leaving black holes
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From: carlip-nospam on 1 Oct 2009 12:39 In sci.astro Bernhard Kuemel <bernhard(a)bksys.at> wrote: [...] > Now if an astronaut orbiting just outside the event horizon of such a > supermassive black hole stuck his hand into the event horizon, would he > not feel any squashing and pulling? Could he still see his hand? Could > he retract it from the event horizon? He would feel his hand ripped off, he would not see it, and he could not retract it. There are two kinds of acceleration here, and it's important not to mix them up: 1. Start with an astronaut in free fall around the Earth. To a good approximation, she doesn't feel any effects of gravity -- she's "weightless" -- because her entire body is experiencing the same acceleration. If she measures more carefully, though, she'll see that she's slightly stretched. If her feet are closer to the Earth, they will accelerate slightly more, since the gravitational field is slightly stronger closer to the Earth. This *relative* acceleration is called tidal acceleration. It is, in fact, the cause of the Earth's tides: the water closer to the Moon is attracted more strongly than the water on the opposite side of the planet, so the oceans are stretched out. For an astronaut orbiting the Earth, this tidal acceleration is tiny. For an astronaut near a black hole, it is much larger. Near a small black hole, the tidal acceleration will tear a freely falling astronaut apart (the technical term is "spaghettification"). Near a supermassive black hole, the tidal acceleration is much less. This seems counterintuitive, but remember that you can get much closer to the center of a small black hole before you cross the event horizon. Tidal accelerations varies roughly as one over the cube of the distance to the center of a gravitating object, and the smaller distance wins. 2. Now imagine that your astronaut is trying to remain outside the horizon of a black hole. She's no longer in free fall; instead, her rocket is using up fuel at a huge rate to keep her from falling. The acceleration she needs to just stay at rest relative to the horizon becomes huge as she approaches the black hole; *at* the horizon, she would need infinite acceleration to remain at rest (or to remain in an orbit outside the horizon). If she now sticks her hand toward the horizon, she's in trouble. Her hand needs a huge acceleration to remain at rest, but it doesn't have a rocket attached; it's accelerating only because it's connected to the rest of her body, which presumably is in her spacecraft. Bones aren't strong enough to transmit a near-infinite acceleration; nothing is. This acceleration -- not the tidal acceleration of a freely falling astronaut, but the acceleration required to remain outside a black hole's horizon -- is the one that's relevant to your question. It goes to infinity at the horizon of any black hole, stellar or supermassive. If your astronaut doesn't want her hand torn off, she will have to jump out of her spacecraft and fall with it through the horizon, so that her whole body is falling at the same rate. (I don't recommend this, though, for obvious reasons.) Steve Carlip
From: dlzc on 1 Oct 2009 13:39 Dear Bernhard Kuemel: On Oct 1, 8:47 am, Bernhard Kuemel <bernh...(a)bksys.at> wrote: > dlzc wrote: > >> Now if an astronaut orbiting just outside the event > >> horizon of such a supermassive black hole > > > The event horizon is at 2M. The lowest stable orbit > > for matter is something like 6M. > > I don't know what M is, but ok. It is a distance in a system of units where g=1, c=1; and is directly related to the mass of the BH. > So let's assume a tough sphere (maybe iron), Iron fails at >6M, for all but large holes. Single iron crystals are small enough (and strong enough) to hold together at 6M. > that withstands the low tidal forces at the > event horizon .... for an ultramassive black hole ... > , passes a black hole (on a hyperbolic or > similar applicable trajectory) so close, that > a small part of it dips into the event horizon. > Is that possbile? No. Essentially, each component of said object is on a different trajectory, and will be shredded by such a passage. Perhaps you have heard of the Rosche limit for gravitationally bound systems? Similar circumstance here, except that very little (if any) of the original body would come out again, and what does will be spread essentially all around the BH travelling radially outwards. Think about "a frog in a blender". > What would happen to the part that dipped > into the event horizon? Would only be seen briefly by the bits that survived to make the trip outwards. > > Depends on the size of the hole. If we > > squashed the Earth into a black hole (to > > satisfy guskz), it would have an event > > hoizon that was something like 4 inches in > > diameter. At our current distance from the > > center, we'd fall normally, not noticing a > > difference in pull between feet and head. > > A size like the one mentioned in the > spaghettification article such that the tidal > forces at the event horizon are not noticable > to humans. .... which is millions of solar masses. Tom Roberts has made a very complete response to you. What is important is that at the event horizon (and "within") there are no vectors that point outwards. The EM contact forces that hold your body together don't propagate "outwards". So you are either going in, or you are losing limbs. Even on a massive hole with a relatively low acceleration. David A. Smith
From: Bernhard Kuemel on 1 Oct 2009 14:26 dlzc wrote: >> that withstands the low tidal forces at the >> event horizon > > ... for an ultramassive black hole ... > >> , passes a black hole (on a hyperbolic or >> similar applicable trajectory) so close, that >> a small part of it dips into the event horizon. >> Is that possbile? > > No. Essentially, each component of said object is on a different > trajectory, and will be shredded by such a passage. Perhaps you have > heard of the Rosche limit for gravitationally bound systems? Actually yes, I have read http://en.wikipedia.org/wiki/Roche_limit before (at least partly). > Similar > circumstance here, except that very little (if any) of the original > body would come out again, and what does will be spread essentially > all around the BH travelling radially outwards. Think about "a frog > in a blender". I see. So even though the tidal forces are weak enough there are other forces that tear apart the body. Is it something like warped space which makes the sphere disintegrate? >> A size like the one mentioned in the >> spaghettification article such that the tidal >> forces at the event horizon are not noticable >> to humans. > > ... which is millions of solar masses. Right. http://en.wikipedia.org/wiki/Supermassive_black_hole : "A supermassive black hole is a black hole with the highest classification of mass, on the order of hundreds of thousands to billions of solar masses. Most, if not all galaxies, including the Milky Way,[2] are believed to contain supermassive black holes at their centers." > Tom Roberts has made a very complete response to you. Yes, but he was talking about thrusters to maintain an orbit or a static hovering posititon. To avoid that I invented the example of a sphere passing by the black hole in free fall at high speed. Bernhard
From: dlzc on 1 Oct 2009 17:43 Dear Bernhard Kuemel: On Oct 1, 11:26 am, Bernhard Kuemel <bernh...(a)bksys.at> wrote: > dlzc wrote: > >> that withstands the low tidal forces at the > >> event horizon > > > ... for an ultramassive black hole ... > > >> , passes a black hole (on a hyperbolic or > >> similar applicable trajectory) so close, that > >> a small part of it dips into the event horizon. > >> Is that possbile? > > > No. Essentially, each component of said object > > is on a different trajectory, and will be shredded > > by such a passage. Perhaps you have heard of > > the Rosche limit for gravitationally bound systems? > > Actually yes, I have read > http://en.wikipedia.org/wiki/Roche_limit > before (at least partly). > > > Similar > > circumstance here, except that very little (if any) > > of the original body would come out again, and > > what does will be spread essentially all around > > the BH travelling radially outwards. Think about > > "a frog in a blender". > > I see. So even though the tidal forces are weak > enough there are other forces that tear apart the > body. No, it is the shape of space. As you get closer, there just aren't as many "choices" of "out" as there are for "in". And finding "out" becomes problematic. > Is it something like warped space which makes > the sphere disintegrate? It is more like a spherical region that even light can't get out of, because *it* can't be accelerated by thrusters. If light can't get out, the binding forces between your molecules fail. Your acceleration inward might be mild, but likewise anything "outward" is even more mild. > >> A size like the one mentioned in the > >> spaghettification article such that the tidal > >> forces at the event horizon are not noticable > >> to humans. > > > ... which is millions of solar masses. > > Right. http://en.wikipedia.org/wiki/Supermassive_black_hole: > "A supermassive black hole is a black hole > with the highest classification of mass, on the > order of hundreds of thousands to billions of solar > masses. Most, if not all galaxies, including the > Milky Way,[2] are believed to contain > supermassive black holes at their centers." > > > Tom Roberts has made a very complete > > response to you. > > Yes, but he was talking about thrusters to > maintain an orbit or a static hovering posititon. > To avoid that I invented the example of a sphere > passing by the black hole in free fall at high speed. And this simply is naieve. He attempted to answer your question as you wanted the answer. Changing the problem only makes what will seem to you to be a completely different answer. You hyperboilically fly around a black hole within (I believe) 3M and you are *never* getting back out again, with finite thrust. And when you stick your hand out, it is essentially in a completely different orbit, and will need to be "tugged" to keep it "precessing" along with you. When curvature is small, this tug is within mechanical limits. At some point, it is not. And this will include regions outside 2M as well. David A. Smith
From: eric gisse on 1 Oct 2009 19:37
dlzc wrote: [...] > You hyperboilically fly around a black hole within (I believe) 3M and Last stable orbit. Arbitrary acceleration can let you get arbitrarily close to r = 2M. > you are *never* getting back out again, with finite thrust. And when > you stick your hand out, it is essentially in a completely different > orbit, and will need to be "tugged" to keep it "precessing" along with > you. When curvature is small, this tug is within mechanical limits. > At some point, it is not. And this will include regions outside 2M as > well. > > David A. Smith |