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The event horizon is the boundary, from where light cannot escape. But can anything else?

Something with real mass will gather inertia while falling into the black hole. First it will mainly increase its speed than approaching the speed of light, its mass will grow, giving it inertia and kinetic energy just enough to be able to climb the gravity well on the other side.

If I let go of a toy car (with insignificantly small wheels, to avoid having to account for the rotational inertia, and no friction) into a valley, it trades gravitational potential energy and kinetic energy to go through the valley, and get out on the other side.

So, my question is: if I let go of my toy spaceship around a black hole in a way that some of its orbit is below the event horizon, why does it not emerge after going inside the event horizon? It has all its inertia racked up while falling inside, it should be just enough to get out!

Does something take its inertia/kinetic energy? Black-hole friction? Does its potential energy converted to some type of energy other than kinetic, which cannot be traded back for potential energy?


EDIT:

I got some pointers, but unfortunately I don't really understand, how these answer my question. So I come up with the following:

Go 3D:

This is a section of a rotationally symmetric object (a well, with oblique lips at the mouth), the symmetry line is the -.-.-.- one (vertically).

\      |      /
 \     .     /
  \    |    /
   |   .   |
   |   |   |
   |   .   |

The topside view of same object is 2 concentric circles:

        ---
       ( o )
        ---

If I let a ball loose from the edge from some inertia, so that the ball does not fall into the vertical drop of the well, it will have a curved path at the slanted lips of the well. This is the same as an object falling towards a black hole, but coming out without touching the event horizon. It speeds up downwards, and looses that speed when coming back up.

\
 \
 (_
   \
    \

If an object falls into the black hole, it is the same as the ball reaching the vertical-walled part of the well, it will keep on accelerating, converting potential energy to kinetic energy, but without the slimmest hope of ever getting back out, spiraling downwards forever.

I don't know, if this is a good explanation or not, though.

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    $\begingroup$ As has been stated here so many times, within the horizon of a (Schwarzschild) black hole, time 'points' toward the singularity. To cross the horizon from within would amount to traveling backwards in time. Have you spent any time searching the existing Q & A here for this (or a related) question? $\endgroup$ – Alfred Centauri Jun 24 '18 at 2:25
  • $\begingroup$ @AlfredCentauri Yes, I know about this fact. Yes, I have spent some time researching it, but I could not find anything that I felt answers my question / that I understood. $\endgroup$ – Zoltan K. Jun 25 '18 at 11:52
  • $\begingroup$ I think that your approach is naive. General relativity figures out non ordinary configurations. Technically the explanation is that the time and radial coordinates change nature and the light cone tilts of $\pi/2$. Any object, matter or light, that crosses the horizon can only proceeds in the direction of the singularity at the center of the black hole. $\endgroup$ – Michele Grosso Jun 25 '18 at 16:18
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if I let go of my toy spaceship around a black hole in a way that some of its orbit is below the event horizon, why does it not emerge after going inside the event horizon?

It can't because beyond the event horizon the flow of time is represented by a decreasing r-coordinate. This means that time-like and light-like geodesics end up in the singularity. You can see this from the metric if you chose r less than the Schwarzschild radius. Then the r- and t-coordinates interchange their role.

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