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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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