The radius of the swept up shell of the observed planetary nebula [closed]

Question

The emission from this shell produces the observed planetary nebula. Assume that the fast wind switches on instantaneously at the end of the red giant phase at $t = 0$. The mass of swept up material is given by: $$ \frac{dM_s}{dt}=A(V_s-V_{RG}) $$ where $V_s=dR_s/dt$ is the velocity of the shell and $A=\dot{M_{RG}}/V_{RG}$ Assuming that the evolution of the interior of the mass-loss bubble is adiabatic, I have to find out the radius of the swept up shell given by $$ R_s(t) = \left ( 2L_w/3A \right )^{1/3} t $$

Answer 1

I have started with \begin{eqnarray} \frac{d}{dt} \int_{shell} \rho d^3x = -\int_{S_2} \rho (v_i -u_i) n_i ds + \int_{S_1} \rho (v_i -u_i)n_i ds \end{eqnarray}

Then finally I got the answer of The mass of swept up material. But I have started with Energy driven and momentum

$$ \frac{d}{dt}\left (M \frac{dR_s}{dt} \right )= \frac{3}{2} \frac{L_w}{v_w} \frac{R_s^2}{R_1^2}= 4\pi p_bR_s^2 $$

I think from here I need find out the required equation. Could some one help ?