Modelling the atmosphere and the barometric law

In this article, Carl discusses the composition of the atmosphere and how the density and pressure can be simply modeled as a function of height through the thin layer of air that makes life on this planet possible…

The extremely thin slab of air that hugs this ball of rock we call home is a complex beast and responsible for all the weather phenomena we see around the globe. Rain, snow, fog, winds are just some of the few things our atmosphere can throw at us. It’s so complex, and so unpredictable, that we can still only calculate probabilities of what weather is coming. The first step to understanding the weather, and the atmosphere, is to look at the physics that underpins it. Today we look at how we can model the density and pressure experienced at different heights above the Earth’s surface.

Modeling the Atmosphere

Our atmosphere is made up of several different layers with different compositions of elements, or species, in each. These layers can be simply modeled, as shown in the figure below, but it must be noted that the boundaries between each of the layers are not that well defined with the heights given below being a round guide to the beginning and end of each layer.

Source: http://www.physicalgeography.net/

You will note that in some layers the temperature, given by the red solid line, increases with height in some layers whereas it decreases with height in others. In the troposphere, the temperature decreases with height because the main heating mechanism is convection and so this layer is mainly heated by the Earth from below. The temperature in the stratosphere increases with height because ultra-violet (UV) solar radiation is strongly absorbed by the Oxygen and Ozone molecules that reside in this layer. The thermosphere also absorbs other high energy solar radiation and so Oxygen atoms dominate here.

The Barometric Law

The barometric law is used to model the variation of pressure, and various other quantities, with height in the atmosphere. To derive this extremely useful relation, we must look at a column of air from the surface (z=0). Taking a slice of this column from a height z and pressure p to height z+dz and pressure p+dp. Here we equate the forces of gravity and pressure acting on this slice.

The force of gravity, F_G, acting downwards is given in the usual form, noting that mass and acceleration due to gravity are both functions of height.

F_G = m(z)g(z) = \rho(z)Adzg(z)

where the mass as a function has been replaced by an expression using the density and volume of the slice of the column. Now, the force due to the pressure of the gas above and below the slab, F_P is given below.

F_P = -Adp

where dp is the change in pressure in the slice and A is the cross sectional area of the column. With expressions for the forces acting on the slice, we can equate these assuming equilibrium. This process is outlined below.

F_G = F_P

\rho(z) A dz g(z) = -A dp

and using the ideal gas law (pV=nRT) we can write the density as \rho(z) = \bar{M}p(z)/RT(z), where \bar{M} is the average mass of the species that comprise our slice. Substituting this expression into our equation for equilibrium we get

\frac{\bar{M}p(z)}{RT(z)}g(z) dz = -dp

Rearranging so that all the dependents of height are on one side and all the dependents of pressure are on the other, we arrive at a differential expression.

\frac{dp}{p(z)} = -\frac{\bar{M}g(z)}{RT(z)}dz

From this, all we have to do is integrate both sides from z = 0 to an arbitrary height z.

\int_{0}^{z}{\frac{dp}{p(z)}} = -\int_{0}^{z}{\frac{\bar{M}g(z)}{RT(z)}dz}

At this stage, we do not know how temperature or the acceleration due to gravity actually depends on height so we cannot do the integral on the right hand side. The left hand side, however, will result in the log of pressure. To get rid of this, all we do is take the exponential of both sides.

\frac{p(z)}{p(0)} = exp[-\int_{0}^{z}{\frac{\bar{M}g(z)}{RT(z)}dz}]

This is the barometric law, but it can be simplified by defining a quantity known as the scale height – H = RT(z)/g(z)\bar{M}.

\frac{p(z)}{p(0)} = exp[-\int_{0}^{z}{\frac{dz}{H}}]

If we now say that H is constant, with temperature, acceleration due to gravity and the mean mass of the species all staying constant, the integral becomes trivial and we arrive at

\frac{p(z)}{p(0)} = exp[-\frac{z}{H}]

This is the barometric law and is useful throughout the study of the atmosphere. It not only tells us about the ratio between pressure at the surface and some height, but also other quantities such as species number density and density – it’s very useful!

Further Your Knowledge

The atmosphere is a complex beast but also a fascinating one. Do some of your own research on the following topics to further your knowledge on this subject!

  • Height of the atmosphere compared to the radius of the Earth?
  • Composition of the different layers?
  • At what height is space defined at?
  • At what height is the positioned?
  • What is the greenhouse effect?
  • What does the greenhouse effect do to the temperature of the Earth?
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