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In Newtonian mechanics, if we throw an object in against direction of gravity with speed $v$ and it achieve max height of $h$. Now if we allow object to fall from that height $h$, it will eventually attain speed $v$ when it reach position where we launch it.

Now applying same idea to a black hole in general relativity. Speed require to escape black hole gravity is greater than $c$, so if we throw something into black hole with almost the speed of light, the object speed will exceed speed of light $c$ before hitting black hole surface! How relativity explain this? Can space-time curvature reduce speed of this freely falling object from attaining speed of light?

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Your analogy is based on Newtonian mechanics, which is not applicable to strong gravitation field around black holes. – Siyuan Ren Apr 24 '12 at 6:01
up vote 10 down vote accepted

To answer your question you need to be clear what co-ordinates you're using. If you use co-ordinates that are co-moving with the rock falling into the black hole then the rock will see the event horizon pass at the speed of light.

External observers, using Schwarzchild co-ordinates, will see the rock slow down as it approaches the horizon, and if you wait an infinite time you'll see it stop.

External observers obviously can't comment on the speed of the rock after it has passed the event horizon because it takes longer than an infinite time to get there. If you use the rock co-moving co-ordinates then you can ask what speed you hit the singularity and ... actually I'm not sure what the answer is. I'll have to go away and think about it.

Incidentally is a fun site describing what happens when you fall into a black hole.

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very informative... thanks – someone_ smiley Apr 24 '12 at 8:55
Although I +1'd this at a quick glance, on second reading, you absolutely need to say that the "slowing down" is a coordinate effect, that you don't actually slow down relative to an outside observer, since the coordinate distance you travel per unit coordinate time (scaled by the metric) is going to c even for the external observer, it's just that the speed c in external coordinates is frozen at the horizon, since the external coordinates are symmetric between white hole and black hole, they do not distinguish the sense of the horizon, so there is no crossing of light through the horizon. – Ron Maimon May 1 '12 at 21:36
@Ron Maimon: So if they don't "actually" slow down, does that mean they even pass the horizon? If so, at what speed are they going then? – mike4ty4 Nov 20 '14 at 0:26
@mike4ty4: The coordinates where the objects freeze are degenerate on the horizon, there is no paradox. The objects freeze because the coordinate time stops, not because their intrinsic velocity is slow. It requires knowing the metric form at a horizon, something you can work out for Rindler space easily, because it's just flat Minkowski space in disguise. – Ron Maimon Nov 21 '14 at 4:15

Nope, it will just fall in a reasonable amount of time (if you go with it, but watch out for tidal forces!), or take forever to fall in (if you are watching from outside).

Also, if I may be so bold as to suggest doing some quantum mechanics instead of kinematics while you are there, you could probably lock down some funding no problem.

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+1 for you could probably lock down some funding :) – Manishearth Apr 24 '12 at 7:18
ohhh yes, that make sense, gravity make time slower, so it would take infinite time for object to hit the black hole. thanks. and well am novice to quantum mechanics. will explore it for more :) – someone_ smiley Apr 24 '12 at 8:31

It will reach the speed of light exactly at the black hole surface.

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protected by Qmechanic Aug 31 '13 at 19:05

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