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(B) For a binary star system, the sum of masses is given by Kepler's Third Law in units of solar masses, astronomical units, and years: $M_1 + M_2 = \

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$25, 5$

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(B) For a binary star system, the sum of masses is given by Kepler's Third Law in units of solar masses, astronomical units, and years: $M_1 + M_2 = \frac{r^3}{T^2}$.Given $r = 30 	ext{ AU}$ and $T = 30 	ext{ years}$, we have $M_1 + M_2 = \frac{30^3}{30^2} = 30 	ext{ solar masses}$ .....$(i)$.By the definition of the centre of mass, $M_1 r_1 = M_2 r_2$. Given that one star is $5$ times farther from the centre of mass than the other, let $r_2 = 5r_1$. Thus, $M_1 r_1 = M_2 (5r_1)$, which implies $M_1 = 5M_2$ .....$(ii)$.Substituting $(ii)$ into $(i)$: $5M_2 + M_2 = 30 \Rightarrow 6M_2 = 30 \Rightarrow M_2 = 5 	ext{ solar masses}$.Then $M_1 = 5(5) = 25 	ext{ solar masses}$.Therefore, the masses are $25$ and $5$ solar masses.

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