# Can light become a satellite of a black hole?

In general, stars can have satellites orbiting around them. Then, can a photon become a satellite of a black hole?

Once a photon enters the Schwarzschild radius, it cannot escape the black hole. However, let's assume that a photon passes very close to the Schwarzschild radius of the black hole. The space-time around the black hole will be distorted due to the gravitational effect of the black hole.

However, outside the Schwarzschild radius, it is not distorted enough to cause light to be sucked into the black hole. However, it could cause light to orbit around the black hole in an elliptical path (or atomic path). In other words, photons that pass very close to the black hole will not be able to enter the center of the black hole or escape, and will become satellites orbiting around the black hole.

After a long time, the black hole will be surrounded by this "light shield". What will happen if the particles emitted by Hawking radiation collide with this light shield? I look forward to your explanations. (I'm not very good at drawing pictures, so please understand.)

No, light cannot form stable, bound orbits around a black hole.

The possible trajectories of light around a Schwarzschild black hole are conveniently parameterised in terms of the "impact parameter" $$b$$, which is the ratio $$cL/E$$, where $$E$$ and $$L$$ are the energy and angular momentum of the light with respect to the centre of the black hole, as measured a long way from the black hole.

If $$b> 3\sqrt{3}r_s/2$$, where $$r_s=2GM/c^2$$ is the Schwarzschild radius of a black hole then the light can approach the black hole, no closer than $$r=3r_s/2$$ and then travels away from the black hole again on an altered trajectory.

If $$b< 3\sqrt{3} r_s/2$$ then the light's trajectory will be curved, it will pass within $$r<3r_s/2$$ but ultimately end up falling to the event horizon of the black hole.

The case of $$b= 3\sqrt{3}r_s/2$$ corresponds to an unstable circular orbit at $$r=3r_s/2$$. In principle, light emitted at $$r=3r_s/2$$ in a direction perpendicular to the radial direction could stay in this orbit. Or light that travels inwards with $$b$$ very close to that value may inject light into an orbit of one or a few times around the black hole before it ultimately falls in or flies away (depending on whether $$b$$ is smaller or bigger than the threshold value).

Here are some example trajectories produced using the GROrbits software. The black circle has a radius of $$r_s$$ and the values of $$b$$ are labelled. Note that even with $$b$$ specified as $$3\sqrt{3}r_s/2$$ to 5 decimal places the incoming light only executes a couple of circuits around the black hole before falling in.

The last diagram shows an example with the light emitted from within $$r<3r_s/2$$. If it has $$b>3\sqrt{3}r_s/2$$ then it falls back in again, if it has $$b<3\sqrt{3}r_s/2$$ it escapes.

For spinning Kerr black holes there also no stable orbits but the situation is much more complex. In addition to two unstable circular orbits (one prograde and one retrograde), with differing radii, in the plane at right angles to the spin vector of the black hole, there are also a continuum of spherical orbits of fixed radius, that oscillate in latitude (see Teo 2003), but all are unstable in the radial direction (Tavlayan & Tekin 2020).

• Neither can I. It's a fact that due to gravitational radiation and other effects there are no stable orbits at all in the context of general relativity. Commented Aug 26 at 21:24
• @LolloBoldo “I can't figure out how this is not the accepted answer” - On this site “accepted” means “most helpful to the OP”. This answer is probably at a higher paygrade and thus not very helpful to the OP. Commented Aug 27 at 5:15
• This is the correct answer, both providing insight and rigor. +1 Commented Aug 27 at 10:27
• @safesphere yes but the accepted answer is built using this one, is at best a comment to this answer, not an answer on his own Commented Aug 27 at 10:31
• @candied_orange still no stable orbits, but a much more complicated answer. Commented Aug 27 at 16:23

The simplest explanation might be that the solution of the 2 body problem, is an elliptic trajectory whose speed is higher on the aphelia that on the perihelia. However, since light has a constant speed, the only trajectory with a constant speed would be a perfect circle which would be unstable : Either it would go too far and would escape the gravity field. Or it would approach too much and would fall into the black hole.

Obviously, this explanation is an oversimplification which doesn't take relativistic consideration in account but it is sufficient to explain why such an orbit could not be stable.