Since the kinetic energy is conserved, unless you cross the event horizon you should come back without any energy loss. In other words, a carefully constructed trajectory would allow an object to flyby the BH even horizon as close as desired and come back (although maybe in different direction from the BH due to GR precession). If the BH large enough the probe even does not need to be afraid of the tidal forces...

Is this true? What a probe can photo shoot at the distance of 1 m from the event horizon?


1 Answer 1


Is it possible to flyby a black hole at a distance of 1 m from event horizon and come back?

No. For a realistic astrophysical black hole any point at 1 m from event horizon would be located deeply inside the photon sphere. For a nonrotating black hole any geodesic trajectory that starts far away from the horizon and enters the photon sphere would also enter the horizon, so a probe coasting along such geodesic would fall inside the black hole.

Mathematical treatment could be found at this Wikipedia page or in many GR textbooks. But we could form some intuition if we remember that the geodesic motion of a particle has an effective potential which typically (for some value of specific angular momentum) looks like this :

Image taken from book Frolov & Novikov 1998

(Image from the book V. Frolov, I. Novikov, Black Hole Physics: Basic Concepts and New Developments, 1998).

A second parameter characterizing given geodesic is specific energy $\tilde E$ and the allowed range of radial coordinate would correspond to the parts of line $\tilde E=\text{const}$ above the effective potential on the plot. A trajectory that starts at infinity and ends at infinity would correspond to the case $\tilde E_2$ in the image, it would have a turning point at some value of radial coordinate larger than the radius of unstable circular orbit ($\tilde E_\text{max}$ in the image), but unstable circular orbits lie within the interval $3/2 r_s < r_\text{uco} < 3 r_s$ (from photon sphere to ISCO).

So for a trajectory to approach within 1 m from event horizon it must be either of the type $\tilde E_3$ or $\tilde E_4$ in the image. $\tilde E_3$ corresponds to the trajectory of the particle falling into the black hole, while $\tilde E_4$ is a geodesics that starts and ends on the horizon.

If the black hole is rotating, then it is possible for unstable circular orbits to be closer to the horizon than $3/2 r_s$, so the turning point of trajectory starting and ending at infinity could also be closer. But in order for such trajectory to be within 1 m of the horizon with the horizon radius of at least $3\text{ km}$ the black hole spin parameter $a$ must be very, very close to unity. See for example the plot in this answer. It is very unlikely that any realistic astrophysical process of black hole formation (and subsequent matter accretion) would allow any black hole to attain such high values of spin.

  • $\begingroup$ Oops, I got this one wrong. Deleted my answer. Thanks for posting a correct answer. $\endgroup$
    – user4552
    Dec 26, 2019 at 22:40
  • $\begingroup$ @BenCrowell What was your answer? $\endgroup$
    – Anixx
    Dec 28, 2019 at 20:32
  • $\begingroup$ I want to understand the plot. If someone is on the orbit E1 and the energy drops, what happens? Also, does the plot indicate that a stable circular orbit can be only at unique radius corresponding to the minimum of the curve? $\endgroup$
    – Anixx
    Dec 28, 2019 at 20:42
  • $\begingroup$ @Anixx The plot is for one specific value of the angular momentum, so yes, only one circular orbit can exist for this one value (at the minimum of the curve). The orbit $E_1$ is elliptical. If the energy drops (due to friction, engines, etc.), the angular momentum would also drop, so the plot would change in such a way that the orbit becomes closer to $E_{max}$. The $E_1$ orbit is conceptually the same around the Earth or a black hole. If the energy on an elliptical orbit drops, the orbit essentially becomes closer and, if the energy keeps dropping, the objects hits $E_{max}$ and goes down. +1 $\endgroup$
    – safesphere
    Dec 29, 2019 at 7:14
  • $\begingroup$ Emax has greater energy than E1 $\endgroup$
    – Anixx
    Dec 29, 2019 at 7:45

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