Two stars A and B travel in circular orbit around their common Centre of mass centre of mass with radii $R_{1}$ and $R_{2}$ respectively as shown in figure. The time for the star B to move through one orbit is 2 days with it orbital radius $R_{2} = 2 \times 10^{9} m$ . The star A has a constant speed of $3.64 \times 10^{4} m s^{- 1}$ in its orbit

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The mass of star B is $1.79 \times {10^{29}}^{} k g .$

The radius of the orbit of the star A, $R_{1} = 1 \times 10^{9} m .$

The mass of star A is $3.57 \times 10^{29} k g .$

According to Kepler’s law $T^{2} \propto r^{3}$ so both have different time periods.

answer is B, C, D.

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Detailed Solution

It is a binary star system, so they have

$ω = \sqrt{\frac{G (m_{1} + m_{2})}{{(R_{1} + R_{2})}^{3}}}$

Will be same for both, hence, equal time period of 2 days.

For A, $T = 2$ days

$\frac{2 π R_{1}}{V_{1}} = T = 2 days \Rightarrow R_{1} = \frac{(2 days) V_{1}}{2 π}$

$= \frac{(2 \times 24 \times 3600 s) (3.64 \times 10)^{4}}{2 π} = 1 \times 10^{9}$

$T = \frac{2 π}{ω} = 2 π \sqrt{\frac{{(R_{1} + R_{2})}^{3}}{G (m_{1} + m_{2})}}$

$\Rightarrow (m_{1} + m_{2}) = \frac{4 π^{2}}{T^{2}} \frac{{(R_{1} + R_{2})}^{3}}{G}$

And $m_{1} ω^{2} r_{1} = F_{G} = m_{2} ω^{2} r_{2}$

$\Rightarrow m_{1} r_{1} = m_{2} r_{2} \Rightarrow \frac{m_{1}}{m_{2}} = \frac{r_{2}}{r_{1}} = 2$

$\Rightarrow m_{1} = 3.57 \times 10^{29} kg and m_{2} = 1.78 \times 10^{29} kg$

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