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I understand that one could theoretically replace the sun with a blackhole without affecting the orbits of the planets. This leads me to believe that a blackhole isn't a "cosmic vacuum cleaner" so much as it is a concentration of mass that has the strongest gravitational field known.

Intuitively, this must mean that there is a threshold beyond which escaping the blackhole's gravity becomes impossible even at light speed, which is the event horizon. Since a Planck length is the smallest unit known at which quantum gravitational effects become apparent, would it be reasonable to assume that a photon that is one Planck length away from the event horizon of a blackhole can escape its gravitational pull?

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    $\begingroup$ FWIW, there is no deep physical significance to the Planck length (or other Planck units), although we do expect quantum gravity effects to become relevant somewhere around that scale. (OTOH, the Hawking-Bekenstein entropy formula does become very simple when expressed in Planck units). $\endgroup$
    – PM 2Ring
    Apr 24, 2019 at 23:14
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    $\begingroup$ It gets very tricky when you mix general relativity (black holes) and quantum mechanics (Planck length). In truth, we don't know how to do this completely. Regardless, if the photon is outside the event horizon, no matter how close, then it will escape. You are correct that black holes are not cosmic vacuum cleaners. $\endgroup$
    – Florin Andrei
    Apr 25, 2019 at 3:52

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It depends on the direction the photon travels.

If the photon is directed straight away from the black hole, it will escape (but will be very redshifted).

Obviously all directions towards the event horizon will lead to capture. But since the spacetime is so curved, this also includes many directions that point away from the hole: the photon trajectory will curve around and hit the event horizon anyway. Closer to the event horizon the cone of directions where photons can escape gets smaller and smaller, until it is essentially just the direction straight away. (See slide 6 in this presentation)

See also this interactive web page; try it for a fixed position looking along the "orbit". The higher up in the picture the closer you get to the horizon. I have found that one can get past the $1.5R_s$ limit (the photon sphere, where exactly half of the sky is directions leading into the hole) by moving to a different browser tab, scrolling, and returning; the effect between 1 and 1.5 $R_s$ is to make the sky look like an ever smaller ball - photons emitted in any other direction will be captured.

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Probably no. It all depends on definition of photon location as photons have both wave and particle properties. In terms of macrocosmos location of photons are well (enough) defines, but as we speak lengths closer to its wavelength (Planck length being in the very extreme of this), it is simply difficult to pinpoint whether the photon is on this or that side.

I would assume that it needs to be at least one wavelength away from the EH in order to be able to escape, but then again - this is highly theoretical as we also need something that’ll emit that photon. I would be very interested in hearing some real professional opinions on this one!

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Yes it should be possible.

Every body has a Schwarzschild radius which defines the radius at which light would not be able to escape:

$$r_s= \frac{2GM}{c^2}$$

By this formula, if a black hole had the mass of the Earth its radius would be about 9mm, but as the Earth has a bigger radius than this we do not see light being captured.

A black hole is an object smaller than its Schwarzschild radius.

Also I found an article on Wikipedia which says that photons will orbit the black hole at $\frac{3}{2}r_s$. Within that photon sphere no stable orbit is possible:

"Any free fall orbit that crosses it from the outside spirals into the black hole. Any orbit that crosses it from the inside escapes to infinity."

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