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Our class is learning about hydrostatic water pressure and we have been told that we can calculate the force of the liquid on an object at any depth using "the density x 9.8 x the depth". However, as the depth increases, wouldn't the density of the liquid increase because of the weight of the liquid above it compressing it? So should't there be something in the equation to account for the varying density? To me, "density x 9.8 x depth" seems like it is saying that the density will be constant...

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I think en.wikipedia.org/wiki/Bulk_modulus might be intresting to you to calculate how little the density of water changes with even pressures of the deep see. –  PlasmaHH Jun 17 at 13:07
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up vote 16 down vote accepted

The density does increase with depth, but only to a tiny extent. At the bottom of the deepest ocean the density is only increased by about 5% so the change can be ignored in most situations.

If you're dealing with these sorts of depths you also need to take temperature into account because the water temperature changes with depth and the density also changes with temperature.

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It's interesting to note that if the Pacific were ten times deeper, there might be ice at the bottom... –  Beta Jun 17 at 21:59
    
@Beta: There is ice on the bottom. en.wikipedia.org/wiki/Methane_clathrate –  Eric Towers Jun 18 at 4:53
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as the depth increases, wouldn't the density of the liquid increase because of the weight of the liquid above it compressing it?

No, it doesn't - or at least only negligibly so. At normal pressures, liquids are essentially incompressible. This table gives the compressibility of some liquids, including water. Note that the units are to be multiplied by $10^{-11}$ per Pascal. For water you get 46ppm per atmosphere. In most cases that is negligible.

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Would you have to take it into account for things like deep sea calculations? Ie when we get 1km deep or more? –  Sam Jun 17 at 12:07
    
At 1km depth the pressure is about 100 atm. Hence water will compress 4600 ppm or 0.46%. In most calculations you will not have to worry about it. –  hdhondt Jun 18 at 2:02
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You said the right word: liquid!

$P=\rho g h$ holds only if the fluid that you are considering is not compressible, that is a liquid. Try to fill a syringe (without the needle) with some water, then close the hole and try to compress it: you will notice that you cannot do much, indeed liquids are not compressible, this mean that the density $\rho=m/V$ does not vary with pressure because the volume stays the same.

However when applying a physics relation it is important to understand its limits of validity. For instance if you compress water a lot you may even force the molecules to pack together and get ice, this will of course break the equation. But with liquids in everyday life that is just fine.

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"For instance if you compress water a lot you may even force the molecules to pack together and get ice" Is that really possible? AFAIK increasing pressure decreases the melting point. –  CodesInChaos Jun 17 at 14:40
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@CodesInChaos Only up to $200~$MPa, then the melting point come back again en.wikipedia.org/wiki/Ice#Phases ;) –  DarioP Jun 17 at 14:59
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