Climate Physics Chapter 2

Hydrostatic Relation

This equation relates atmospheric pressure to altitude. This relation is what allows us to work in pressure coordinates, instead of altitude.

We consider a column of air with pressure p, height z, and cross-sectional area A. By looking at the force on a small horizontal slice of the column, assuming the acceleration due to gravity, g, remains relatively constant throughout the column, we can say:

\begin{equation} dm=-\frac{dp}{g}\\ \end{equation} \begin{equation} \frac{dp}{dz}=-\rho g\\ \end{equation}

An important consequence is that this allows us to find the mass per unit of planet surface area of an atmosphere.

\begin{equation} m=\frac{p_{s}}{g}\\ \end{equation}

Where \(p_{s}\) is the surface pressure. To find the mass of a specific substance A, multiply by the mass specific concentration \(q_{a}\). To find the mass from a partial pressure \(p_{A}\) in a well mixed mixture of gasses, replace \(p_{s}\) with \(p_{A}\frac{M_{A}}{\overline{M}}\), where \(M_{A}\) is the molecular weight of A and \(\overline{M}\) is the mean molecular weight of the mixture.

Example Problem: Finding mass of atmosphere

Venus has a surface pressure of 92 bar, and a surface gravity of \(8.87\textrm{ m}/{\textrm{s}}^{2}\). 3.5 percent of the atmosphere by mole fraction consists of \(N_{2}\). Compute the mass of \(N_{2}\) per unit surface area of Venus.
Mass per square meter = \(mq_{a}=q_{a}\frac{p_{s}}{g}=3.6\cdot 10^{4}\textrm{ kg}\)

We can also use integrate the hydrostatic relation to get the pressure as a function of altitude.

\begin{equation} p(z)=p_{s}\exp\left(-\frac{g}{R\overline{T}}z\right)\\ \end{equation}

\(\overline{T}\) is the harmonic mean temperature, and since it is measured relative to absolute zero it is relatively constant.

A graph of the altitude versus pressure of the atmosphere as calculated by the integrated hydrostatic relation. As expected, we see pressure decrease as we go to higher altitudes.