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The limiting inward departure from the marginal Paczyński-Wiita circular orbit at $r=3a$ conserves the displayed specific angular momentum and specific orbital energy. In the Paczyński-Wiita potential, its inward speed satisfies $v_r^2=GM(3a-r{)}^3/[4a{r}^2(r-a)]$. The exactly marginal circular orbit does not depart without a perturbation; the formula describes the separatrix limit of inward releases.

Let $r_S=2GM/c^2$. In the Paczyński-Wiita potential, circular orbit balance and the radial epicyclic frequency formula giveThus the innermost stable circular orbit lies at $r=3r_S$. Its specific orbital energy is $-c^2/16$, giving a zero-torque radiative efficiency of black-hole accretion of $1/16$. The model reproduces the Schwarzschild inner orbit exactly but approximates its relativistic energy and frequency. Abramowicz's derivation explains the potential's relation to the Schwarzschild effective potential.

Write $s=(r^2+z^2{)}^{1/2}$. The Paczyński-Wiita potential is a spherical Newtonian gravitational potential with ${\Phi }^{\prime }(s)=GM/(s-a{)}^2$ and ${\Phi }^{\prime \prime }(s)=-2GM/(s-a{)}^3$. At $z=0$, $s=r$, soSubstituting into the orbital formulas givesChoosing positive orbital and vertical frequencies, $\Omega ={\Omega }_z=\sqrt{GM}/[\sqrt{r}(r-a)]$. For $r>3a$, $\kappa =\Omega \sqrt{(r-3a)/(r-a)}$. For $a<r<3a$, ${\kappa }^2<0$ and radial disturbances grow at rate $\sqrt{-{\kappa }^2}$, even though the vertical motion remains stable. Thus circular orbits are radially unstable for $a<r<3a$, and $r=3a$ is the marginal orbit. This recovers the innermost stable circular orbit of the Paczyński-Wiita circular orbit model.

For an equatorial circular orbit in the Paczyński-Wiita potential, radial force balance givesIts specific angular momentum is $j=R^2\Omega =\sqrt{GM}{R}^{3/2}/(R-R_S)$. Perturbing the circular radius at fixed $j$ gives the radial epicyclic frequency squared ${\kappa }^2=R^{-3}d(j^2)/dR$. ThereforeCircular motion is radially stable for $R>3R_S$, marginal at $3R_S$, and unstable between $R_S$ and $3R_S$. Thus the inner edge of the nearly circular thin accretion disk is $R_{\mathrm{in}}=3R_S$, joining a plunging region of a black-hole accretion disk inside the innermost stable circular orbit. This conclusion concerns the circular thin-disk model, not an assertion that all gas disappears inside this radius.

Take the inward mass accretion rate $\dot{M}$ as positive, and write the positive outward viscous torque in an accretion disk asSteady conservation of angular momentum makes the net inward flux $C=\dot{M}j-\mathcal{G}$ independent of radius. Impose a zero-torque inner boundary condition at $3R_S$, appropriate to an approximately stress-free transition to plunging motion. ThenThis is the steady accretion disk with arbitrary rotation law angular-momentum balance.

The local viscous dissipation integrated through the full disk thickness is $D=\nu \Sigma (R{\Omega }^{\prime }{)}^2=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)$. It is the power per unit planar area summed over both radiating faces. Sincesubstitution gives the two-face dissipation of a Paczyński-Wiita disk:In particular $D$ vanishes at the stress-free inner edge and is nonnegative outside it because $j$ increases there. A one-face radiative flux is $D/2$; the total accretion luminosity is therefore $L={\int }_{3R_S}^{\infty }2\pi RDdR$, with no additional factor of two.

For a direct evaluation put $x=R/R_S$. ThenThe first integral is $7/4$, using $(3x-1)/(x-1{)}^3=3/(x-1{)}^2+2/(x-1{)}^3$. For the second,so it equals $2/[\sqrt{3}(3-1)]=1/\sqrt{3}$. HenceThis is the radiative efficiency of black-hole accretion within this pseudo-Newtonian model.

Independently, the specific orbital energy of a circular particle isThus $E(\infty )=0$ and $E(3R_S)=-GM/(8R_S)=-c^2/16$. The energy released per unit accreted mass is exactly $c^2/16$, confirming the integrated accretion luminosity. The Paczyński-Wiita circular orbit reproduces the Schwarzschild inner radius but gives an approximate binding energy; the efficiency derived here is for the stated potential.

Put $r_S=2R_g$. On the midplane, circular orbit balance gives $r{\Omega }^2={\partial }_r\Phi =GM/(r-r_S{)}^2$. The specific angular momentum, specific orbital energy and orbital shear parameter of a Paczyński-Wiita circular orbit are thereforeFor a small radial displacement at fixed specific angular momentum, the effective potential stability criterion gives the radial epicyclic frequencyHere ${\kappa }_r$ is the question's epicyclic $\kappa$. Its square is negative for $2R_g<r<6R_g$, so a radial displacement grows instead of undergoing epicyclic motion. The orbit at $r_{\mathrm{in}}=6R_g$ is marginal, and the outer orbits are stable to small radial displacements. At this inner edge,This efficiency belongs to the Paczyński-Wiita potential; it is an approximation to the relativistic black-hole result.

For the Paczyński-Wiita potential, circular balance givesThereforeThe squared radial epicyclic frequency isCircular orbits change from stable to unstable where ${\kappa }_r^2$ vanishes, so

For a steady accretion disk with arbitrary rotation law and a zero-torque inner boundary condition at $R=3R_S$, the outward viscous torque in an accretion disk is $\mathcal{G}=\dot{M}(j-j_{\mathrm{in}})$. The dissipation per unit planar area summed over both disk faces is $D=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)=\nu \Sigma (R{\Omega }^{\prime }{)}^2$. Integrating $2\pi RD$ gives $L=\dot{M}GM/(8R_S)=\dot{M}{c}^2/16$ in the Paczyński-Wiita potential. The radiative flux from one face is $D/2$, preventing a factor-of-two error in the luminosity.