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

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 $$R_s(t) = \left ( 2L_w/3A \right )^{1/3} t$$

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## closed as too localized by Manishearth♦Dec 29 '12 at 21:28

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