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

Let $z$ increase upwards and let $g>0$ denote the downward gravitational acceleration. The environment obeys hydrostatic equilibrium, $dp/dz=-\rho g$. Displace a small parcel by $\zeta$ while it remains in pressure balance with its surroundings and exchanges negligible heat. For fixed composition and adiabatic index $\gamma$, its density change at constant specific entropy satisfies $d{\rho }_{\mathrm{par}}/dp=\rho /(\gamma p)$. At the displaced position, its density excess over the environment is thereforeThe buoyancy acceleration is $\ddot{\zeta }\simeq -g\Delta \rho /\rho$. HenceStability requires $N^2>0$, equivalentlyAn upward-displaced parcel is then denser than its surroundings and falls back; a downward-displaced parcel is lighter and rises. Equality gives neutral stability in this parcel model. For a perfect gas with constant composition, $ds=c_{v}d\ln (p/{\rho }^{\gamma })$ gives $N^2=(g/c_p)ds/dz$, so a rising specific entropy is stable. This is the parcel form of the Schwarzschild criterion. Rapid pressure equilibration, negligible heat diffusion during displacement and no composition-driven buoyancy are the assumptions behind this test.

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]

Immediately upstream, free fall from rest at infinity gives $u_1^2=2G{M}_0/r_0$. Treat the surface shock as locally planar and stationary, with negligible upstream pressure. The Strong-shock Rankine-Hugoniot conditions for a monatomic gas, $\gamma =5/3$, give ${\rho }_2/{\rho }_1=4$ and $u_2=u_1/4$. momentum conservation then givesUse the source's gas-constant convention $p=\rho (R^{\ast }/\mu )T$. The immediate postshock temperature isHere adiabatic means that the jump does not lose appreciable heat to the surroundings; the shock is nevertheless irreversible and raises entropy. The gas still has velocity $u_1/4$ immediately downstream, so setting all upstream kinetic energy equal to downstream internal energy would give the wrong coefficient. Later settling, radiative cooling or changing ionization can alter the temperature; they are outside this immediate, fixed-composition perfect-gas jump.