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I know that the escape velocity exceeds the speed of light but, what stops a rocket which can thrust constantly from returning, like, to escape earth we don't have to exceed escape velocity if we have constant thrust.

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That's why the term "escape velocity" is misleading for black holes.

Within the event horizon all possible trajectories are inwards, towards the singularity. There are no possible trajectories leading outwards of the black hole. Hence it is not possible to return after crossing the horizon. See graphic on Wikipedia.

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This is a very good question which seems to be obvious.

The point is that - if you look at the Schwarzschild metric - the curvature factor (1-2M/r) becomes negative if the r-coordinate is less than the Schwarzschild radius 2M. This means that r- and t-coordinate interchange their sign (more precisely timelike coordinates change to spacelike coordinates)! Consequently as time passes the r-coordinate decreases inevitably and anything inside the black hole will reach the singularity.

This is also true for null geodesics. A photon emitted upwards will "fall" towards the singularity.

That's why constant thrust doesn't help.

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A rocket oriented with its thrust trying to travel in the outwards direction, and with constant thrust in the sense of constant proper acceleration, will approach more and more closely to the edge of a light cone whose origin is somewhere inside the black hole. Such a rocket is doing its best to "catch up with light". But the light cone itself does not cross the horizon. You can think of this in terms of the speed of light if you like, but perhaps it is better to think of it in terms of geometry. It is saying that spacetime is becoming infinitely "steep" at the horizon, in the sense that in order to make progress in the radial direction more and more time has to elapse according to a system of measuring time that could be employed by ordinary clocks outside the horizon.

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  • $\begingroup$ @safesphere Yes, just as trying to cross a lightcone in ordinary flat spacetime would also amount to going backwards in time in some frame or other. $\endgroup$ Feb 13, 2021 at 10:13
  • $\begingroup$ @safesphere No, my answers are not based on physical intuition; they are based on thorough study of general relativity. There are timelike lines which avoid the Kerr singularity, and in the case of Schwarzschild singularity different timelike geodesics passing through any given event have different spatial directions. Those that remain at fixed $\phi,\theta$ in Schwarzschild coordinates are suitably called "radial"; of these, the one with decreasing $t$ coordinate (which is now spacelike) is "inward", the other is "outward". ... $\endgroup$ Feb 13, 2021 at 17:39
  • $\begingroup$ @safesphere On a Penrose diagram or a diagram in Kruskal-Szekeres coordinates there are two null lines passing through any given event; one is "inward", the other "outward". If the cone is after the horizon then both these lines hit the singularity, but the language of spatial direction is still well-defined. $\endgroup$ Feb 13, 2021 at 17:40

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