance between the centers of the Earth and the Sun, and $M_{S}$ is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period $n T$ , where $T$ is the period of Earth’s revolution around the Sun. The value of $n$ is

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Solution

The time period of earth revolving around the sun is given by

$T = 2 π \sqrt{\frac{R^{3}}{G M_{S}}}$

Centre of mass of two stars lies at $x$ from $3 M_{S}$ , which is

$x = \frac{(3 M_{S} \times 0) + (6 M_{S} \times 9 R)}{(3 M_{S} + 6 M_{S})} = 6 R$

Let $T^{'}$ be the time period of star having mass $3 M_{S}$ and $v$ be its velocity.

$\frac{3 M_{S} v^{2}}{6 R} = \frac{G (3 M_{S}) (6 M_{S})}{(9 R)^{2}}$

$v = \frac{2}{3} \sqrt{\frac{G M_{S}}{R}}$

$∴ T^{'} = \frac{2 π (6 R)}{v} = \frac{18 π R \sqrt{R}}{\sqrt{G M_{S}}} = 9 \times 2 π \sqrt{\frac{R^{3}}{G M_{S}}}$

$\Rightarrow T^{'} = 9 T$

$∴ n = 9$

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