How much water could a black hole suck?

Suppose you took a black hole with a mass of, say, 1000 metric tons and threw it into a sufficiently large swimming pool.

At what rate (in liters per second) would it suck up water?

You can’t get a black hole that small, in terms of mass.

it would suck as much water as a black hole could suck if a black hole could suck water

Aside from the fact that it’s debatable whether you could or could not have a black hole that small, bear in mind that you can’t put a black hole of any size in a swimming pool anyway.

It’s a hypothetical question, obviously. Given a black hole of arbitrary mass m, what equation will give you the rate r at which it would suck up water or some other material around it?

One might imagine that it would be the same rate as if the black hole were just a drain with the same area. The problem is that the rate at which water goes down a drain is limited by the fact that the water just inside the drain gets in the way, which isn’t a problem for a black hole.

Suppose you have a blackhole with an effective gravitational radius r. It can suck anything within pi*r[sup]2[/sup] in. r is a function of mass - the thing is, as a blackhole sucks up matter, it increases in mass, thus increasing r. IIRC, such an increase will apporach a limit depending on initial mass.

Also note that small black holes (much larger than 1000 m in diameter, however) are very fiercely radioactive and don’t last very long. I wouldn’t be surprised if there were some very firm limits on just how small a black hole can actually be.

Of course, if you begin to ignore those properties of black holes, you’re essentially postulating an imaginary entity to which you can attribute any properties whatsoever.

(In other words, if you make shit up you get to make the rules.)

A black hole of 1000 meters mass, not diameter…
the diameter of such a thing would be smaller than a proton.
It would therefore have a slight difficulty sucking anything in at all.
this would have to be artificially created, and would radiate energy by Hawking radiation so wouldn’t last long…
what you would have in fact would be a rapidly evaporating artificial nugget of stored energy, which couldnt suck a single proton in because it was too small, but would rapidly convert the swimming pool to steam…

That should read 1000 metric tonnes mass, not metres diameter…

Here are the numbers for a black hole of 1000 metric tons…

Radius: 10[sup]-21[/sup]m. Much smaller than a proton.
Luminosity: 10[sup]20[/sup]W. About 6-7 orders of magnitude less than the luminosity of the sun.
Mass loss rate due to Hawking radiation: 1.5 tons/sec
Lifetime: 220 seconds.

Also, although a black hole of mass 1000 metric tons is fiercely radioactive, a black hole of radius 1000 m is not. In general, all black holes give off Hawking radiation which has a wavelength about 25 times larger than the radius of the black hole itself. Radiation of wavelength 25 km is nothing to worry about. A black hole this size is about one-third the mass of the sun.

Now, the question in the OP is a little trickier. I’ll have to make a lot of assumptions to get any sort of an answer, but here goes. Suppose you had a black hole embedded in an infinite medium of water. It would begin to accrete (suck up) matter onto it and come into equilibrium at some accretion rate. This is not exactly the same thing as the question in the OP, because, among other things, you’re not throwing anything at the black hole - you’re letting it pull it in itself. Bondi (1952) gives the mass accretion rate (how much mass it sucks up per second) as:

A = 4 pi lambda × (GM)[sup]2[/sup]/c[sup]3[/sup] × rho[sub]infinity[/sub]

This is equivalent to a volume accretion rate of:

VAR = 4 pi lambda × (GM)[sup]2[/sup]/c[sup]3[/sup]

I don’t know what a good value lambda is, unfortunately, but it’s probably about 1. So putting in M = 10[sup]6[/sup]kg, we get

VAR = 2 × 10[sup]-30[/sup] liters/s.

This is probably a lot slower than you’d expect, but you have to realize that being a black hole does not give an object more gravity, and 1000 metric tons is not all that much mass. At one centimeter away from this black hole, you’d feel less than one-tenth the gravity you feel on Earth. At one meter away, you’d feel less than one 100,000th Earth’s gravity.

Anyway, if you take Hawking radiation into account, then it’s losing mass at a much faster rate than it’s gaining it from accretion. According to my calculations, this will be true of any black hole smaller than 10[sup]11[/sup] metric tons (one ten-billionth the mass of the Earth).

Perfect answer, Mr Achernar.
Such a luminious object would flash the swimming pool into an expanding sphere of glowing gas, and continue to do so for 220 seconds- quite a handy weapon, except it would need an impossibly large array of accelerators to provide the collison to create this object.
#Unless anybody has any other ideas.

Bah. That’s what you get for skimming through the units part of the OP.

Achernar, great answer. :slight_smile:

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Actually, the mass isn’t important. The real question is, given a black hole of mass m, how do you figure out the rate r at which water will disappear from the pool into the black hole?

The reason I wasn’t considering Hawking radiation was mostly because I was interested in the aspects of the question dealing with “if you had a boundary beyond which things would disappear, how fast would stuff go into that boundary?”

I was kind of expecting a statistical mechanics answer, because I figure that it’s like asking a question like this:

“Suppose you have a box with a divider, and half the box is full of water. At t=0 the boundary disappears. At t=0, how fast is that gas going to the other half of the box?”

Of course, there’s some question as to how accurate such an approximation would be, and it gets harder when you use water, particularly because once the water at the surface of the hole disappears, Earth’s gravity alone will pull more water in to replace it, just like it pulls water into a drain. (And, I might add, even if the drain is in the side of the pool, since the weight of the water in the pool is pushing water out the drain.)

A black hole of the appropriate size adds an additional (but, I think, computationally pretty simple) complication of the hole’s gravity sucking in the water.

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This is, indeed, the question I’m trying to ask.

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What do all these variables mean? When I get a chance, I’ll try to look up your cite and see how this was derived.

Actually, the real issue isn’t the gravity of the hole at this point, since the Earth’s gravity will force the water through a drain.

The real issue, I think, is that a hole that small occupies so little volume that it becomes a question of (like I said) statistical mechanics: what is the probability that, within 1 second, a molecule of water will hit the hole and disappear?

Once you have a black hole whose size is comparable to an actual drain (say 1cm in diameter) the gravity of the hole becomes a major issue, and I think that in that size regime Bondi’s equations are more appropriate. (Although they certainly generate believable figures as is.)

Actually, I think that Bondi’s accretion solution addresses your concern quite nicely. The volume accretion rate r (or as I called it, VAR) is (within a factor of 2) equal to the area of the black hole event horizon times c, the speed at which stuff could be falling in.

That’s a good question, and, I admit, not the one I answered. I assumed the driving force on the water was gravitation, but it sounds like you want the driving force to be the thermodynamic motion of the the fluid. I don’t know enough to answer that, but it doesn’t seem so hard. One thing, though, the black hole you specified is much smaller than a water molecule, so the problem as stated is probably not what you wanted to ask.

Yes, but like I said, Earth-like gravity only acts over a very very small volume for a tiny black hole. So at any appreciable distance, the water has very little weight. Thus, in the case of a black hole, the pressure on the water due to what’s above it will be very small.

The reference is “On Spherically Symmetric Accretion”, by H. Bondi, Monthly Notices of the Royal Astronomical Society, Vol. 112, p.195. 1952. Equation 10. Here are what all the letters mean:
G and c are universal constants.
A is the accretion rate.
M is the mass of the accreting object.
rho[sub]infinity[/sub] is the density of the medium (water) if you don’t take the effects of the accreter into account. Since rho is mass / volume, to convert from mass accretion rate to volume accretion rate, I just divided both sides by rho.
lambda is given as “the non-dimensional parameter determining the accretion rate”. I don’t see much about it. I don’t think think it’s a constant you can derive; it has to be measured. And I’m not even sure it’s a constant.

I think that’s a good way of looking at it.

Ah, I see- you plug the Schwarzschild radius into 4pi*r^2, and multiply by c.

My problem with this is that it’s not clear to me why you should assume a speed of c for the infalling matter unless the matter started out an infinite distance from the hole. I’m wondering if that’s where the lambda comes in- an empirical fudge factor that corrects for the speed of the infalling matter. Has the speed been measured empirically, by Doppler measurements on accretion disks?

The Bondi accretion problem does not assume that the object accretes at c, but it does assume that the medium starts at an infinite distance. This isn’t all that bad of an assumption, though. For an accreter of mass 1000 tons, the escape speed at a distance of only 10 meters is about 0.003 m/s. The orbital energy that it has as this distance is some 20 orders of magnitude less than when it’s accreting, so the initial distance is wholly insignificant.

Shouldn’t the escape velocity be c at the event horizon? I thought that meant that material starting at infinity with velocity=0 would be travelling at c when it hit the event horizon.

That’s roughly correct. However, if you want a precise answer, you’d have to think in terms of GR.

Ah, thanks.

The manual for my Sears Quantum Singularity X-5 Wet/Dry Vac says that it’s rated to suck 4.766 x 10[sup]8[/sup] metric tons of water per second. Does that help?

Peace.

Hand me that quantum singularity, please.