Is it possible for a star orbit at any distance from the black hole and what would happen if, while in elliptical orbit, it partially passed through the event horizon?


1 Answer 1


As already pointed out in the comments, there exists an innermost stable circular orbit (ISCO) which is the stable orbit with the smallest possible radius and thus distance from the black hole.

As explained in the Wikipedia article linked above, for non-rotating (Schwarzschild) black holes,

$$r_{\text{isco}}=6{\frac {GM}{c^{2}}}=3r_{S}$$

where $r_S$ is the Schwarzschild radius which for a non-rotating black hole coincides with the event horizon. Thus, any object (including stars) can stably orbit a (non-rotating) black hole if the distance is $\ge r_{\text{isco}}=3r_S$.

If the orbit has a radius between $2$ and $3r_s$, an orbit is still possible, but unstable – as far as I know, the orbiting object will eventually end up falling into the black hole.

For rotating (Kerr) black holes, the equations are more complicated though it turn out that


and an unstable orbit is only possible for radii larger than about $2.9r_S$.

Note that in all these cases, the orbital radius is larger than the event horizon. It is thus impossible for an orbiting object to "dip" into the event horizon or even get very close to it. If a non-orbiting star "touches" the event horizon however, it will "fall" below it.

TL;DR: No, only for orbits larger than the ISCO. Anything (partially) passing the event horizon will fall towards the singularity and is unable to "get back out".

  • $\begingroup$ nice anser! I wonder how the ISCO is derived, didn't find it on the wikipedia site. Do you know any site for more Info or a derivation? $\endgroup$ Commented Aug 12, 2021 at 6:47
  • $\begingroup$ @CharlesTucker3 Thanks! I could find this question on astronomy SE: Why is the Innermost Stable Circular Orbit (ISCO) of a non-rotating black-hole 3 times the Schwarzschild radius? which provides some links. $\endgroup$
    – jng224
    Commented Aug 12, 2021 at 9:23
  • $\begingroup$ Many thanks for the link! $\endgroup$ Commented Aug 12, 2021 at 9:39
  • $\begingroup$ @Jonas Your last two paragraphs are incorrect. In the coordinate system of any external observer, anything approaching the horizon will stay there forever due to linear frame dragging and never cross the horizon. $\endgroup$
    – safesphere
    Commented Aug 13, 2021 at 5:16

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