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Assuming that Hawking radiation cause black holes to become less massive over time, it should follow that the event horizons of black holes should shrink over time as well.

In this case, what would happen if an object is orbiting the singularity within the event horizon, will the event horizon eventually recede below the orbit of the object, thus freeing it? Or would the object’s orbit shrink along with the event horizon? If the latter, what would be the cause?

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  • $\begingroup$ I think anything that crosses the "point of no return" is doomed! maybe unharmed but still doomed.😝 $\endgroup$
    – user6760
    Commented Mar 17, 2018 at 0:59

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It isn't possible to orbit the singularity within the event horizon.

If an object is inside the event horizon, it's going to hit the singularity within a rather short amount of proper time, even if it's accelerating outward. How long that will take increases with the black hole's mass, but it's within a matter of not that many minutes even in the case of a supermassive black hole. Interestingly, if an object accelerates too much away from the singularity in an attempt to avoid hitting the singularity, that can actually cause the object to hit the singularity in a shorter amount of proper time, due to time dilation effects.

In contrast, black holes evaporate via Hawking radiation over an extremely long time scale, unless the black hole is very small. A normal-sized black hole, say one with the mass of the sun, will have an evaporation time that's vastly longer than the current age of the universe.

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  • $\begingroup$ What causes objects to descend towards the singularity within the event horizon, and why does it seem to differ from the way most other orbits work? $\endgroup$
    – user289661
    Commented Mar 17, 2018 at 19:18
  • $\begingroup$ @user289661 The Newtonian model of gravity is only an approximation, that only works well in regions of spacetime that are close to being flat. Near and inside a black hole, spacetime isn't at all close to being flat, and the elliptical orbits predicted by Newtonian gravity become wildly inapplicable. According to general relativity, the more accurate model of gravity, inside of the event horizon of a black hole, all "orbits" (i.e., all timelike geodesics) lead toward the singularity. $\endgroup$
    – Red Act
    Commented Mar 17, 2018 at 21:55
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Another way of seeing the causal structure of an evaporating black hole is to look at a Penrose diagram, showing the structure of spacetime but distorting it so it fits into a finite diagram (but keeping angles right: light moves at $\pm$45 degree slant). Backreaction has a nice overview comparing the full diagram, the one with an actual imploding star, and finally an evaporating black hole.

Penrose diagram of a black hole forming from a collapsing star and eventually evaporating.

Penrose diagram of a black hole forming from a collapsing star and eventually evaporating. From The Geometry of General Relativity by Tevian Dray (licensed under a Creative Commons Attribution-NonCommercial-NoDerivs License.)

The spacetime inside the blue patch is a regular curved interior solution not doing anything odd, the rest is empty space. The dashed lines are just symmetry lines of no importance. The right-hand lines represent past and future "lightlike infinity". The thick $r=0$ line is the singularity. Note that it is spacelike: if you enter the hole you will have to move inside a 45 degree lightcone upward, so you will meet it at some point.

Things get way more mind-boggling for a rotating Kerr black hole that evaporates, but as far I can remember there are no good ways of loitering inside.

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If information is never lost when it goes into a black hole. Your information will come out quantum encoded in the hawking radiation.

Although you cannot orbit the singularity once within the event horizon as all possible path moves you towards the singularity, you will come out as hawking radiation eventually. You are scrambled beyond recognition but theoretically, you and everything that went into the black hole can be reconstructed from that information.

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    $\begingroup$ Hawking radiation is thermal and cannot encode information. $\endgroup$
    – gabe
    Commented Mar 16, 2018 at 23:16
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    $\begingroup$ This is the subject of debate and, more importantly, is only tangentially related to the question. $\endgroup$
    – user4552
    Commented Mar 17, 2018 at 1:53

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