what happens to particle in unstable circular orbit in schwarzschild metic when it is pushed outwards? Does its geodesic change into an ellipse with around stable radius corresponding to its angular momentum or does it fly off to infinity?


In Schwarzschild metric (static and spherically symmetric black hole) a geodesic for a massive particle is described by:
$E = \frac{1}{2} (dr/d\tau)^2 + V_{effective}(r)$
$E = (\epsilon^2 - 1)/2$ fictitious energy
$\epsilon$ energy per particle unit rest mass
$r$ radial coordinate
$\tau$ proper time
$V_{effective}(r) = -M/r + l^2/(2 r^2) - Ml^2/r^3$
$c = G = 1$ natural units
$M$ black hole mass
$l$ orbital angular momentum per particle unit rest mass

The effective potential has a maximum and a minimum if $l/M \gt \sqrt{12}$. That gives two circular orbits, of which the one closer to the horizon is unstable.

If so, we may have:
$\sqrt{12} \lt l/M \lt 4$, fictitious energy $E$ negative
$l/M = 4$, fictitious energy $E$ zero
$l/M \gt 4$, fictitious energy $E$ positive

If a particle on the unstable circular orbit is pushed outwards, depending on the sign of the fictitious energy $E$ we have:
$E$ negative: it will trace a bound precessing orbit as a negative $E$ will cause the radial velocity $dr/d\tau$ to stop.
$E$ zero: it will fly off to infinity as a zero $E$ will cause the radial velocity $dr/d\tau$ to stop only at infinity.
$E$ positive: it will fly off to infinity as a positive $E$ will not allow for the radial velocity $dr/d\tau$ to stop.

Note: In any case the bound orbits are not ellipses, because they are subject to precession.


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