Consider a horizontal layer of air having unit surface area and infinitesimal thickness 'dz'. If it is in static equilibrium the force of gravity acting downwards is exactly balanced by the pressure forces. This implies that the difference in pressure between the top and bottom of the air layer is

dP = - Dg dzEquation 1

where D and g are, respectively, the density of air and the acceleration due to gravity at the altitude of the air layer, and the minus sign ensures that pressure decreases when the altitude increases (positive dz). This is the **hydrostatic** **equation** for a fluid.

At the usual temperatures and pressures normally present in the atmosphere, to an excellent approximation air behaves as an ideal gas, and therefore obeys the law

PV = RTEquation 2

where one mole of air occupies the volume V at pressure P and absolute temperature T, and R is the universal gas constant. The density of air is

D = M/VEquation 3

where M is the molecular weight of air. Combining Equation 1 and Equation 3 we get

dP = - (Mg/V)dzEquation 4

and using Equation 2 to replace V,

dP/P = -Mg/RTEquation 5

The acceleration due to gravity ('g') is the sum of two terms: (i) the attractive force per unit mass as given by Newton's universal law of gravitation, and (ii) the "repulsive" centrifugal force caused by the earth's rotation. The latter effect is very small, contributing no more than 0.35% at sea level. The former drops off inversely as the square of the distance from the center of the earth. but even at an altitude of 10,000 meters it is only 0.3% smaller than its value (g0) at sea level. Except near the ground, where the concentration of carbon dioxide shows significant local variation due to industrial activities, fires, photosynthesis, and exchange with the oceans, the composition of dry atmospheric air is remarkably constant up to an altitude of 100 km. Therefore, the only factor on the right hand side of Equation 5 that shows a significant variation with altitude is the temperature.

The atmospheric temperature profile is approximately as shown in the figure below. In the troposphere, where our "weather" occurs, it decreases linearly with altitude until it reaches -55 Celsius at the tropopause (~ 11 km), where it starts to increase because of the solar heating of the stratospheric ozone. The drop in air temperature per unit increase in altitude, i.e. the **lapse rate**, is B=6.5 degrees K /km.

Accordingly, below the tropopause

T = T0 - BzEquation 6

where T0 is the sea level air temperature and z is the altitude in km. Substituting this into Equation 5,

dP/P = - (Mg/R)/(T0 -Bz)Equation 7

Integrating both sides from sea level (z=0) to z, we get

P(z)/P0 = [1 - z(B/T0)]^(Mg/BR)Equation 8

where the symbol ^ denotes exponentiation. Using the definition of the standard atmosphere adopted by the International Civil Aviation Organization, we have: M=28.9644 (carbon-12 scale), T0=15 deg C = 288.15 deg K, g=g0=9.80665 m/s^2, R=8.314 Joules/gram-mole/deg K, P0=1013.25 bar (1 atmosphere). Substituting into Equation 9 we have, finally,

P(z)/P0 = (1 - 0.02255z)^5.256Equation 9

z(km) P/P00 1 0.5 0.942 1.0 0.887 1.5 0.834 2.0 0.785 2.5 0.737 3.0 0.692

* J. V. Iribarne, "Atmospheric Thermodynamics" (D. Reidel Publishing Company 1973), Chapter VII.*

* Arthur Beiser, "Earth Sciences" (McGraw-Hill 1975), Chapter 2.*