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Constants and radial motion. Write $m=GM/c^2$ and $f(r)=1-2m/r$. A Schwarzschild black hole has spherical symmetry, so its conserved angular momentum fixes an orbital plane through the centre. A rotation of the angular coordinates puts any nonradial orbit in the equatorial plane. A radial orbit can also be placed in such a plane. This is a choice of coordinates, not a restriction on the possible orbital inclinations.

The quadratic geodesic Lagrangian, with a dot denoting differentiation with respect to the satellite's proper time, isThe last equality is four-velocity normalization, imposed after deriving the Euler-Lagrange equations. The cyclic coordinates $t,\varphi$ give independent conserved specific Killing energy and angular momentum:Here $e$ is the conserved energy per unit rest energy and $\ell$ has dimensions of angular momentum per unit mass. Substituting them in the normalization condition givesThe two signs of $\dot{r}$ distinguish outgoing and ingoing branches; the expression must be nonnegative on the allowed radial interval.

For a nonradial orbit, put $u=1/r$. Since $\dot{r}=-\ell du/d\varphi$, the radial equation becomesDifferentiating and extending continuously through isolated radial turning points givesA circular solution can equivalently be obtained directly from the radial geodesic equation, avoiding division by $du/d\varphi =0$ everywhere on that special orbit.

Circular launch and the station's clock. For $u=1/R$ the orbit equation givesThe static astronaut measures circumferential proper length $Rd\varphi$ and station proper time $d{\tau }_s=\sqrt{f(R)}dt$. Thus the Schwarzschild circular-orbit local speed and return time areThis is a local speed measured with the astronaut's clock, not the satellite's proper velocity $\gamma v$. After one azimuthal revolution the circular satellite meets the fixed station again. For comparison, the satellite's own elapsed proper time would instead be $2\pi \sqrt{R^3/(GM)}\sqrt{1-3m/R}$.

A static station requires $R>2m$; a timelike circular satellite requires $R>3m$, since the local circular speed must be below $c$. The circular orbit is stable only for $R>6m$, marginal at $6m$, and unstable for $3m<R<6m$. Thus a request for a robust near-circular trajectory implicitly needs more than merely placing the station outside the horizon.

Slightly changing the launch speed. In the stable, bound regime, a small tangential increase makes the launch point a pericentre; a small decrease makes it an apocentre, provided the perturbed orbit remains outside its plunge separatrix. In either case the orbit is a precessing rosette, not an exactly closed Newtonian ellipse. For an explicit first-order description write $A=A_c+\delta A$ and $u=1/R+\delta u$. ThenThe tangential launch gives $\delta u(0)=\delta u^{\prime }(0)=0$. At the station, $\ell =R\gamma v$, so $\delta \ell /{\ell }_c={\gamma }_c^2\Delta v/v_{\mathrm{circ}}$ and $\delta A=-2A_c{\gamma }_c^2\Delta v/v_{\mathrm{circ}}$. ThereforeThis perturbative expression also requires the resulting radial excursion to be small compared with $R$; close to the innermost stable orbit, $|\Delta v|\ll v_{\mathrm{circ}}$ alone does not ensure that.

The Schwarzschild launch-orbit recurrence needs care when deciding whether a collision is inevitable. The satellite returns to radius $R$ once per radial cycle, after an azimuthal advance approximately ${\Phi }_r=2\pi /{\kappa }_{\varphi }$. It returns to the station's exact point only if an integer number of radial cycles also gives an integer number of azimuthal revolutions. Rational ${\Phi }_r/(2\pi )$ gives an exact return; an irrational ratio gives arbitrarily close approaches but no exact post-launch return for ideal point objects. At $R=12m$, for example, the linearized ratio is $\sqrt{2}$, illustrating why precession alone does not prove an exact collision. For generic bound orbits the dense recurrence eventually threatens any station and satellite of nonzero size; a resonant closed rosette returns exactly. A small speed adjustment is therefore not a robust permanent collision-avoidance manoeuvre for a bound satellite.

There is no unconditional collision conclusion without specifying the regime. An unstable circular orbit cannot support the small bounded oscillations used above. In particular, near $R=3.5m$ the circular specific energy is already $e_c>1$; a small positive speed increment can send the satellite outward on an unbound escape orbit, with no subsequent return to $R$. Such an escape is a different way to avoid a future encounter. Radiation reaction and manoeuvres have been excluded throughout the ideal test-particle calculation.

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 isThe 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 obeyThere 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:Timelike circular orbits require $r>3\mu$. They are radially stable for $r>6\mu$, so the marginally stable inner limit is $r_{\mathrm{ISCO}}=6GM/c^2$. The Newtonian potential instead has one stable minimum at $r=h^2/(GM)$, without an innermost stable circular orbit.

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 $v$, $rd\varphi /d\tau =\gamma v$. Circular normalization givesThus $v=c/2$ 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 $y=\mu /r$ givesTake the positive root for a future-directed exterior orbit. It approaches $m{c}^2-GMm/(2r)$ at large radius. The marginally bound circular Schwarzschild orbit solves $k=1$, giving $y(4y-1)=0$ and hence $r_{\mathrm{mb}}=4GM/c^2$ at finite radius. It is unstable; strictly bound circular orbits have $r>4\mu$, with $E<m{c}^2$. 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,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 $\mu$, $E/(m{c}^2)\to 1/\sqrt{3}$, and the corresponding binding-energy efficiency tends to 42.3%. Retrograde rotation has the opposite trend: the extremal limit is $r_{\mathrm{ISCO}}=9\mu$, $E/(m{c}^2)=5/(3\sqrt{3})$, giving about 3.8%. The prograde limiting coordinate radius does not identify a massive orbit with the null horizon generator.^[ref]

Finally $\dot{t}=k/(1-2\mu /r)=1/\sqrt{1-3\mu /r}$. Dividing $\dot{\varphi }=h/r^2$ by this givesIts 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 $G{M}_{\odot }\simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$,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 $d\varphi /dt$. 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.