Atmospheric pressure at any altitude represents the total weight, per unit area, of the air column above that altitude. The pressure, therefore, decreases with altitude as there is less air above pressing down. To find the rate of decrease of pressure with height, consider a vertical column of air with unit cross-sectional area, having pressure \( p\) at height \( z $. At height $ z + dz\) the pressure has decreased to some value \( p-dp\), and the pressure difference \( dp\) is equal to the weight of the slice of the vertical air column having thickness \( dz\) (at height \( z\)). Assuming \( dz\) is small enough that the air density, \( \rho\), and the acceleration due to gravity, \( g\), can both be considered constant within the slice:

$$ dp = -g \rho dz $$

Here \( g\) is the acceleration of gravity that transforms the areal mass element \( \rho dz\) into an areal weight element \( g\rho dz\).

This equation is called the hydrostatic equation. It is commonly written as:

$$ \frac{\partial p}{\partial z} = -g \rho $$

This expresses that the upward vertical pressure-gradient force is balanced by the downward gravitational force, a situation usually referred to as hydrostatic equilibrium.

With increasing altitude, the density of air also decreases (e.g. from an average of 1.2 kg/m^3^ at the surface to an average of 0.7 kg/m^3^ at 5 km); hence the rate of change of pressure with height decreases with height. Air temperature also affects density (in a way described by the equation of state), and therefore the rate of pressure decrease with altitude.

Substituting in the equation of state allows us to integrate the hydrostatic equation to obtain an expression for \( p(z)\). The equation of state for dry air is:

$$ p = \rho R T $$

where \( R = R^{*}/m\) is the individual gas constant for dry air, and \( R^{*}\) and \( m are the universal gas constant and the molecular weight of dry air, respectively. (The equation of state for moist air can be obtained by replacing $ T\) with \( T_v\), which is the virtual temperature. It can be shown that moist air is lighter than dry air of the same temperature and pressure, because the water vapour is lighter than the dry air it replaces; so that in cases where only the density of air is important, dry air of slightly higher temperature may be substituted for moist air. Virtual temperature is the fictitious temperature to which dry air must be raised to have the same density as the moist air in question, and of course it depends on the moisture content as well as the pressure.)

Making this substitution gives:

$$ \frac{\partial p}{\partial z} = -\frac{g}{RT}p $$

and hence:

$$ \frac{dp}{p} = \frac{g}{RT}dz $$

By integration:

$$ p = p_0 \exp \left( -\frac{g}{R} \int_{z_0}^z \frac{1}{T} dz' \right) $$

(assuming \( g\) is constant with height).

In the special case where the temperature is constant with height, the pressure profile is:

$$ p = p_0 \exp \left( -\frac{g(z-z_0)}{RT} \right) $$

The temperature in the atmosphere varies by a factor of two, whereas the pressure varies by six orders of magnitude.

Model atmospheres

The following is question 3.5 from A Short Course in Cloud Physics, by Rogers and Yau.

Two model atmospheres often used in theoretical work are the homogeneous atmosphere:

\( \rho(z) = \rho_0 $\)

and the exponential atmosphere:

\( \rho(z) = \rho_0 e^{-z/H} $\)

where \( \rho_0\) is the density at the surface and H is called the scale height of the atmosphere. The top of the homogeneous atmosphere is defined as the altitude where the pressure falls to zero. Prove that the height of the top of the homogeneous atmosphere equals the scale height of the exponential atmosphere.

For this problem we need the equation of hydrostatic balance:

$$ \frac{\partial p}{\partial z} = -\rho(z) g $$

which describes the equilibrium achieved in a fluid under gravity, where the upwardpressure gradient force balances the opposing force due to gravity.

First consider the homogeneous atmosphere; using this equation it is very simple to calculate the height at which the pressure falls to zero because, since the density is constant, the gradient of pressure with height is also constant:

$$ \frac{\partial p}{\partial z} = -\rho_0 g $$

That is, the pressure falls off linearly:

$$ p(z) = \left( -\rho_0 g \right) z + p_0 $$

where \( p_0\) is the pressure at the surface (\( z = 0\)). From this equation we see that the pressure falls to zero when:

$$ z|_{p=0} = \frac{p_0}{\rho_0 g} $$

Now consider the exponential atmosphere; we can determine H in the following way. First, we integrate the equation for hydrostatic balance from some height \( z = z'\) up to \( z = \infty\) to obtain the pressure at height \($ z'\):

$$ \int_{z'}^{\infty} \frac{\partial p}{\partial z} dz = -\rho_0 g \int_{z'}^{\infty} e^{-z/H} dz $$

If we assume that the pressure tends to zero as height tends to infinity, and we set \( z' = 0\), then:

$$ p_0 = \rho_0 g H $$

Consequently:

$$ H = \frac{p_0}{\rho_0 g} $$

Hence, as long as \( p_0\) has the same value in both the homogeneous and exponential atmospheres, the height of the top of the homogeneous atmosphere is equal to the scale height of the exponential atmosphere.

[![Pressure (mb) vs. height (m) for two model atmospheres](https://plot.ly/~0x7df/11.png)](https://plot.ly/~0x7df/11/ "Pressure (mb) vs. height (m) for two model atmospheres")

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