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.
Allure
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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.
timm
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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.
Andrew Steane
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