A binary star consists of two stars $A$ (mass $2.2 M_{s}$ ) and $B$ (mass $11 M_{s}$ ), where $M_{s}$ is the mass of the sun. They are separated by distance $d$ and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star $B$ about the centre of mass is

6

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2

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Solution

The correct option is A $6$

Given $M_{1} = 2.2 M_{s}$ and $M_{2} = 11 M_{s}$ . Let $R_{1}$ and $R_{2}$ be their respective distances from the center of mass. The total angular momentum about the center of mass is

$L_{T o t a l} = (I_{1} + I_{2}) ω$

and the angular momentum of B is

$L_{2} = I_{2} ω$

or

$\frac{L_{T o t a l}}{L_{2}} = \frac{I_{1} + I_{2}}{I_{2}} = 1 + \frac{I_{2}}{I_{1}}$

$= 1 + \frac{M_{1} R_{1}^{2}}{M_{2} R_{2}^{2}}$

$= 1 + \frac{M_{1} v_{1}^{2}}{M_{2} v_{2}^{2}}$ (as $v = r ω$ )

$= 1 + (\frac{M_{1} v_{1}}{M_{2} v_{2}})^{2} \times \frac{M_{2}}{M_{1}}$

$1 + 1 \times \frac{11}{2.2} = 6$ ( $M_{1} v_{1} = M_{2} v_{2}$ )

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