The distance between two stars of masses 3 M_ | vedclass

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(D) For the Earth revolving around the Sun,the period $T$ is given by Kepler's Third Law: $T = 2 \pi \sqrt{\frac{R^3}{G M_S}}$.For two stars of masses

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$9$

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(D) For the Earth revolving around the Sun,the period $T$ is given by Kepler's Third Law: $T = 2 \pi \sqrt{\frac{R^3}{G M_S}}$.For two stars of masses $m_1 = 3 M_S$ and $m_2 = 6 M_S$ separated by a distance $d = 9 R$,they orbit their common center of mass with a period $T'$ given by: $T' = 2 \pi \sqrt{\frac{d^3}{G(m_1 + m_2)}}$.Substituting the given values: $T' = 2 \pi \sqrt{\frac{(9 R)^3}{G(3 M_S + 6 M_S)}} = 2 \pi \sqrt{\frac{729 R^3}{G(9 M_S)}} = 2 \pi \sqrt{\frac{81 R^3}{G M_S}} = 9 	imes 2 \pi \sqrt{\frac{R^3}{G M_S}}$.Since $T' = n T$,we have $n T = 9 T$,which implies $n = 9$.

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