Orbital Period of Binary Star System Relative to Earth

To find the value of $n$, we analyze both the Earth-Sun system and the binary star system using Newton's law of gravitation and circular motion.

1. Earth-Sun System

Assuming the mass of the Earth is negligible compared to the mass of the Sun ($M_S$), the centripetal force required for Earth to maintain a circular orbit of radius $R$ is provided by the gravitational force between Earth and the Sun: $\frac{G M_S M_E}{R^2} = M_E \omega_E^2 R = M_E \left(\frac{2\pi}{T}\right)^2 R$

Solving for the orbital period $T$ of the Earth around the Sun: $T = 2\pi \sqrt{\frac{R^3}{G M_S}}$

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2. Binary Star System

Let the two stellar bodies have masses $m_1 = 3M_S$ and $m_2 = 6M_S$, and let their separation distance be $d = 9R$.

The total mass of the binary system is: $M_{\text{total}} = m_1 + m_2 = 3M_S + 6M_S = 9M_S$

The center of mass of the two stars serves as the center of their circular orbits. The distance of star $m_1$ from the center of mass is given by: $r_1 = \left(\frac{m_2}{m_1 + m_2}\right) d$

The gravitational force of attraction between the two stars provides the necessary centripetal force for $m_1$ to revolve in a circular path of radius $r_1$ with angular speed $\omega$: $\frac{G m_1 m_2}{d^2} = m_1 \omega^2 r_1$

Substituting $r_1 = \frac{m_2 d}{m_1 + m_2}$: $\frac{G m_1 m_2}{d^2} = m_1 \omega^2 \left(\frac{m_2 d}{m_1 + m_2}\right)$

Canceling $m_1 m_2$ from both sides, we get: $\omega^2 = \frac{G(m_1 + m_2)}{d^3}$

The orbital period $T_{\text{binary}}$ of the binary system is: $T_{\text{binary}} = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{d^3}{G(m_1 + m_2)}}$

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3. Relation Between $T_{\text{binary}}$ and $T$

Substitute $d = 9R$ and $m_1 + m_2 = 9M_S$ into the formula for $T_{\text{binary}}$: $T_{\text{binary}} = 2\pi \sqrt{\frac{(9R)^3}{G(9M_S)}} = 2\pi \sqrt{\frac{729 R^3}{9 G M_S}} = 2\pi \sqrt{81 \cdot \frac{R^3}{G M_S}} = 9 \cdot \left(2\pi \sqrt{\frac{R^3}{G M_S}}\right)$

Since $T = 2\pi \sqrt{\frac{R^3}{G M_S}}$, we have: $T_{\text{binary}} = 9T$

Given that the orbital period is $nT$, comparing $T_{\text{binary}} = 9T$ with $nT$ gives: $n = 9$