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Inside the event horizon is a singularity and it has a surface. Given that even light can't escape the pull once the event horizon is crossed, does that mean that anything that crosses the event horizon will be accelerated and will be moving at, at least, $c$ after doing so? Due to relativism, how long would it take for the object "hit" the singularity?

In other words: I have an indestructible ship.
It has a not-so-miraculous engine that allows me to control my approximation to the black hole in a manner that my speed to the EH is 1m/s at all times.
Once I cross the EH, my engine is out of fuel.
When I cross the horizon I will start my clock. What will be the number on the clock when I hit the singularity?

EDIT 1:
Details: Lets assume the black hole is not rotating, or it is, whatever is the simplest and easiest to answer. Same thing to the charged.

The ship crossed the EH perpendicularly, like a ball balanced on a pin.

Forget the speed of the ship, imagine that the ship is being kept at a distance of 1 meter of the EH by a magical cable that will be cut whenever we want

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    $\begingroup$ This is underspecified. Is the hole rotating? Charged? Does the ship cross the horizon radially, or at some angle? In what coordinate system is the ship moving at 1 m/s? $\endgroup$
    – G. Smith
    Mar 24 at 20:09
  • $\begingroup$ @G.Smith I provided such details... let me know if you need more $\endgroup$
    – Leonardo
    Mar 24 at 20:16
  • $\begingroup$ Here is the calculation for various black holes up to the biggest one known: physics.stackexchange.com/questions/618006/… $\endgroup$
    – safesphere
    Mar 25 at 5:50
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What will be the number on the clock when I hit the singularity?

For a Schwarzschild black hole, a radial geodesic for a massive object satisfies

$$\left(\frac{dr}{d\tau}\right)^2=\frac{E^2}{m^2}-1+\frac{2M}{r}\tag1.$$

Here $M$ is the mass of the hole, $m$ is the mass of the falling object, $r$ is the object’s radial Schwarzschild coordinate, $\tau$ is the proper time measured by the object, and $E$ is a constant of the motion, identifiable as the object’s energy. Units here are such that $G=c=1$. The event horizon is at the Schwarzschild radius $r=2M$, and the singularity is at $r=0$.

If the clock falls from rest at the horizon, then one can set $r=2M$ and $\frac{dr}{d\tau}=0$ to evaluate the constant as $E=0$, so (1) simplifies to

$$\left(\frac{dr}{d\tau}\right)^2=-1+\frac{2M}{r}\tag2.$$

One can separate the $\tau$-dependence on one side and the $r$-dependence on the other. For an infalling clock, choose the negative square root:

$$d\tau=-\frac{dr}{\sqrt{\frac{2M}{r}-1}}\tag3.$$

To get the freefall time, integrate from $r=2M$ to $r=0$ to get

$$\Delta\tau=-\int_{2M}^{0}\frac{dr}{\sqrt{\frac{2M}{r}-1}}=2M\int_0^1\frac{du}{\sqrt{\frac{1}{u}-1}}\tag4.$$

The integral evalutes to $\pi/2$, so the result is

$$\Delta\tau=\pi M=\frac{\pi GM}{c^3}\tag5$$

where the final step restores factors of $G$ and $c$.

For a one-solar-mass hole, this is 15.5 microseconds. Supermassive black holes can have ten billion solar masses, which makes the freefall time a day or two... perhaps enough time to conduct some experiments on the interior of the hole (the results of which cannot be communicated to us) before you get spaghettified.

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  • $\begingroup$ I didn't actually needed the math, but its beautiful $\endgroup$
    – Leonardo
    Mar 24 at 21:36
  • $\begingroup$ Where the does the Schwarzschild cöordinate $t$ show up? You refer to this coördinate after the geodesic . $\endgroup$ Mar 24 at 21:50
  • $\begingroup$ I've edited that out. It isn't used. $\endgroup$
    – G. Smith
    Mar 24 at 21:51
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When you free fall towards the singularity, never special happens to you if you cross the EH. To you. For an observer far away from the BH it seems as if you do all thing in slow-motion. If he's able to see you at all because the wavelengths of photons you emit (which make him see) are red-shifted for the observer. This redshift approximates infinity when you approach the EH.
What will you see? Dependent on the size of the BH, it will take a finite time to reach the singularity. In a supermassive BH, the distance to the singularity will be larger than the distance in a small one.
The material out of which your rocket is made must be very tough though. And so must you. The tidal forces grow to a value that will tear you and your spaceship apart before being swallowed by the singularity.
So you can better ask what will happen to an elementary particle. It can't get torn apart by the tidal forces. Near the singularity, spacetime curvature goes to infinity. Space and time are curved. Timewillalmost come to stop, while space is stretched beyond imagination. But the particle will reach the singularity (how else could it have come into being?). Even if space is intensively stretched. This stretching is compensated for by true time dilation (so not only a relative one).

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