Ratio of Total Angular Momentum to Star Angular Momentum in Binary Star System

Let the masses of star A and star B be denoted by $m_A$ and $m_B$, respectively:
$m_A = 2.2\,\text{M}_\text{s}$
$m_B = 11\,\text{M}_\text{s}$

The two stars orbit their stationary centre of mass (CM) with a common angular velocity $\omega$.

Let the distance of star A and star B from the centre of mass be $r_A$ and $r_B$, respectively. By the definition of the centre of mass:
$m_A r_A = m_B r_B \implies \frac{r_A}{r_B} = \frac{m_B}{m_A}$

The angular momentum of star A about the centre of mass is:
$L_A = m_A r_A^2 \omega$

The angular momentum of star B about the centre of mass is:
$L_B = m_B r_B^2 \omega$

The total angular momentum of the binary star system is:
$L_{\text{total}} = L_A + L_B$

The ratio of the total angular momentum of the system to the angular momentum of star B is:
$\frac{L_{\text{total}}}{L_B} = \frac{L_A + L_B}{L_B} = \frac{L_A}{L_B} + 1$

Substituting the expressions for $L_A$ and $L_B$:
$\frac{L_A}{L_B} = \frac{m_A r_A^2 \omega}{m_B r_B^2 \omega} = \frac{m_A}{m_B} \left(\frac{r_A}{r_B}\right)^2 = \frac{m_A}{m_B} \left(\frac{m_B}{m_A}\right)^2 = \frac{m_B}{m_A}$

Substituting the given values of the masses:
$\frac{L_A}{L_B} = \frac{11\,\text{M}_\text{s}}{2.2\,\text{M}_\text{s}} = 5$

Therefore, the required ratio is:
$\frac{L_{\text{total}}}{L_B} = 5 + 1 = 6$