Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2016 / Paper 1 / 1 / b / Solution

For a Newtonian circular orbit, $v_{\varphi }^2=GM/r$, so the specific orbital energy is $e=v_{\varphi }^2/2-GM/r=-GM/(2r)$. Half of the released potential energy remains in circular orbital motion. The binding-energy estimate between the specified edges is thereforeUsing $M_{\odot }=1.98847\times 10^{30}\mathrm{kg}$ and one Julian year $=3.15576\times 10^7\mathrm{s}$ gives $\dot{m}=6.30\times 10^{22}\mathrm{kg}{\mathrm{s}}^{-1}$. Thus the radiative luminosity scale is $L\sim 1.9\times 10^{38}\mathrm{W}$, approximately $1.9\times 10^{45}\mathrm{erg}{\mathrm{s}}^{-1}$ or $4.9\times 10^{11}{L}_{\odot }$. The black-hole mass cancels because both edge radii scale with $M$. The specified inner edge is $6R_s$; it must not be silently replaced with a different relativistic orbit radius.

For a finite viscous disc, the precise radiated power also depends on boundary work: torque transport of energy in a steady accretion disk means that orbital binding-energy release need not all be radiated inside the two stated edges. If the intended model is a radiatively efficient steady Keplerian accretion disk with a zero-torque inner boundary condition, integrate the one-face flux $F=3GM\dot{m}[1-\sqrt{r_{\mathrm{in}}/r}]/(8\pi r^3)$ over both faces. It givesThe difference from $L_{\mathrm{bind}}$ is the outward torque power at the outer edge. The question supplies neither boundary torque nor the external material-supply energy budget, so the two calculations state their assumptions explicitly: about $2\times 10^{38}$ W is the binding-energy estimate, while $1.37\times 10^{38}$ W is the finite-disc zero-inner-torque dissipation result. Both neglect radiation advection, winds and relativistic corrections as required for this model.^[ref]