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September 13[edit]

Teleporting an uncharged black hole 3-20 feet from a human[edit]

Is there a mass range that would not affect him much for a few milliseconds (or at least not kill him) but still cause a very strong hurricane to Mach 1 force wind push towards the black hole after 3 to 20 divided by 1100 seconds? What other effects would occur? Enough radiation to kill you in a heartbeat? Visible (in slow motion) polar jets or accretion disc? A "boom" noise? At what distance would you tear apart? How close before your electrons come off? What would happen to a building you were in? (it's teleported with zero velocity relative to the building instead of typical astronomy speeds) It would then fall to your antipode I guess and go back and forth. How long before Earth is destroyed? If it didn't have to fall then at what distance would it feel like a breeze of wind (before it grows much)? Sagittarian Milky Way (talk) 01:43, 13 September 2016 (UTC)[reply]

Since teleportation is already magic, and you've already violated the basic laws of the universe to do so, feel free to invent your own physics to answer these questions anyway you want to. --Jayron 32 01:47, 13 September 2016 (UTC)[reply]

There's an exceedingly infinitesimal chance of a black hole of arbitrary size appearing anywhere as a quantum fluctuation is there not? So not magic. Sagittarian Milky Way (talk) 04:19, 13 September 2016 (UTC)[reply]

It is fairly straightforward, if time consuming to calculate some of this yourself. I believe our articles on tidal force, Schwarzchild radius, and Newton's law of universal gravitation provide more than enough to let you figure out the tidal force on the human for a given black hole. Someguy1221 (talk) 01:50, 13 September 2016 (UTC)[reply]

I don't know enough physics to know if a hole small enough to have bearable tidal forces would simply not evacuate air fast enough to cause a wind that would blow you in before the hole fell in the Earth. Sagittarian Milky Way (talk) 02:07, 13 September 2016 (UTC)[reply]

I believe you want a large black hole to make the tidal forces bearable (outside the event horizon), as then the gravitational attraction is spread out more evenly. See spaghettification#Inside_or_outside_the_event_horizon. StuRat (talk) 02:37, 13 September 2016 (UTC)[reply]

That says a 1 meter 1 kilogram rope will break if it points to a 10 solar mass black hole about 320 km away. If you were near that black hole which is 3,000,000 times the mass of the Earth but instead of Earth's radius away you were only like 10 miles away then you'd fall in ludicrously fast (the escape velocity at the event horizon is the speed of light) but the tidal force would still be big enough to kill you. A hole the mass of a car would have as much tidal force at several meters from the center as a car but the Hawking radiation would be as bright as a star and it'd evaporate in nanoseconds so if this mass range exists it'd have to be higher than that. Sagittarian Milky Way (talk) 03:17, 13 September 2016 (UTC)[reply]

If I did my math right the g force 20 feet from a black hole who's Hawking radiation is as bright as sunlight at the top of Earth's atmosphere should be a bit stronger than a roller coaster. Unless there's some weird relativity thing that prevents you from using the surface gravity and inverse square law without adjustment like it's Newtonian. If a relativity adjustment is not needed you should only fall in about a centimeter before any sound reaches you. Sagittarian Milky Way (talk) 03:35, 13 September 2016 (UTC)[reply]

Tidal forces might be about 1g per foot. If the radiation that reaches you has a lot of ionizing and gamma radiation then it might be quickly fatal. I can't tell whether you'd get a chance to be sucked in much before dying. Sagittarian Milky Way (talk) 04:08, 13 September 2016 (UTC)[reply]

Look, it's great that you're stretching your imagination, but the science reference desk is here to help you find encyclopedic references. Are you looking for a book on black holes? A good book to start with is The Universe in a Nutshell, which will help realign your imagination with real, actual science. This book is aimed at a general audience and is not overly-technical. Nimur (talk) 15:04, 13 September 2016 (UTC)[reply]

I wasn't imagining, I was using the numbers from [1] that I linked earlier. This Newton's law of universal gravitation calculator gives over 15 g's and over 1.4 g's per foot which I assume is the correct value since 50 trillion black hole radii away isn't extreme enough to make Newton very wrong I hope). I still don't know whether the hole would actually cause an explosive decompression effect or how quickly the Hawking radiation would injure or kill you. I just thought it'd be cool if a size of black hole existed that would not injure or move you much if it appeared in the room until something resembling explosive decompression reaches you at the speed of sound. Maybe it doesn't exist. Sagittarian Milky Way (talk) 16:56, 13 September 2016 (UTC)[reply]

There's a reasonable math question in this. Let's try a rephrase: A small black hole is surrounded by a sphere of vacuum, which is enclosed in, say, a 1 m diameter hard shell, which is surrounded by air at 1 atm. The shell is caused to shatter. Assuming an unlimited supply of atmosphere far from the hole, how quickly will air flow past the 1 meter mark? Which is a very basic accretion disk type question of genuine relevance (at least in the general case) to astrophysics. I think as air rushes toward the hole, there will be back pressure; there will be a certain rate of consumption anyway though. I asked a fairly similar question and got an answer with detailed math here, but I was never really that sure the approximation used there was valid, though I certainly don't know enough to dispute it.

Figuring out where a black hole is deadly is much easier, at least if you don't need a right answer: you basically need to consider where the weight of your butt pulling away from the weight of your trunk is going to pull something you don't want pulled. Like, if the top of you is at 5 g and the bottom at 10 g then you have your butt-weight times five pulling it down away from your trunk. NASA does the math here (slightly differently I think; by all means use their version). Wnt (talk) 21:05, 13 September 2016 (UTC)[reply]

Low pressure environment: does the oxygen move in reverse direction?[edit]

In a low pressure environment, for example, after an aircraft decompression, would the oxygen move from the blood through the lungs to the environment? --Hofhof (talk) 15:30, 13 September 2016 (UTC)[reply]

The movement of gases across heterogeneous phases is determined by Henry's law. --Jayron 32 15:32, 13 September 2016 (UTC)[reply]

It isn't even necessary to have a low total pressure, just a low enough partial pressure of oxygen - pure nitrogen at normal atmospheric pressure for example. See inert gas asphyxiation. --catslash (talk) 22:19, 13 September 2016 (UTC)[reply]

[EC]That does not really answer the question, oxygen is not just "dissolved" in the blood, it's actually bound to hemoglobin so I don't think low pressure alone is enough to make it come out of solution. Nitrogen is dissolved in the blood, along with small quantities of other inert gasses that can come out of solution like in Decompression sickness. In that case, it just happens where ever the blood and the "low pressure" is, as far as I am aware there is nothing to cause the gas to move from the blood to the lungs, our article states "bubbles can form in or migrate to any part of the body".Vespine (talk) 22:30, 13 September 2016 (UTC)[reply]

Oxygen IS dissolved in the blood; the concentration is controlled from both directions: binding to hemoglobin and absorbtion from the air in the lungs. There are equilbria in both directions O_{2 (g)} <--> O_{2 (aq)} <--> O_{2 (bound to hemoglobin)}. The hemoglobin in the red blood cells doesn't absorb oxygen directly from the air, it absorbs it from the liquid matrix of the blood. Actually, there are SEVERAL steps along the way, including dissolving in the plasma of the blood, passing through the cell wall of the RBC, dissolving in the cytosol of the RBC, and finally being bound to the iron in the hemoglobin. --Jayron 32 12:42, 14 September 2016 (UTC)[reply]

I've re-read it a few times and I think I misunderstood the question, "would the oxygen move from the blood through the lungs to the environment" is a very peculiar way of describing "exhaling" :) which happens regardless whether the pressure is low or high. What stops happening then the pressure is too low is not enough oxygen comes BACK through the lungs into the blood. Vespine (talk) 22:45, 13 September 2016 (UTC)[reply]

Also I still think the "oxygen" coming out of your lungs will be in carbon dioxide, not just "oxygen". Vespine (talk) 22:49, 13 September 2016 (UTC)[reply]

Exhaling is just expelling air that's in the lungs. What I was asking is whether we would lose the oxygen that's already in the blood. That appears not to be the case, according to your comment above. I also wonder whether we would be able to hold our breadth in a low pressure environment. Hofhof (talk) 08:16, 14 September 2016 (UTC)[reply]

Just to be clear: I have no expertise in this area, but just took on faith what I read on the inert gas asphyxiation page (After just two or three breaths of nitrogen, the oxygen concentration in the lungs would be low enough for some oxygen already in the bloodstream to exchange back to the lungs and be eliminated by exhalation.). It may need to be corrected.

According to blood#oxygen_transport, deoxygenated blood returning to the lungs is still 70 to 78% oxygen-saturated. According to Oxygen–hemoglobin dissociation curve, blood of that saturation would be in equilibrium with a partial pressure of oxygen of about 32 to 52mm of Hg (taking into account the range of curves shown). If 1 atmosphere is 760mm of mercury, and is 21% oxygen, then the normal partial pressure of oxygen is 160mm of Hg. So deoxygenated blood will lose what oxygen it has to the lungs when the concentration of oxygen in the lungs is 1/5 to 1/3 of that in the normal sea-level atmosphere. You can check the pressure altitude page, but I believe such conditions would prevail if you were exposed to the atmosphere at between 38 and 27 thousand feet. --catslash (talk) 21:19, 14 September 2016 (UTC)[reply]

But (at the risk of appearing to give medical advice), don't hold your breath during decompression (see barotrauma). After decompression, check your altimeter to decide whether breathing would be positively harmful. Even below the altitudes calculated above, there may be other advantages to holding your breath: the consequent build-up of carbon dioxide in your body may prompt it to make the best use of what oxygen it has left. Also the air outside might be harmfully cold. --catslash (talk) 21:37, 14 September 2016 (UTC)[reply]

Dyson ring around a black hole ?[edit]

Find a black hole which isn't predicted to "feed" for millions of years. Then imagine you build a ring far enough out that the structure can withstand the tidal forces. The ring would also rotate to cancel the pull of the black hole. Collect the Hawking radiation as an energy supply to stabilize the orbit of the ring, say with ion engines for orbital station-keeping. So, would this be more practical than a Dyson ring about a star ? That is, could it be closer in and not have to worry about mass coronal ejections, etc ? StuRat (talk) 17:16, 13 September 2016 (UTC)[reply]

For any non-microscopic black hole, the Hawking radiation emits an entirely trivial amount of energy. For example, a black hole the mass of the sun emits Hawking radiation with a temperature of less than one billionth of a Kelvin. You can use this calculator to try different values: [2]. To get Hawking radiation with the temperature of the sun (5800 K), the black hole would need to be only 2 x 10^19 kg, less than 0.000004 times the mass of the Earth, and would have a radius of 31 nanometers. As far as is known, such microscopic black holes do not exist in nature. CodeTalker (talk) 18:09, 13 September 2016 (UTC)[reply]

And even though the effective temperature would be high, the total amount of energy given up would still be very small, due to the small emitting surface. You would need about 100 million of these to power a decent light bulb. --Stephan Schulz (talk) 18:23, 13 September 2016 (UTC)[reply]

Thanks for the responses. So we would need a different energy source. How about if we drop mass into the black hole at a constant rate ? We could use the gravitational potential energy of that mass, along with any energy that escapes as it nears the event horizon. We would need a non-rotating ring to drop the mass from, and we could also strategically drop the mass from different points on the ring for station-keeping. Of course, this requires that we have lots of mass to work with. StuRat (talk) 18:57, 13 September 2016 (UTC)[reply]

A couple of points. First, if gas is fed to a black hole, it will produce radiation as the molecules collide with each other and lose energy. The maximum energy that can be produced in this way is called the Eddington luminosity, and for even a modest sized black hole can be very large. A black hole with the mass of the moon can emit energy equivalent to one Chicxulub impact every second. But the second point is that gravity in the black hole regime becomes non-Newtonian. An object revolving around a black hole emits gravitational waves, which cause it to spiral gradually inward. The rate of spiraling depends on parameters, but can be pretty rapid. The widely reported first observation of gravitational waves in February depended on that effect. Looie496 (talk) 20:12, 13 September 2016 (UTC)[reply]

The structure of the ring should prevent spiraling inward, if there is an equal gravitational force on each side of the ring. However, some form of energy would be needed for station-keeping. StuRat (talk) 20:32, 13 September 2016 (UTC)[reply]

There are a number of ways that have been proposed to extract energy from a black hole, such as the Penrose process which extracts rotational energy (almost all black holes are assumed to be rapidly rotating), the Blandford–Znajek process which also uses rotational energy as well as the magnetic field, or simply finding a black hole that has an accretion disk, which would be naturally emitting copious amounts of energy. The last one, unlike the first two, wouldn't require particularly advanced technology. CodeTalker (talk) 20:22, 13 September 2016 (UTC)[reply]

If you can build your own Ringworld I imagine you can arrange to create an accretion disk around it. I mean, take black hole, add a few human-infested planets, it's a win-win for the cosmos. Wnt (talk) 21:07, 13 September 2016 (UTC)[reply]

You sound like V'Ger: "I've been programmed by V'Ger to observe and record normal functioning of the carbon-based units infesting U.S.S. Enterprise." [3]. StuRat (talk) 22:08, 13 September 2016 (UTC) [reply]

One relevant and interesting fact which I'm not sure has been stated is that a black hole with the mass of the sun, will in effect, behave pretty much exactly the same as the sun. As in, if you replace the sun with a black hole the mass of the sun, all the planets in the solar system would behave just as they do now. There's nothing magical about the gravity of a black hole. Mind you a black hole the mass of the sun will be about 6 km in diameter. So if you can build a dyson ring around the sun, you can build one around a solar mass black hole, i guess it depends on how "close" you want to get to the black hole. Also, i think the energy you could collect out of a solar mass star, is clearly a LOT more than you could collect from a solar mass black hole, I don't think anything you do, throw mass in etc, is going to counter that balance the other way. Vespine (talk) 02:11, 14 September 2016 (UTC)[reply]

Oh no, quite the contrary, if the black hole has an accretion disk. Our accretion disk article says "Accretion process can convert about 10 percent to over 40 percent of the mass of an object into energy as compared to around 0.7 percent for nuclear fusion processes." An accretion disk around a black hole is the most efficient process known for converting matter into energy. CodeTalker (talk) 02:44, 14 September 2016 (UTC)[reply]

Ok well you're making more than a couple of assumptions. Black holes with accretion disks are basically quasars, good luck building a ring around one of those. Vespine (talk) 04:39, 14 September 2016 (UTC)[reply]

A quasar is a supermassive black hole. Ordinary stellar-mass black holes can have accretion disks too. Of course you'd need to make your Dyson ring large enough to be outside the region of dangerous radiation, no matter what kind of central object it surrounds. CodeTalker (talk) 04:50, 14 September 2016 (UTC)[reply]

Right, i was not aware that there were non massive black holes with accretion disks, i guess it makes complete sense. So the elephant in the room remains; harness energy from a jet of x-rays so powerful that it would no doubt obliterate anything in it's wake, perhaps that's not such a great problem to a civilization that can build a ring a million kilometers in diameter. Vespine (talk) 06:22, 14 September 2016 (UTC)[reply]

Would it really need to be that big ? The idea was that you avoid the problems with building near a star, the coronal mass ejections and such. And, if it's more efficient at converting mass into energy, then you can use a much less massive black hole to get the same amount of energy, hopefully also allowing you to build closer in. As for the construction method, I suggest initially placing separate satellites in the same orbit, then linking them up. The earlier comment that any orbit around a black hole will decay would mean some type of station-keeping would be needed at that stage, perhaps something like solar sails, but using radiation from the accretion disk ? StuRat (talk) 11:20, 14 September 2016 (UTC)[reply]

Ah, well, if you're trying to avoid the instability of the environment around a star, using an black hole accretion disk is probably not the best solution. The energy output from accretion disks is highly variable since it's dependent on complicated dynamic frictional effects. Google "accretion disk variability" for a number of papers that seem to indicate that this process is not well understood, although it is clearly observed. You may need to go back to using the Penrose process or other artificial methods like that. Also, Vespine, note that accretion disk jets are strongly directional, emitted from the poles of the system. So possibly it would be safe to reside in the equatorial region around a black hole that is emitting jets, even if crossing the jets themselves would be fatal. CodeTalker (talk) 17:59, 14 September 2016 (UTC)[reply]

Or how about if you use a black hole with such a tiny accretion disk that the jets are manageable and you can collect energy from them ? StuRat (talk) 20:47, 14 September 2016 (UTC)[reply]