Pete's pretty much got the right of it, yeah. For a 'traveller' falling into a black hole, and an 'observer' sitting stationary at a relatively safe distance, the whole experience appears quite different. The observer sees the traveller's clock slowing down, and their image redshifting, as they fall towards the horizon, until at the very moment they would reach it, the traveller's clock seems to stop as their image is redshifted beyond observability.
The traveller, of course, sees none of this. To him, his clock just keeps on ticking, and while there are probably some pretty spectacular optical effects (dealing with gravity bending light in amusing ways), he'll never really even know when he passes the event horizon. Long before he reaches the singularity (or whatever they have inside them) he'll get spaghettified.
The disruptive tidal effects of black holes come from the gravitational gradient of the mass of the black hole. For low-mass black holes, you have to get very close to them before you cross the horizon, and the gravity gradient becomes quite steep: across even the small distance from your head to your feet (assuming you're falling feet-first), the force of gravity increases radically... you get stretched, and squished thin at the same time... spaghettified.
For larger-mass black holes, the horizon can have very large diameters, well beyond the point where the gravitational gradient becomes deadly. So you get to disappear from the universe
before you get mangled horribly.
It all comes from the inverse-square behavior of the force of gravity, really.
For the purposes of this example, let's chop the traveller into two halves, and make those halves point-masses separated by one meter (just to keep things simple... and you thought the black hole was dangerous!
). Now, my math isn't terribly precise here, and I'm sure as heck not gonna bother with Relativity (which really would apply at these energies), so I'll keep things in the nice, simple, but not exactly accurate Newtonian realm. But the gist of this should be close enough for jazz.
For a one-solar mass black hole, the event horizon has a radius of:
Rs = 2GM/c^2
= 2950m (almost three kilometers)
So let's say our traveller starts out stationary, with their 'feet' at the horizon, and their 'head' one meter above it.
So for their feet, at radius R (2950m) the acceleration due to gravity is:
a(feet) = GM / R^2
= 1.5252e13 m/s^2
And for their head, at radius R+1 (2951m)
a(head) = GM / (R+1)^2
= 1.5242e13 m/s^2
Those don't look very much different, do they? But let's subtract the second from the first...
delta-a = a(feet) - a(head)
= 1e10 m/s^2
Which is a difference of
ten billion meters per second squared, over the distance of one meter. That will kill you dead. Oh yeah.
Now if I do the same for a one million solar mass black hole, with an event horizon radius of:
Rs = 1.475e9 m (nearly one and a half million kilometers)
a(feet) = 6.1009135e7 m/s^2
a(head) = 6.1009135e7 m/s^2
Now, those are pretty much equal down to an error level of one meter per second, so the gravity gradient is pretty gentle out there, despite the fact that the traveller is right at the point of no return.
Weird, huh? That's why big black holes aren't as immediately lethal as their smaller cousins. At least when you're at the event horizon, that is.
Dr. Skeptic: not all black holes are necessarily surrounded by an accretion disk. I'd probably guess that the majority of stellar-mass black holes
aren't surrounded by such, at least not for very long, unless they happen to have a companion star upon which to draw matter. But you're definitely right... an accretion disk would be an hideously unpleasant sort of environment.
