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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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