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Well for example we put a ball filled with water, with a density of $1\cdot 10^3 kg/m^3$, in space.

How to calculate the pressure the water is exposed to?

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  • $\begingroup$ Do you mean you put a sphere of water in a vacuum, or in an atmosphere but zero-G like the ISS? $\endgroup$ – John Rennie Feb 12 '14 at 10:31
  • $\begingroup$ @JohnRennie the latter $\endgroup$ – sqd Feb 13 '14 at 9:56
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Consider a ball of water floating in zero G, as demonstrated on the ISS. Ignoring for the moment the surface tension of the water (I'll come back to that) the pressure inside the water is the same as the pressure of the air around it. This is simply because without any forces, like gravity, acting on the water there is nothing to cause a pressure gradient.

Well the pressure isn't quite the same. In my first paragraph I mentioned that the air-water interface has a surface tension, and this acts a bit like the elastic of a balloon. It compresses the water inside the drop and increases the pressure slightly. The Hyperphysics site has an article deriving the pressure and the result is that the extra pressure is given by:

$$ \Delta P = \frac{2\gamma}{r} $$

where $\gamma$ is the surface tension, which is about 0.07 N/m for pure water, and $r$ is the radius of the ball of water. If we take a radius of 1 cm we get $\Delta P = 14$ Pa. This is negligable compared with the atmospheric pressure of $101,325$ Pa, so it's a very good approximation to say the pressure inside the ball of water is the same as the pressure of the air around it.

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  • $\begingroup$ The link listed appears to use your equation for a droplet. I typically think of a droplet as having a side compressed against a surface. In this case does it mean a solid ball, where a bubble would represent a hollow ball? $\endgroup$ – Reid Erdwien Mar 13 '14 at 4:36
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    $\begingroup$ Yes, the word droplet means a ball of liquid. The difference between a droplet and a bubble is that the bubble has two surfaces, inner and outer, so the pressure inside is twice as high as the droplet. $\endgroup$ – John Rennie Mar 13 '14 at 7:18
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    $\begingroup$ Of course, for a large enough ball of water (something more on the scale of a pond then a droplet) self-gravitation begins to play a leading role. (Still assuming the system is not subject to significant perturbation.) $\endgroup$ – dmckee Mar 24 '17 at 4:40
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You can have a whole range of pressures inside a ball filled with water. It depends on how much you have pressurized when you fill the ball with water. The internal pressure results in material expansion of the ball. The stresses in the shell balance the hydrostatic pressure in water contained in the ball. For a more relatable example, consider a rubber balloon filled with water. You can achieve different pressures depending on the expansion of the balloon.

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