How do I derive the formula for lithostatic (overburden) pressure?

Question

The title pretty much says it. I have the formula: $P = \\rho g h$ where $\\rho$ is the density, $h$ is how deep the pressure is in the Earth and $g$ is the gravitational acceleration(?). I don't get the units either. If I substitute the units for each term I get this: $$P = \\frac{\\mathrm{kg}}{\\mathrm{m}^3} \ imes \\frac{\\mathrm{m}}{\\mathrm{s}^2} \ imes \\mathrm{m} = \\frac{\\mathrm{kg}}{\\mathrm{m}*\\mathrm{s}^2}$$ Shouldn't it be something like $P = \\dfrac{\\mathrm{kg}}{\\mathrm{m}^2}$? As far as I'm concerned that's the unit for pressure. Where do I get this formula from and how do I derive the unit of measurement?

Answer 1

Start with the definition of pressure:

P = F/A

Imagine a vertical column of rock with cross-sectional area A and thickness h. Its volume is Ah, so if its average density is ρ, its mass is

m = ρAh.

The downward force produced by the weight of that rock is

F = mg = ρAhg.

Dividing the weight by the area gives the pressure at the base of the column:

P = F/A = ρAhg/A = ρgh.

The units also reduce correctly:

(kg/m³)(m)(m/s²) = kg/(m·s²) = N/m² = Pa.

So kg/m² is not the pressure unit because the factor of gravitational acceleration converts mass into force. The formula P = ρgh assumes constant density and gravitational acceleration. If density changes with depth, the more general form is an integral of ρg over depth.