Why does the stone stop before reaching the surface at the other end?

[Bjarne]

Imaging a hole the whole way through the Earth, and we will drop a stone into the hole. Imaging there were no resistances at all.

Now the stone would reach the opposite at the Earth and speed here would be exactly zero at the surface at the other end..

Now we will repeat the experiment but this time “Relativistic Resistances” would still count.

It requires more and more energy to get a diminishing increase in speed. – This means that when the stone would reach the center it would not have achieved that speed it is necessary to escape to the surface at the opposite end.

So the stone would now stop before reaching the surface at the other end. (?)
Is that false or correct?

If it is false, - why ?

 

[A.T.]

Imaging a hole the whole way through the Earth, and we will drop a stone into the hole. Imaging there were no resistances at all.

Now the stone would reach the opposite at the Earth and speed here would be exactly zero at the surface at the other end..

Now we will repeat the experiment but this time “Relativistic Resistances” would still count.

It requires more and more energy to get a diminishing increase in speed. – This means that when the stone would reach the center it would not have achieved that speed it is necessary to escape to the surface at the opposite end.

So the stone would now stop before reaching the surface at the other end. (?)
Is that false or correct?

If it is false, - why ?

The red part. It seems you are trying to use relativistic mechanics for the descent but then argue classically about the ascent based on speed. You assume that the "speed necessary to escape to the surface" would be the same in classic or relativistic mechanics.

No matter if classic or relativistic mechanics: The stone converted all potential energy from surface to center into kinetical energy, so it can convert it back to get to the same height. But the speed it has with maximal kinetical energy might be different.

You can also treat the problem geometrically. Since the space time curvature is symmetric, the stone will arrive at the same opposite height. The space time curvature for this case is visualized here:

http://www.adamtoons.de/physics/gravitation.swf

Set initial position to -1 or 1 and initial velocity to 0 for a drop from the surface. As you see the object swings between the surfaces, simply by following a geodesic path in curved space time.