The distance between two stars of masses $$3 M_S$$ and $$6 M_S$$ is 9𝑅. 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 𝑛𝑇, where 𝑇 is the period of Earth’s revolution around the Sun.
The value of n is ___.

For a system of two bodies revolving around their common centre of mass in circular orbits, the period $$P$$ of revolution is given by Kepler’s third-law form

$$P^{2}= \frac{4\pi^{2}a^{3}}{G\,(m_{1}+m_{2})}$$

where
$$a$$  = separation between the bodies (for a circular orbit, the semi-major axis equals the distance between their centres),
$$m_{1},m_{2}$$ are their masses, and $$G$$ is the gravitational constant.

For the Earth-Sun system (taking Earth’s mass negligible compared to the Sun’s mass $$M_S$$), the orbital period is the known value $$T$$, and the orbital radius is the mean Earth-Sun distance $$R$$. Hence

$$T^{2}= \frac{4\pi^{2}R^{3}}{G\,M_S} \quad -(1)$$

Now consider the given binary-star system:
  Masses $$m_{1}=3M_S$$, $$m_{2}=6M_S$$
  Separation $$a = 9R$$

Their common period $$P$$ therefore satisfies

$$P^{2}= \frac{4\pi^{2}(9R)^{3}}{G\,(3M_S+6M_S)}$$

Simplify the numerator and denominator:

$$P^{2}= \frac{4\pi^{2}\,729R^{3}}{G\,9M_S}= \frac{4\pi^{2}\,81R^{3}}{G\,M_S} \quad -(2)$$

Divide $$(2)$$ by $$(1)$$ to relate $$P$$ and $$T$$:

$$\frac{P^{2}}{T^{2}} = \frac{81R^{3}}{R^{3}} = 81$$

Taking square roots:

$$P = 9T$$

Thus the period of the binary system is nine times the Earth’s orbital period, i.e. $$n = 9$$.

Final answer: 9