Reading 5: Algebraic Form of the Geostrophic Equation
in Natural Coordinates: Atmosphere and Ocean
Newton's Second Law of Motion can be solved for the acceleration. In that form, Newton's Second Law simply states that the acceleration, say, of a fluid parcel (air or ocean) is directly related to the sum of the forces per unit mass acting upon the parcel. Forces per unit mass are accelerations.
I. Geostrophic Flow in the Atmosphere
For the atmosphere, for example, the horizontal accelerations are pressure gradient, Coriolis and friction. Let's apply Newton's Second Law to the frictionless portion of the atmosphere, say, at 18000 feet. In that case, the total acceleration experienced by an air parcel is only related to pressure gradient and Coriolis.
|Total Horizontal Acceleration Experienced by Air Parcel||DUE TO||Pressure Gradient Acceleration||AND/OR||Coriolis Acceleration|
Let's rewrite this as a mathematical statement. For this exercise and discussion, you can assume that the derivation is done for you. We'll just worry how to interpret each term. Here is the equation of motion for the frictionless part of the atmosphere that is involved in non-curved flow, in natural coordinates.
where z is the height of the given isobaric surface. ∆z/∆s and ∆z/∆n are the height differences along and at right angles to the flow. Those terms are analogous to pressure gradients. Equation (1) states that in the first instant of motion, air accelerates in the direction of lower pressure (lower heights). In that first instant, there is no Coriolis (because there is no V) and since the air parcel is moving directly towards lower pressure (heights), ∆z/∆n, is also zero. Thus, initially the air parcel (or fluid parcel) moves across isobars (height contours) towards lower values.
As will be demonstrated on the board, as soon as the air parcel has a large enough velocity, the Coriolis term deflects its motion to the right of its path. At some point, the air parcel's motion will be along the isobars because there is an equal an opposite balance of the pressure gradient and Coriolis accelerations. When that happens, the air parcel's velocity will cease to change....there is no further net acceleration. It turns out that in the middle troposphere, this disapperance of net acceleration is very characteristic.
Returning to equation (1), if the total acceleration is small, the left side is zero. In addition, if the air parcel is moving parallel to the isobars, there is no ∆z/∆s. Thus, you can solve the equation for the velocity, now termed the geosotrophic wind (or geostrophic flow); and there is an implied "equal and opposite" balance of the two accelerations on the right hand side.
At a given latitude, the geostrophic wind speed is directly proportional to the height (pressure) gradient (indirectly proportional to the spacing of the isobars/height contours). At 40 degrees latitude f has a value of approximately 10.0 X 10-5 s-1. Since g is a constant, 9.8 m s-2, you can see that Equation (2) allows you to easily calculate the geostrophic wind speed from a map of heights. That's what you will be doing in Homework Assignment31.
II. Geostrophic Flow in the Oceans
The following was extracted from:
Marshall, J., and Plumb, R. A, 2007: Atmosphere, ocean, and climate dynamics: an introductory text. Academic Press, 2007
ISBN 0125586914, 9780125586917, 319 pp.
Note the following definitions:
M' = Geostophic Mass Transport = Vg delta n (3.5)
delta z = - (rho) delta h (3.6)
where rho is seawater density (greek letter rho) and delta h is steric height
Put (3.5) and (3.6) into (3.4) and you get
Look familiar? In atmospheric geostrophic flow, the usual explanation is that air parcels begin "rolling" down the height gradient towards lower values (due to gravity) and then get deflected to the right by Coriolis acceleration. In the diagram below, you can see the same balance (except the author called the pressure gradient acceleration shown in red "gravity" but it is really a pressure gradient accleration since gravity acts at right angles to the surface.)