# Black hole why wouldn't this work

1. Apr 17, 2012

### vdub

I read that there is no way to escape a black hole once you enter the no-escape-boundary, no matter what you do.
Your ship travels directly towards the center point of the black hole so that it doesn't orbit it. After it passes the no-escape-boundary, you thrust in the opposite direction with a slightly greater force than the gravity pulling it in. So you slowly move from the black hole's center and escape the black hole. Why wouldn't this work???

2. Apr 17, 2012

### dipole

Because it would require an infinite amount of acceleration. Photons can not even escape the black hole - as they approach the event horizon they slow down and simply hover near it, never actually passing it. If a photon can't escape, nothing with mass has any chance to.

3. Apr 17, 2012

### vdub

I could see if the ship were orbiting the black hole. But what if it were traveling towards it like a 1-dimensional line. The force pulling you in is finite, if you push back with a slightly greater force you should still get sucked in?

4. Apr 18, 2012

### K^2

That's basically where you go wrong. Force required to keep an object stationary at event horizon is infinite.

5. Apr 18, 2012

### tampora

This bugs me also. I keep envisioning an Earth with a hole running through it. After falling in, I continue to fall until I reach ground-level on the opposite side. At this point, all of my kinetic energy has been converted to potential energy again before falling back into the Earth. If at any time during my fall, I were to hit my jet pack accelerator, I would be able to get much higher than just ground-level.

Now, crank up the density of the planet until ground-level becomes an event horizon. My potential energy while standing on the surface has increased dramatically, but so will the amount of kinetic energy gained from my fall.

I can only suspect that falling beyond the event horizon would want to accelerate the object beyond the speed of light, which is impossible. Instead, energy is transferred to the object by a different means than kinetic energy, such as mass. Now that the object has increased mass at the expense of original potential energy, it no longer has enough kinetic energy to reach the event horizon, and will remain trapped.

So, yeah.. it bugs me that I should know better.

6. Apr 18, 2012

### Staff: Mentor

There are nice coordinates where this problem can be shown.

If you happen to be at the event horizon, it appears to you that this is moving away from you with the speed of light (lines in a 45°-angle are "moving" with the speed of light) and that the singularity moves towards you even faster than the speed of light. Note that the event horizon and the singularity does not actually move around - it is more like "the space moves towards it". And you are limited to the speed of light relative to this moving space. Once you are inside, there is no way out. Every direction of movement is towards the singularity.

7. Apr 18, 2012

### phinds

HUH ???? Where did you get this idea? Can you site any references?

8. Apr 18, 2012

### PAllen

Phinds is right to question this. As shown in mfb's picture (based in Kruskal coordinates), any light emitted inside the event horizon soon meets the singularity.

9. Apr 18, 2012

### dipole

Well no, not necessarily. In general a light signal will intercept the singularity but one which starts exactly at the event horizon will neither propagate inward or out, it will remain stationary forever.

So I guess what I said was misleading, which is my fault, but the point I was making that even in the best possible scenario photons can't escape, so no spaceship could either.

10. Apr 18, 2012

### PAllen

Please read what you quoted: "any light emitted inside the event horizon soon meets the singularity".

Is light emitted at the event horizon the same a light emitted inside the event horizon? There is no exception to my statement in the black hole region of SC geometry.