= Solution
\b[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>, is
$$
\mathcal L=\frac12\left[-fc^2\dot t^2+f^{-1}\dot r^2+r^2\dot\phi^2\right]=-\frac{c^2}2.
$$
The last equality is <four-velocity> normalization, imposed after deriving the <Euler-Lagrange equations>. The cyclic coordinates $t,\phi$ give independent conserved specific <Killing energy> and <angular momentum>:
$$
\boxed{e=f\dot t,\qquad \ell=r^2\dot\phi,\qquad \dot t=\frac e f,\qquad \dot\phi=\frac\ell{r^2}}.
$$
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 gives
$$
\boxed{\dot r^2=c^2e^2-f\left(c^2+\frac{\ell^2}{r^2}\right)
=c^2(e^2-1)+\frac{2GM}r-\frac{\ell^2}{r^2}+\frac{2GM\ell^2}{c^2r^3}}.
$$
The 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 $\dot r=-\ell\,du/d\phi$, the radial equation becomes
$$
\left(\frac{du}{d\phi}\right)^2=\frac{c^2(e^2-1)}{\ell^2}+\frac{2GM}{\ell^2}u-u^2+2mu^3.
$$
Differentiating and extending continuously through isolated radial turning points gives
$$
\boxed{\frac{d^2u}{d\phi^2}+u=\frac{GM}{\ell^2}+\frac{3GM}{c^2}u^2,\qquad A=\frac{GM}{\ell^2},\quad B=\frac{3GM}{c^2}}.
$$
A circular solution can equivalently be obtained directly from the radial <geodesic equation>, avoiding division by $du/d\phi=0$ everywhere on that special orbit.
\b[Circular launch and the station's clock.] For $u=1/R$ the orbit equation gives
$$
\ell_c^2=\frac{GMR}{1-3m/R},\qquad e_c=\frac{1-2m/R}{\sqrt{1-3m/R}},\qquad \Omega=\frac{d\phi}{dt}=\sqrt{\frac{GM}{R^3}}.
$$
The static astronaut measures circumferential proper length $R\,d\phi$ and station <proper time> $d\tau_s=\sqrt{f(R)}\,dt$. Thus the <Schwarzschild circular-orbit local speed> and return time are
$$
\boxed{v_{\rm circ}=\frac{R\Omega}{\sqrt{f(R)}}=\sqrt{\frac{GM}{R-2GM/c^2}},\qquad
\Delta\tau_s=2\pi\sqrt{\frac{R^3}{GM}}\sqrt{1-\frac{2GM}{c^2R}}}.
$$
This 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.
\b[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 $u=1/R+\delta u$. Then
$$
\delta u''+\kappa_\phi^2\delta u=\delta A,\qquad \kappa_\phi^2=1-\frac{6m}R,\qquad
\delta u=\frac{\delta A}{\kappa_\phi^2}[1-\cos(\kappa_\phi\phi)].
$$
The tangential launch gives $\delta u(0)=\delta u'(0)=0$. At the station, $\ell=R\gamma v$, so $\delta\ell/\ell_c=\gamma_c^2\Delta v/v_{\rm circ}$ and $\delta A=-2A_c\gamma_c^2\Delta v/v_{\rm circ}$. Therefore
$$
\boxed{r-R\simeq\frac{2(R-2m)}{1-6m/R}\frac{\Delta v}{v_{\rm circ}}\left[1-\cos(\kappa_\phi\phi)\right]}.
$$
This perturbative expression also requires the resulting radial excursion to be small compared with $R$; close to the innermost stable orbit, $|\Delta v|\ll v_{\rm circ}$ 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 $\Phi_r=2\pi/\kappa_\phi$. 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 $\sqrt2$, 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.
\Image[/past-exam-of-the-general-relativity-and-cosmology-course-of-the-university-of-oxford/2016-schwarzschild-launch-rosettes.png]
{title=Bound Schwarzschild trajectories after small positive and negative tangential launch-speed changes}
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