Wind-powered mass loss from a steady accretion disk (${\dot{M}}^{\prime }=2\pi RS,\nu \Sigma =R{\dot{M}}^{\prime }/(9\pi f)$)

For a Keplerian disk, let $S$ be the total loss through both faces per unit projected area and $\dot{M}>0$ the inward accretion rate. Annular mass conservation gives ${\dot{M}}^{\prime }=2\pi RS$. A circular orbit has specific orbital energy $-GM/(2R)$, so a marginally escaping wind requires additional energy $GM/(2R)$ per unit lost mass. Equating this sink to a fraction $f$ of the two-face viscous dissipation $(9/4)\nu \Sigma {\Omega }^2$ gives the displayed relation. If the wind carries the local specific angular momentum $h=\sqrt{GMR}$, conservation of angular momentum gives $(\dot{M}h-\mathcal{G}{)}^{\prime }=h{\dot{M}}^{\prime }$, with viscous torque in an accretion disk $\mathcal{G}=3\pi \nu \Sigma h$. Eliminating the torque yields $R^2{\dot{M}}^{\prime \prime }+\frac{3}{2}R{\dot{M}}^{\prime }=\frac{3}{2}f\dot{M}$.