Binary star system consists of $2$ stars such that mass of one star is twice that of another. Which of the statement is correct?

One of the stars has twice the angular momentum about common COM as compared to the other one

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Both have same angular momentum about centre of mass

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Both stars have same linear speed

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Both have same kinetic energy

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Solution

The correct option is A One of the stars has twice the angular momentum about common COM as compared to the other one

$r_{1} : r_{2} = m_{2} : m_{1} = 2 : 1$

$L_{1} : L_{2} = m_{1} ω_{1} r_{1}^{2} : m_{2} ω_{2} r_{2}^{2}$
Centripetal force is provided by gravitational force. So,
$m_{1} ω_{1}^{2} r_{1} = \frac{G M \times 2 M}{{(r_{1} + r_{2})}_{1}^{2}} = m_{2} w_{2}^{2} r_{2}$

$m_{1} ω_{1}^{2} r_{1} = m_{2} ω_{2}^{2} r_{2}$

$m w_{1}^{2} 2 r = 2 m w_{2}^{2} r$

$⟹ ω_{1} = ω_{2}$

$L_{1} : L_{2} = m ω {2 r}^{2} : 2 m ω r^{2}$
$⟹ L_{1} : L_{2} = 2 : 1$

$(K E)_{1} : (K E)_{2}$
$\frac{1}{2} m_{1} v_{1}^{2} : \frac{1}{2} m_{2} v_{2}^{2}$
$\frac{1}{2} m (ω \times 2 r)^{2} : \frac{1}{2} 2 m (ω \times r)^{2}$
2: 1
$\Rightarrow (K E)_{1} : (K E)_{2} :: 2 : 1$

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