A gas atmosphere in static equilibrium must support the weight of the gas above each level. Because gas density decreases as pressure decreases, atmospheric pressure does not fall linearly with height as it would for an incompressible fluid.

Let $z$ increase upward. Hydrostatic equilibrium over a thin horizontal layer gives

$$\boxed{\frac{dp}{dz}=-\rho g}.$$

For an ideal gas of molar mass $M$ at constant absolute temperature $T$,

$$pV=nR_uT$$

and $n=m/M$. Since $\rho=m/V$,

$$\boxed{\rho=\frac{pM}{R_uT}}.$$

Substituting into the hydrostatic equation gives

$$\frac{dp}{dz} =-\frac{Mg}{R_uT}p.$$

This is a first-order differential equation in which the rate of pressure decrease is proportional to the pressure itself.

Exponential pressure profile

Choose a reference height $z_0$ where the pressure is $p_0$. Solving the differential equation gives

$$\boxed{p(z)=p_0\exp\left[-\frac{Mg(z-z_0)}{R_uT}\right]}.$$

Define the atmospheric scale height

$$\boxed{H=\frac{R_uT}{Mg}}.$$

Then the result becomes

$$\boxed{p(z)=p_0e^{-(z-z_0)/H}}.$$

After rising one scale height,

$$p(z_0+H)=\frac{p_0}{e}\approx0.368p_0.$$

The scale height therefore sets the characteristic vertical distance over which pressure changes substantially.

Density profile

For constant $T$ and constant gas composition,

$$\rho=\frac{pM}{R_uT},$$

so density has the same exponential dependence:

$$\boxed{\rho(z)=\rho_0e^{-(z-z_0)/H}}.$$

The pressure gradient becomes progressively smaller in magnitude with altitude because there is progressively less mass per unit volume to support.

Worked example: approximate scale height of Earth's air

Approximate dry air by

$$M=0.0290,\mathrm{kg/mol},$$

with

$$T=288,\mathrm K$$

and constant

$$g=9.81,\mathrm{m/s^2}.$$

Then

$$H=\frac{(8.314)(288)}{(0.0290)(9.81)} \approx8.42\times10^3,\mathrm m.$$

Thus

$$\boxed{H\approx8.4,\mathrm{km}}.$$

In this simplified isothermal model, the pressure at $5.0,\mathrm{km}$ above the reference level is

$$\frac{p}{p_0}=e^{-5000/8420}\approx0.552.$$

So the pressure is about $55%$ of the reference pressure.

Why the constant-density formula fails for a deep atmosphere

For a liquid whose density changes negligibly, hydrostatics gives an approximately linear relation

$$p=p_0-\rho g(z-z_0).$$

A gas is compressible. As altitude increases, pressure falls, which lowers density; the smaller density then makes the pressure decrease more gradually. Combining this feedback with the ideal-gas law produces the exponential profile.

Limits of the isothermal model

A real planetary atmosphere is not generally isothermal. Temperature varies with altitude, composition can change, humidity may matter, and $g$ eventually changes with distance from the planet. If $T=T(z)$, the hydrostatic equation still applies locally, but the pressure profile must use the actual temperature variation rather than one constant scale height.

The barometric formula is therefore an idealized but powerful example of how a local force balance and an equation of state combine to determine a macroscopic spatial profile.