What would the pressure be like in a body of water in free fall?

Submitted by OneTreePhil t3_yx9z3q in askscience

I was thinking about the giant free "ponds" in Larry Niven's Integral Trees. On a planet, pressure as a function of depth is fairly straightforward. But if the water is held together only by cohesive forces, how does it change with "depth"? In the book they were free floating and various sized

Water is less dense as a solid so pressure causes it to melt, so I'm not thinking there could be a solid core by pressure. I'm wondering, if there was a drop of water 14 miles across (so the center is as far from the surface s the Challenger Deep), what would the pressure be at the center? Could we scuba all the way through?

Maybe it would help to imagine a spherical "drop" of water of radius (depth) x, in free fall, with enough of a "balloon" around it to prevent heat loss and evaporation but not add and "squeeze pressure." How would pressure change through the pond? How big would it have to be to noticeably influence itself by gravitational effects?

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cowox93112 t1_iwp22tg wrote

Free fall is equivalent to being in a region of space without gravity. Let's also assume a vacuum around it such that we can neglect external air pressure etc.

There is one force left, namely cohesion. It will attempt to minimize the droplet surface, morphing the it into a spherical shape. The pressure at the center is given by P= 4 gamma/R with R being the droplet radius. It is tiny for any sizeable droplet. It also gets smaller for larger droplets. until the droplet becomes massive enough to generate its own sizeable gravity.

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OneTreePhil OP t1_iwq33c1 wrote

That's a combined curve I'd like to see graphed. I hear excel calling me now!

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mfb- t1_iwp25p9 wrote

The blob of water is just floating in space (probably surrounded by some shell to avoid evaporation)? You would only get some pressure from gravity of the water. With a surface gravity of 4 mm/s^2 we get a pressure difference of 28 kPa or 0.28 Earth's atmospheric pressure between surface and center (equivalent to 2.8 m of water on Earth). You need some pressure at the surface to keep water liquid, going to the center won't change the total pressure much.

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ShoulderToCryOn00 t1_iwp58zp wrote

On the same issue, What would happen if a giant balloon (say 100m across of water) was put in space and then the ballooncarefully cut away? Would the water maintain a sperical shape untill it all evaporates? And how would the evaporation look like, would the vapour radiate ourwards like the energy of a star? Since there is no up or down in space.

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mfb- t1_iwp8462 wrote

I don't think you can remove it carefully. At low pressure water boils, which increases its volume a lot and cools it until you are left with a mixture of water vapor and ice. That could be a pretty explosive process if you suddenly remove the source of pressure.

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ShoulderToCryOn00 t1_iwq4hmt wrote

Thanks man. How could I forget about pressure lol. I had not even considered it for some reason.

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Adventurous-Quote180 t1_iwq9qzf wrote

And what if we would do this in a hypothetical space that has no gravity, but has a similar atmosphere to earth? (Like maybe we would have a giant room filld with air, frloating in space, that has walls made out of some nearly weightless but strong material. And inside this box would we have tha previously mentiond water balloon experiment)

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mfb- t1_iwqjfhj wrote

Then we get a ball of water with uniform pressure.

Or almost uniform, if we use 100 m and gravity.

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OneTreePhil OP t1_iwqri5u wrote

How did you get 4mm/s2?

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mfb- t1_iwqtws6 wrote

I misread the size and used a radius of 14 km, but reducing it to 7 miles only changes the acceleration to 3 mm/s^(2). Calculation.

The mass is 4/3 pi r^3 rho, the acceleration is a = Gm/r^2 = 4/3 pi r rho G. That's not limited to the surface, it applies everywhere inside the sphere, so we get a linear relationship. The pressure gradient is a*rho, integrated over the radius it's 1/2 r a_surface rho = 2/3 pi r^2 rho^2 G = 18 kPa with the fixed radius.

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colcob t1_iwp2c3o wrote

A big drop of water floating in space could only have such pressure internally as would be caused by its own gravity. A rough calculation says that surface gravity on the mini water planet would be about 1/10’000th of earth gravity, so pretty small, but which would develop about 1.5 psi of pressure under 7 miles of water.

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OneTreePhil OP t1_iwq3boa wrote

Wow so you would actual suffer from decompression if you just "beamed" there

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Ok_Construction5119 t1_iwre4op wrote

Yeah, the pressure of your 14 mile droplet is noticeably less than atmospheric pressure

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Chemomechanics t1_iwqj6cd wrote

>Water is less dense as a solid so pressure causes it to melt, so I'm not thinking there could be a solid core by pressure.

Liquid water is less dense than ice-Ih (the familiar crystal structure) but not less dense than ice-VI, into which liquid water would transform at room temperature given enough pressure (and a large enough ball of water). It's an interesting exercise to estimate the required radius.

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Ok_Construction5119 t1_iwrdsz1 wrote

The surface tension of water is negligible compared to the other forces at work here. The "pressure" from water is due to the weight, not the mass, making it entirely dependent on gravity (P = rho * g * h). Thus, if g = 0, as in the case of freefall, P is also 0.

If you are asking about the pressure changing due to the water itself, it would have to be much larger than 14 miles to have enough mass to exert gravity that would be detectable by human senses.

As stated earlier, surface tension is negligible compared to these forces. According to wikipedia, the surface tension of water is 72.8 mN/m, which is again so small you could barely feel it. This is what is known as an intensive property, meaning it is not dependent on the amount of a substance.

As gravity is directly proportional to mass, it would need to have the same mass as the earth to have the same pressurization capabilities due to gravity. See newton's law of universal gravitation if you are still curious.

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OneTreePhil OP t1_iwrfycp wrote

So, since the earth is 5.51 g/cm3 overall, I would need a "drop" of water 5.51 times the volume of earth to get the same pressure vs depth relationship?

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OneTreePhil OP t1_iwrh5ak wrote

I get about 11300 km radius water drop to get the same g as earth. So the preassurevat Challenger Deep depth would be the same?

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Ok_Construction5119 t1_iwriglk wrote

Yes, your density to mass conversion is correct.

V,h2o [m^3 ] * rho,h2o [kg/m^3 ] = V,earth [m^3 ] * rho,earth [kg/m^3 ]

With the density of water, volume of earth, and density of earth constant, you can use this equality to find the required volume, which will be proportional to the ratio of densities.

As you have already done, multiplying volume by density yields mass, and with constant density, volume is directly proportional to mass.

Lots of words to say yes, you are correct in terms of volume required to give the same gravity given the above assumptions.

This next paragraph relies on many invalid assumptions: However, due to the different volume and identical gravity, the pressure would increase more slowly with depth than on earth. This is because water is in fact slightly compressible, and under such enormous forces it would in fact be more dense as you approached the core. If you took the density as an average density, which is inaccurate, you would see a pressure/depth change similar to how it is on earth, with a similar pressure to the earth's core at the core of your big water droplet, and a similar pressure at the depth of the challenger deep that you would see on earth.

Again, p = rho g h, so if g = G,earth, and rho = rho,h20, then p will increase linearly with the 'height' of the water above you

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OneTreePhil OP t1_iwrmb1x wrote

Thanks for the awesome response I'm pretty much exactly where I was hoping to be with this.

Now all we have to do is find one to check!

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