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We know the pressure can be determined in a water by: $p = ρ*g*h$.

The whole reason of trying to bring h in the formula is that we are aware that pressure changes with the depth of water and we include h to account for it.

The problem with this in my head is that it's still not correct. density is different at different heights and the bigger the height we choose, the more wrong our calculated pressure will be there.

What's the point of trying to bring h in the formula if the pressure formula will still be wrong due to density in the equation ?

Qmechanic
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Matt
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  • Many fluids are (practically) incompressible, for instance water. That means the density is (practically) constant over large ranges of pressure. See e.g. here – Marius Ladegård Meyer Apr 28 '23 at 14:57
  • Yes, but if the deeper you go, the bigger the pressure is due to bigger force from top due to bigger mass. If the pressure is bigger the deeper you go, that means density will be definately bigger as per volume, more molecules are squashed than it would be at the top. – Matt Apr 28 '23 at 15:00
  • No, that is not correct. As per the link I gave, 1 extra Pa of pressure changes the volume of water by a factor of roughly $4.5 \times 10^{-10} = 0.00000000045$. That is practically no change at all. – Marius Ladegård Meyer Apr 28 '23 at 15:02
  • Of course I agree that the pressure increases, but that is precicely what the $h$ factor accounts for! – Marius Ladegård Meyer Apr 28 '23 at 15:04
  • For pretty much all deviations of STP, the density is constant w.r.t. pressure (cf my answer here). It is not until pressures well above 10 MPa (100x standard atm) that density deviates significantly from 1 g/cm$^3$. – Kyle Kanos Apr 28 '23 at 15:05
  • Let's forget about the above formula and look at P = (m * g) / A . So the deeper you go, the mass becomes bigger, hence pressure is bigger. If you only think logically, if mass is bigger pushing downard, that means the molecules should be more squashed at depth than at the top, and if they're more squashed, then density must be bigger at depth. – Matt Apr 28 '23 at 15:24
  • Does this answer your question? Why can, or can not, a perfectly incompressible fluid exist? – Farcher Apr 28 '23 at 15:33
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    You say "they should be more squashed", and we are telling you that they are, but to an extremely small amount that can be neglected in practically all cases. Again, you will need millions of Pa to change the volume (and therefore density) by even 1%. – Marius Ladegård Meyer Apr 28 '23 at 15:37
  • What's Pa mean ? – Matt Apr 28 '23 at 15:40
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    It's the SI unit of pressure, named after Pascal. 1 Pa is equal to 1 Newton of force per square meter of area. – Marius Ladegård Meyer Apr 28 '23 at 15:41
  • When object falls down, we know that at some point, it will get a constant speed due to a = 0. This is due to the fact that downward and upward force get equal. Upward force is air resistance. Air resistance increases when object falls down as there're more air molecules per volume(density increases). but per your explanation, density increase will also be negligible here. Then why air resistance force become equal at some point and not immediatelly ? the increase speed of the object won't be the only factor here and I believe density increase plays a big role. – Matt Apr 28 '23 at 15:44
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    Air resistance (also called drag) is somewhat complicated, but a formula that is often very close to accurate is $ F = 1/2 \rho v^2 C A$. Here $\rho$ is the constant density of the fluid that the object falls through (air in your example). The reason the drag force becomes bigger and bigger and eventually cancels the downward force of gravity is that the velocity increases until $v^2$ is big enough. – Marius Ladegård Meyer Apr 28 '23 at 15:50
  • So, it's only because of increase of the speed and not because the air density ? I don't mean the formula, I'm asking logically. air density increase + speed increase both make me believe it's why air resistance gets equal to downward force. are you saying air density increase plays no role ? – Matt Apr 28 '23 at 15:57
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    I'm saying speed increase plays a tremendously bigger role that density increase, so much so that in many practical applications we can completely ignore the density altogether. That being said, air is certainly much more compressible that water is, by orders of magnitude. – Marius Ladegård Meyer Apr 28 '23 at 16:04
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    So if you drop an object from an altitude of 10 km for instance, the density there is certainly much less that down at the surface. But after falling for only one second, the speed of the object has increased by around 10 m/s, and the density has virtually not changed. Wait one more second and the speed is now a little less than 20 m/s, so the air resistance has quadrupled over 1 second, and the density is still more or less the same (the object has moved only around 20 meters). – Marius Ladegård Meyer Apr 28 '23 at 16:06
  • Why does it not continue increasing the air resistance force when the force becomes equal to downward force ? I know it seems silly, but in terms of velocity and dencity, what's the reason – Matt Apr 28 '23 at 16:18
  • Let us continue this discussion in chat. – Marius Ladegård Meyer Apr 28 '23 at 16:34

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