A binary star consists of two stars A (mass 2.2 MS) and B (mass 11 MS), where MS 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 a star B about the centre of mass is

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Center of Mass

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Angular Momentum and Conservation of Angular Momentum

Question

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 a star B about the centre of mass is

$\frac{m_{1} v_{1}^{2}}{r_{1}} = \frac{{Gm}_{1} m_{2}}{d^{2}}$

$\frac{v_{1}^{2}}{r_{1}} = \frac{{Gm}_{2}}{d^{2}}$

$v_{1} = \sqrt{\frac{{Gm}_{2} r}{d^{2}}} = \sqrt{\frac{{Gm}_{2}}{d} (\frac{5}{6})}$

A binary star system consists of two stars orbiting their common center of mass, which remains stationary due to no external forces. The center of mass (COM) is the point where the system's mass is balanced. For two masses, its position along the line joining them is determined by $r_{1} = \frac{m_{2} d}{m_{1} + m_{2}}$ and $r_{2} = \frac{m_{1} d}{m_{1} + m_{2}}$ , where $r_{1}$ and $r_{2}$ are distances from COM to masses $m_{1}$ and $m_{2}$ respectively, and $d$ is the separation.

Given: Mass of star A, $m_{A} = 2.2 M_{S}$ , mass of star B, $m_{B} = 11 M_{S}$ .

Step 1: Find distances from COM to each star.

Let $r_{A}$ be distance from COM to star A, $r_{B}$ from COM to star B.

Using COM formula: $r_{A} = \frac{m_{B} d}{m_{A} + m_{B}} = \frac{11 M_{S} d}{2.2 M_{S} + 11 M_{S}} = \frac{11 d}{13.2} = \frac{5 d}{6}$

$r_{B} = \frac{m_{A} d}{m_{A} + m_{B}} = \frac{2.2 M_{S} d}{13.2 M_{S}} = \frac{2.2 d}{13.2} = \frac{d}{6}$

Step 2: Angular momentum of each star about COM.

Angular momentum $L = m v r$ for circular motion, where $v$ is orbital speed.

Both stars have same angular velocity $ω$ about COM, so $v = ω r$ .

Thus, $L_{A} = m_{A} (ω r_{A}) r_{A} = m_{A} ω r_{A}^{2}$

$L_{B} = m_{B} ω r_{B}^{2}$

Step 3: Total angular momentum $L_{t o t a l} = L_{A} + L_{B} = ω (m_{A} r_{A}^{2} + m_{B} r_{B}^{2})$

Step 4: Find the ratio $R = \frac{L_{t o t a l}}{L_{B}} = \frac{ω (m_{A} r_{A}^{2} + m_{B} r_{B}^{2})}{m_{B} ω r_{B}^{2}} = \frac{m_{A} r_{A}^{2} + m_{B} r_{B}^{2}}{m_{B} r_{B}^{2}} = \frac{m_{A} r_{A}^{2}}{m_{B} r_{B}^{2}} + 1$

Substitute values: $m_{A} = 2.2 M_{S}$ , $m_{B} = 11 M_{S}$ , $r_{A} = \frac{5 d}{6}$ , $r_{B} = \frac{d}{6}$

$\frac{m_{A} r_{A}^{2}}{m_{B} r_{B}^{2}} = \frac{2.2 M_{S} \times {(\frac{5 d}{6})}^{2}}{11 M_{S} \times {(\frac{d}{6})}^{2}} = \frac{2.2 \times 25}{11 \times 1} = \frac{55}{11} = 5$

So, $R = 5 + 1 = 6$

Final Answer: The ratio is 6.

Formulae

Center of Mass for two masses: $r_{1} = \frac{m_{2} d}{m_{1} + m_{2}}$ , $r_{2} = \frac{m_{1} d}{m_{1} + m_{2}}$

Angular Momentum in circular motion: $L = m r^{2} ω$

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