METEO 300
Fundamentals of Atmospheric Science

2.2 The Atmosphere’s Pressure Structure: Hydrostatic Equilibrium

The atmosphere’s vertical pressure structure plays a critical role in weather and climate. We all know that pressure decreases with height, but do you know why?

cube showing forces that determine hydrostatic balance as described in the text below
Air parcel at rest with three forces in balance

The atmosphere’s basic pressure structure is determined by the hydrostatic balance of forces. To a good approximation, every air parcel is acted on by three forces that are in balance, leading to no net force. Since they are in balance for any air parcel, the air can be assumed to be static or moving at a constant velocity.

There are 3 forces that determine hydrostatic balance:

  1. One force is downwards (negative) onto the top of the cuboid from the pressure, p, of the fluid above it. It is, from the definition of pressure,

    F top = p top A This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

    [2.11]
  2. Similarly, the force on the volume element from the pressure of the fluid below pushing upwards (positive) is:

    F bottom = p bottom A This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

    [2.12]
  3. Finally, the weight of the volume element causes a force downwards. If the density is ρ, the volume is V, which is simply the horizontal area A times the vertical height, Δz, and g the standard gravity, then: 

    F gravity =-ρVg=-ρgAΔz This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

    [2.13]

By balancing these forces, the total force on the fluid is:

F= F bottom + F top + F weight = p bottom A p top AρgAΔz This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[2.14]

This sum equals zero if the air's velocity is constant or zero. Dividing by A,

0= p bottom p top ρgΔz This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[2.15]

or:

p top p bottom =ρgΔz This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[2.16]

Ptop − Pbottom is a change in pressure, and Δz is the height of the volume element – a change in the distance above the ground. By saying these changes are infinitesimally small, the equation can be written in differential form, where dp is top pressure minus bottom pressure just as dz is top altitude minus bottom altitude.

dp=ρgdz This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

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The result is the equation:

dp dz =ρg This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.@

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This equation is called the Hydrostatic Equation. See the video below (1:18) for further explanation:

Click for Transcript for Hydrostatic Equation Video

Using the Ideal Gas Law, we can replace ρ and get the equation for dry air:

dp dz =g p R d T         or     dp p = g R d T  dz=  Mg R * T  dz  This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[2.19]

We could integrate both sides to get the altitude dependence of p, but we can only do that if T is constant with height. It is not, but it does not vary by more than about ±20%. So, doing the integral,

p=  p o e z H      where  p o =surface pressure and H=  R * T ¯ M air g

[2.20]

H is called a scale height because when z = H, we have e-1. If we use an average T of 250 K, with Mair = 0.029 kg mole-1, then H = 7.2 km. The pressure at this height is about 360 hPa, close to the 300 mb surface that you have seen on the weather maps. Of course the forces are not always in hydrostatic balance and the pressure depends on temperature, thus the pressure changes from one location to another on a constant height surface.

From the hydrostatic equation, the atmospheric pressure falls off exponentially with height, which means that about every 7 km, the atmospheric pressure is about 1/3 less. At 40 km, the pressure is only a few tenths of a percent of the surface pressure. Similarly, the concentration of molecules is only a few tenths of a percent, and since molecules scatter sunlight, you can see in the picture below that the scattering is much greater near Earth's surface than it is high in the atmosphere.

NASA image to show scattered light near earth's surface.
Scattered light near Earth's surface.
Credit: NASA

Quiz 2-2: Harnessing the power of the hydrostatic equation

This quiz will give you practice using the hydrostatic equation to learn interesting and useful properties and quantities of the atmosphere.

  1. I strongly suggest that you do all your calculations on this quiz using the Excel workbook.
  2. There is no Practice Quiz 2-2. However, you have extra time to take this quiz.
  3. When you feel you are ready, take Quiz 2-2 in Canvas. You will be allowed to take this quiz only once. This quiz is timed, so after you start, you will have a limited amount of time to complete it and submit it. Good luck!