= Solution
The wind removes $2\pi rW(r)\,dr$ of mass per unit time from an annulus; $W$ is already the surface-density loss term in the supplied <mass conservation> equation, so there is no additional two-face factor. Integrating the <photoevaporation> profile gives
$$
\dot M_w=2\pi W_0r_g^{5/2}\int_{r_g}^\infty r^{-3/2}dr
=\boxed{4\pi W_0r_g^2}.
$$
The convergence at infinity is important: the loss is concentrated near the <photoevaporative gravitational radius>. Using $r_g=1.4\times10^{14}\,\mathrm{cm}$ yields $\dot M_w\simeq2.46\times10^{17}\,\mathrm{g\,s^{-1}}$, or about $3.9\times10^{-9}\,M_\odot\,\mathrm{yr^{-1}}$. The initial disk mass is $2\times10^{30}\,\mathrm g$, so the wind-only depletion time is
$$
t_{\mathrm{disp}}\sim\frac{M_{\mathrm{disk}}}{\dot M_w}
\simeq8.1\times10^{12}\,\mathrm s=\boxed{2.6\times10^5\,\mathrm{yr}}.
$$
This estimate treats the heated area and wind normalization as fixed and neglects additional removal through stellar accretion. Once the disk shrinks, its wind rate and geometry need not remain constant.
Back to article page
