 Solution

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.