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I know that hydrostatic pressure $P$ is directly proportional to density $\rho$ and depth $h$. Why does $P$ depend on gravitational acceleration $g$ ?
$P=\rho g h$
Is it because the force of the water columns is related to the weight? But then where would the surface be in the formula?
$P = h \rho g$ ($\rho$ is density) is an approximation that can only be used when both $\rho$ and $g$ are constant over the height change $h$.
The accurate relationship is proven in most standard texts on the subject and indeed on the wikipedia page for hydrostatic equilbrium, by considering the equilibrium of a thin slab of thickness $\Delta h$ and area $A$.
What is found is that $$\frac{dP}{dh} = - \rho g,$$ where both $\rho$ and $g$ may be a function of $h$. The simple formula is obtained by assuming they are constant and integrating from a depth $-h$ to zero.
The downward force exerted by a column of water due to the gravitational attraction of the Earth is called the weight of the column of water. If the column is of unit cross sectional area the mass of the column is height X density and its weight is mass X gravitational field strength (g) = height X density X g and because this weight is acting on unit area, this is the pressure exerted by the column of liquid. If there is a pressure on the surface of the liquid then you would add that to the pressure due to the column of liquid to find the total pressure.
Old post, but in case someone else comes across it, consider the units. Density is mass (not weight) per unit volume. You must have acceleration in the equation for the result to be weight (force) per unit area. Likewise, a tank in an airplane will experience 4 times the hydrostatic pressure if the airplane pulls up (or turns) at 4g.
