= Solution
Take overdots in this question to mean differentiation with respect to the orbiting particle's <proper time> $\tau$. Write $\mu=GM/c^2$, $x=r/\mu$ and $\ell=hc/(GM)$, with $h$ the <specific angular momentum>. Then the dimensionless timelike effective potential is
$$
u(x)=\frac{U}{c^2}=-\frac1x+\frac{\ell^2}{2x^2}-\frac{\ell^2}{x^3}.
$$
The last term is the relativistic correction to the Newtonian centrifugal-plus-gravitational potential. At large $r$ the two agree. Newtonian <angular momentum> gives an infinite inner <centrifugal barrier> for nonzero $h$; the relativistic correction lowers that barrier and permits plunge. The exterior problem starts at $r=2\mu$, where $U=-c^2/2$; extrapolating the potential towards $r=0$ is not a static exterior description.
Circular orbits obey
$$
U'=0\quad\Longleftrightarrow\quad GMr^2-h^2r+3\mu h^2=0,
\qquad h^2=\frac{GMr^2}{r-3\mu}.
$$
There are two extrema when $h^2>12(GM/c)^2$: an inner maximum representing an unstable circular orbit and an outer minimum representing a stable one. At equality they merge at $r=6\mu$; below it neither exists. More explicitly, evaluate the <curvature> at a circular orbit:
$$
\left.U''\right|_{\rm circ}=\frac{GM(r-6\mu)}{r^3(r-3\mu)}.
$$
Timelike circular orbits require $r>3\mu$. They are radially stable for $r>6\mu$, so \b[the marginally stable inner limit is $r_{\rm ISCO}=6GM/c^2$]. The Newtonian potential instead has one stable minimum at $r=h^2/(GM)$, without an <innermost stable circular orbit>.
\Image[/past-exam-of-the-relativistic-astrophysics-and-cosmology-course-of-the-university-of-cambridge/2012/paper-1-effective-potential.png]
{title=Schwarzschild and Newtonian potentials}
For the <Schwarzschild circular-orbit local speed> in the question, $\gamma$ is the <Lorentz factor> relative to a local <Schwarzschild static observer>, not $dt/d\tau$. That observer's clock obeys $dT=\sqrt{1-2\mu/r}\,dt$. Since the particle passes with local speed $v$, $r\,d\phi/d\tau=\gamma v$. Circular normalization gives
$$
\frac{(r\dot\phi)^2}{c^2}=\frac{\mu}{r-3\mu},\qquad
\gamma^2=1+\frac{(r\dot\phi)^2}{c^2}=\frac{r-2\mu}{r-3\mu},
\qquad \frac{v^2}{c^2}=\frac{\mu}{r-2\mu}.
$$
Thus \b[$v=c/2$ at the <innermost stable circular orbit>]. Using <Schwarzschild time> in place of the local clock would give a different speed.
For the conserved particle <energy> measured at infinity, substitute $h^2$ in the circular <energy> equation. Setting $y=\mu/r$ gives
$$
k^2=1-2y+\frac{y(1-2y)}{1-3y}
=\frac{(1-2y)^2}{1-3y},
\qquad\boxed{E=mc^2\frac{1-2\mu/r}{\sqrt{1-3\mu/r}}}.
$$
Take the positive root for a future-directed exterior orbit. It approaches $mc^2-GMm/(2r)$ at large radius. The <marginally bound circular Schwarzschild orbit> solves $k=1$, giving $y(4y-1)=0$ and hence \b[$r_{\rm mb}=4GM/c^2$] at finite radius. It is unstable; strictly bound circular orbits have $r>4\mu$, with $E<mc^2$. The phrase smallest bound orbit here denotes this marginal circular limit, not a prohibition on a bound-energy trajectory plunging below it.
At the <innermost stable circular orbit>,
$$
\boxed{E_{\rm ISCO}=\sqrt{\frac89}\,mc^2\simeq0.94281mc^2},\qquad
\eta=1-\sqrt{\frac89}\simeq\boxed{0.0572}.
$$
A radiatively efficient <thin disc> with a <zero-torque inner boundary condition> can therefore release about 5.7% of the accreted <rest energy> before the plunge. <energy> advection, stress at the inner edge and <photon> capture can alter the escaping <luminosity>, so this is a model efficiency rather than a universal emitted fraction.
For the <Kerr ISCO spin limits> of an equatorial prograde disc around a rapidly rotating <Kerr black hole>, the <innermost stable circular orbit> moves inward and its specific <energy> decreases. In the ideal extremal limit its Boyer-Lindquist radius tends to $\mu$, $E/(mc^2)\to1/\sqrt3$, and the corresponding binding-energy efficiency tends to 42.3%. Retrograde rotation has the opposite trend: the extremal limit is $r_{\rm ISCO}=9\mu$, $E/(mc^2)=5/(3\sqrt3)$, giving about 3.8%. The prograde limiting coordinate radius does not identify a massive orbit with the null horizon generator.<Rotating Black Holes: Locally Nonrotating Frames, Energy Extraction, and Scalar Synchrotron Radiation by James M. Bardeen and others>[§II, Eqs. 2.20–2.23 and Fig. 1, pp. 351–353]{ref}
Finally $\dot t=k/(1-2\mu/r)=1/\sqrt{1-3\mu/r}$. Dividing $\dot\phi=h/r^2$ by this gives
$$
\boxed{\Omega_\infty=\frac{d\phi}{dt}=\sqrt{\frac{GM}{r^3}}}.
$$
Its algebraic form is exactly Newtonian, but $r$ is the <Schwarzschild spacetime> <areal radius> and $t$ <Schwarzschild time>, normalized at infinity. Local clock rates, orbit stability and <binding energy> remain relativistic. With $GM_\odot\simeq1.3271\times10^{20}\,\mathrm{m^3\,s^{-2}}$,
$$
\boxed{f_{\rm ISCO,\infty}=\frac{c^3}{2\pi6^{3/2}G(10M_\odot)}\simeq0.220\,\mathrm{kHz}},
\qquad\boxed{\frac{f_{\rm ISCO}}{f_{\rm mb}}=\left(\frac46\right)^{3/2}\simeq0.544}.
$$
The marginally bound circular-orbit <frequency> is about $0.404\,\mathrm{kHz}$. These <Schwarzschild ISCO orbital frequencies> and marginally bound <frequencies> are rotation <frequencies> per distant stationary observer's time, as intended by using $d\phi/dt$. A static observer at finite radius $r_o$ instead divides both <frequencies> by $\sqrt{1-2\mu/r_o}$, leaving their ratio unchanged. Identifying an observed <quasi-periodic oscillation> with the <innermost stable circular orbit> orbital <frequency> is an additional astrophysical assumption.
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