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The increasing water pressure as you go deeper is generally explained in terms of the weight of the water column above the observation point pressing down. The question, then, is what would happen if you had a big blob of water in free fall, say 100m in diameter-- not big enough to produce large gravitational forces on its own-- and swam to the center of it. I'm imagining some sort of contained environment, here-- a giant space station, or the planet-sized envelope of air in Karl Schroeder's Virga novels, that sort of thing-- so you don't worry about the water boiling off into vacuum or freezing solid, or whatever.

Would there be any pressure differential between the surface of the blob and the center of the blob? At first glance, it would seem not, or at least not beyond whatever fairly trivial difference you would get from the self-gravity of the water. But it's conceivable that there might be some other fluid dynamics thing going on that would give you a difference.

For that matter, what difference would you expect between the pressure inside the water and outside the water? We know from shots of astronauts goofing around that water in free fall tends to stay together in discrete blobs. This is presumably some sort of surface tension effect. Does that lead to a higher pressure inside the water than out? How much of a difference would that be? Or would you not particularly notice a change from sticking your head into a blob of free-falling water? (Other than, you know, being wet...)

(This is just an idle question, brought on by thinking about the equivalence principle, and thus free-falling frames. It occurred to me that somebody here might know something about this kind of scenario, so why not post it?)

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You may be interested to read The Integral Trees and its sequel, which feature exactly these types of ponds floating around in freefall – Brian Gordon Sep 28 '11 at 2:14

indeed there would be a (very small) and homogenous pressure within the blob, coming from surface tension. This pressure is calculated by the Kelvin Equation and is significant in small droplets (reason for small droplets to have a higher vapour pressure than bulk liquid) In Your 100 m blob, this extra pressure is negligible of course. There is another thing in liquids the so called internal pressure, caused by the cohesion forces. (more theoretical) But this You cannot sense in Your blob, because Your body per se always has this internal pressure.

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The surface tension is the only pressure that may be relevant here, indeed. The surface tension of water in contact with air is approximately $0.07 N/m = 0.07 J/m^2$. The total energy stored in the surface tension is a product with $R^2$, and the pressure is the energy density per volume, so it goes like $1/R^3$. To estimate the pressure in $Pa=J/m^3$, just divide $0.07$ by $100$ meters to get $0.7 mPa$ - not too much relatively to $10 kPa$ from a one-meter-tall column of water. The big droplets in free fall barely hold together. – Luboš Motl Feb 5 '11 at 15:22
""The big droplets in free fall barely hold together."" Right, the pictures I remember in early days of space flight, showed strong but very slow "vibrations" of such droplets. – Georg Feb 5 '11 at 15:33

The same pressure throughout.

By the same argument used to show that horizontal bands under a gravitational field must be at equal pressure: if they weren't there would be a net flow.

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I disagree, I think the pressure is the same at equal distances from the center, ie the new horizontal, but increases slightly from the edge to the center. – Jim Graber Feb 5 '11 at 15:27
@Jim: That is true insofar as the blob is a self-gravitating object, but for the size that Chad's talking about this negligible. By the time you are swimming in there you re creating larger pressure gradients with every twitch of your body. – dmckee Feb 5 '11 at 15:33

I know less about this than you do, and I think you answered your own question correctly;
The only significant force is the self gravity of the water.
Just for fun, I calculated the acceleration at the surface and got approximately 10^-6 g.
Therefore the pressure at the center should be approximately 10^-5 atmospheres.

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If you actually tried to swim in to the water, it would clearly disrupt things significantly, absent some other restraint – Jim Graber Feb 5 '11 at 15:32
'Course, the fun part would be to make the blob bigger. At ten kilometers we have an enormous volume available, and the pressure at the center is still quite reasonable (circa 1/7 atm). Intelligent sharks. With laser beams. – dmckee Feb 5 '11 at 16:38
When that "sharking" becomes boring, one can freeze that ball and dig that famous tunnel right down through the center and out again to the antipode point. – Georg Feb 5 '11 at 18:11

I'm not sure how it would behave if the internal pressure is less than the vapor pressure of water at that temperature. If the absolute pressure is lower than that thermodynamics would tend to favor the formation/growth of steam bubbles. You bubble would be (nonlinearly) unstable to boiling. I used the term nonlinear, because surface tension increases the internal pressure in bubbles, and only bubbles that exceed a certain critical size would grow. So your liquid very much resembles soda water, provide some surfaces or bubbles on which the vapor phase can grow and it woul start fizzing.

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""So your liquid very much resembles soda water, provide some surfaces or bubbles on which the vapor phase can grow and it woul start fizzing."" What would happen to the 90 % water in Chads body? Did You read this in Chads question: ""I'm imagining some sort of contained environment, here-- a giant space station, "" ? – Georg Feb 5 '11 at 16:24
The vapor pressure of water in standard conditions is very roughly around 1% of atmospheric pressure. I would hope his space station could apply that much pressure. I would also hope Chad's skin could do likewise (at a somewhat higher pressure, because his body is warmer than STP). – Omega Centauri Feb 5 '11 at 19:30
Have a look into some junior textbooks, Omega – Georg Feb 6 '11 at 12:24

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