 Solution

A geometrically thin accretion disk around a central mass $M$ has nearly Keplerian orbits, with ${\Omega }^2\simeq GM/r^3$. Its vertical hydrostatic equilibrium gives $H\sim c_s/\Omega \ll r$. The specific angular momentum $\ell =r^2\Omega \propto r^{1/2}$ decreases inward, so accreting matter must transfer angular momentum to other matter. A viscous stress couples neighboring annuli rotating at different angular velocities.

Writing $\nu$ for effective kinematic viscosity and $\Sigma$ for surface mass density, the outward viscous torque in an accretion disk isSteady mass conservation defines $\dot{M}=-2\pi r\Sigma u_r>0$. If the inner radius $r_{\mathrm{in}}$ has negligible applied torque, angular momentum conservation gives $\mathcal{G}=\dot{M}(\ell -{\ell }_{\mathrm{in}})$, henceWell outside that boundary, $u_r\simeq -3\nu /(2r)$ and the inward drift time is of order $r^2/\nu$, much longer than an orbital period in a thin disk with modest effective stress. Dissipation of orbital energy heats the disk. If that heat escapes locally from two faces, the radiative flux from either face isIntegration over both faces gives $L=GM\dot{M}/(2r_{\mathrm{in}})$ in this Newtonian zero-torque model. Material can lose angular momentum while the disk's outer parts spread outward: outward transport does not imply that most of its mass moves outward. In real ionized disks, magnetorotational instability and magnetic stresses often provide the effective transport; molecular viscosity alone is generally insufficient.