Calculating underwater pressure at altitude

That equation gives the so-called gauge pressure, which is the pressure above atmospheric pressure. The absolute pressure is $$P=P_0+rho g h$$where $P_0$ is the atmospheric pressure at the surface and h is the depth below the surface.

The pressure due to a liquid in liquid columns (or beaker) of constant density or at a depth within a substance is represented by the following formula:

$$p=rho gh$$

where:

$ p$ is liquid pressure,

$ g$ is gravity at the surface of overlaying material,

$ ρ$ is the density of the liquid,

$h$ is the height of a liquid column or depth within a substance.

If atmosphere pressure is $P_0$ then total pressure by the liquid will be given by $$P=P_0+rho gh$$

It's so-called gauge pressure. Now this formula works unless you go very deep in the water where other factors may overcome.

First, we need to take care of the effective gravity : $$g_{eff}=g_0left(1-frac{d}{R}right)$$

Another factor is the density of water, It may possible that you are in an ocean where the density of water varies due to salt and other chemicals so we take density to the function of $h$ so that $rho=rho(h)$.

So the formula can be modified as $$dp=rho(h)g(h)dh=rho(h)g_0left(1-frac{d}{R}right)dh$$ or $$P=P_0+int_0^hrho(h)g_0left(1-frac{d}{R}right)dh$$ Once you know the How the density is changing ie. $rho(h)$, you can compute pressure.

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Let's compute so OP is in the lake ( for ocean density will vary. see here) so we can consider that density will not vary appreciably and thus $rho(h)=rho_0$. For $2000$m below the sea, the value of gravity will be

$$g_{eff}approx g_0$$ as the factor $d/R$ is off the order of $10^{-6}$. So that factor too doesn't change much.

So We can conclude Atleast for OP, it's safe to go with gauge formula.

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