An astronaut launched into space
can you imagine the look on his face
When his fuel was gone
and he thought he was done
with his universe travelling days!
We have an astronaut located at a shell near a black hole. He just launched from this shell with the hopes of getting away from it when he suddenly ran out of fuel. After some brief panic (and a couple of breathing exercises), he decides to do the math. Is he actually doomed, or will he be driving home for Christmas after all?
When we are near a black hole we have a potential. You may be familiar with potential energy, and they should not be confused. A potential is an objects ability to do work. Low potential means low ability to do work, and the other way around. We want to look at the gravitational force's ability to do work on our poor astronaut to see if he can escape the black hole. In the case of gravitational force, it only works one way, and that is inwards. Bad news if you're too close!
Let's look at the expression for potential:
\(V_{eff}(r) = \sqrt{\left(1 - \frac{2M}{r} \right) \left [1 + \frac{(L/m)^2}{r^2} \right]}\)
It may look ugly at first, but we are only going to talk about the general concepts and let the astronaut do the math behind the scenes. \(M\) is the mass of the black hole, \(m\) is the mass of the astronaut, \(r\) is the astronaut's distance to said black hole, and \(L\) is the spin. (Which we definitely will be doing behind the scenes)
For our astronaut to know whether he is safe, he needs to know where the potential is at its lowest and highest. We need the extreme points of the function, which we find by differentiating \(V_{eff}(r)\). We then find the points where \(V_{eff}'(r)=0\). (This should be familiar math) We get the points
\(r_{extremum} = \cfrac{\cfrac{L^2}{m^2} \pm \sqrt{\cfrac{L^4}{m^4} - \cfrac{12L^2M^2}{m^2}}}{2M}\)
This also looks messy, but it's nothing but a bunch of numbers that the astronaut will have to punch into his calculator. So what does he do now that he knows where the potential has its extreme points?
He needs to compare his position to the extreme points. If he is at the lowest point, he is safe. If he is at the critical point, he is in a very precarious position, as he can no longer use his engine. The good news is that even if he dips past the critical point, it may not be too late, as long as he gets his engine working! As long as he is just outside the event horizon (which is located at \(r=2M\)), he may yet save himself. As the astronaut is very far away from us, we didn't receive his position before the deadline of publishing this post, but we hope he is further away than \(r_{crit}\) and makes it home in time for Christmas.
So let's think of the worst case scenario while we wait! If he does fall into the black hole, he won't feel a thing at first. The event horizon isn't a magical place where time and space suddenly breaks down for an observer within it. (For his loved ones outside of the event horizon it will look like he is moving slower and slower until he completely stops) Unfortunately, as he gets nearer the singularity of the black hole, he will definitely start to feel something.
Let's say he exits his spaceship (because that makes the illustration look much cooler) and falls into the black whole feet first. The gravitational pull will be much stronger on his lower body than his upper, and he will be ripped to pieces. (Or really elongated, depending on who you're asking. Smarter people than us are still debating this.)
If he, for some reason, wore one clock on his big toe and one on his wrist, he would observe that time passes differently on his toes compared to his wrist. This is because time passes differently in an inhomogeneous gravitational field, which a black hole definitely is, as the forces (or "forces") works radially inwards.