Two stars, A of mass 2M and B of mass M, where M = 1.0 × 10³⁰ kg, move in circular orbits about…

Step 1. Radii. The centre of mass is fixed: $2M\,r_A = M\,r_B$ and $r_A + r_B = d$. So $r_A = \dfrac{d}{3} = 5.0 \times 10^{10}\ \text{m}$ and $r_B = \dfrac{2d}{3} = 1.0 \times 10^{11}\ \text{m}$.

Step 2. Period. For star B: $\dfrac{G(2M)M}{d^2} = M\omega^2\,\dfrac{2d}{3}$, so $\omega^2 = \dfrac{3GM}{d^3}$ (star A gives the same). Then

Step 3. Speeds. $v_A = \omega r_A = 12.2\ \text{km/s}$ and $v_B = \omega r_B = 24.3\ \text{km/s}$, in opposite directions. Their relative speed is $v_{rel} = v_A + v_B = \sqrt{\dfrac{3GM}{d}} = 36.5\ \text{km/s}$.

Step 4. After the shell leaves. A now has mass $m$, B still has $M$, they are still $d$ apart, and their relative speed is still $v_{rel}$. The motion of the two about their centre of mass has energy $\tfrac12\mu v_{rel}^2 - \dfrac{GmM}{d}$, with $\mu = \dfrac{mM}{m + M}$. The pair stays bound if this is negative:

So A must keep more than $\dfrac{M}{2} = 5.0 \times 10^{29}\ \text{kg}$. It can lose at most $1.5 \times 10^{30}\ \text{kg}$, three quarters of its mass.