 Solution

Steady mass conservation gives $\dot{m}=-2\pi R\Sigma u_R$. With specific angular momentum $l=R^2\Omega$, multiply the angular-momentum equation by $2\pi$ and define the signed viscous torque in an accretion diskThenIntegration from $R_1$ to $R_2$ gives

The viscous power generated in an annulus is $2\pi R{F}_{\mathrm{diss}}dR=\mathcal{G}d\Omega$. Taking the inner torque to vanish and writing $l_{\mathrm{in}}=l(R_{\mathrm{in}})$ gives $\mathcal{G}=-\dot{m}(l-l_{\mathrm{in}})$, henceIntegration by parts yieldsFor steady circular force balance, $de=\Omega dl$; equivalently, the stated equality of gravitational- and rotational-potential differences makes the integral $e_{\mathrm{out}}-e_{\mathrm{in}}$. Therefore

When $R_{\mathrm{out}}\gg R_{\mathrm{in}}$, the outer energy and boundary term vanish. For an approximately Keplerian inner orbit, $e_{\mathrm{in}}=-R_{\mathrm{in}}^2{\Omega }_{\mathrm{in}}^2/2$, soComparison with part (c) givesA Keplerian inner flow generates the standard thin-disk power. A pressure-supported sub-Keplerian slim disk generates less through shear, and its emergent luminosity can be smaller still because photon trapping in an accretion flow carries part of that generated energy through the inner edge.