INTRODUCTORY
OCEANOGRAPHYINTRODUCTORY
OCEANOGRAPHY
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In this part, we go further by showing how we can estimate the actual currents in the ocean indirectly by determing the differences in density between two hydrographic stations in the ocean and estimating the slope and the current that results.
Most
of our knowledge of ocean currents comes not from the direct
measurement of the circulation, but from an indirect calculation of
the slope of the sea surface based on differences in vertical density
between two or more points in the ocean separated by 50 to 100 km.
The flow of water that results from these ocean slopes are called
Geostrophic currents. Geostrophic currents (literally,
'earth-turning') exist when the Coriolis effect, acting perpendicular
to the direction of motion, balances the down-slope component of
gravity (or the pressure gradient).
Note in Fig. 9.7a to the right that this current (over the extent of the North Atlantic Basin) actually follows a curved path. Below, in my discussion about estimating a geostrophic current, I focus on the current having a straight path. In reality, we can assume it is straight only over a relatively short distance of a few hundred kilometers. Note finally, in Fig. 9.7b, that the actual height of the sea surface (as determined by satellite data) shows this 'hill' of water on the western side of the North Atlantic Basin. This 'hill' will be discussed in Part 3 of this lesson when we examine why the the western boundary currents in all our ocean basins are intensified.
To fully understand this balance, and the examples given below to explain how the balance is achieved, I need to remind you of Newton's first law of motion for a moving mass: "A mass in motion will remain in motion with a constant velocity (constant speed and direction) so long as the forces acting on it are balanced". The corollary, therefore, is "A mass in motion will have a curved path, or will accelerate, if the forces acting are it are not balanced."
We will first model Geostrophic Flow by considering the motion of a ball down an inclined frictionless plane and then extend what we learn to the movement of water down a sea surface slope. There are audio and video files at the end of this discussion section that may be very helpful in learning how to model geostrophic flow.
Consider the motion of a ball rolling down an inclined frictionless plane under two sets of conditions: (1) the plane is fixed to a non-rotating platform; and (2) the plane is fixed to a platform that is rotating counterclockwise. This motion is shown on the cross-sectional view (i.e., looking at it 'from the side") and plan view (i.e., looking 'down from the top') shown below for each of the two sets of conditions.
When the ball is released it will roll down the inclined plane with a velocity (v) and along a path that is perpendicular to the lines of constant height (the slope gradient), because the only force acting on the ball is the down-slope component of gravity (shown as a force vector, G).
Now the inclined plane is placed on a rotating platform that is turning counterclockwise (the same direction that the earth rotates when viewed from above the North Pole). When the ball is released it begins to run down the inclined plane as before, with G acting always perpendicular to the lines of constant height, but because we have introduced the Coriolis effect (shown as a force vector, CE), the path of the ball begins to be deflected to the right. CE will always remain perpendicular to the velocity vector v so, unlike G, it will change direction as the motion is deflected. According to Newton's first law of motion, so long as CE and G are not balanced, the path will continue to curve. Eventually, the curvature will bring the path parallel to a line of constant height and CE (now pointing directly up-slope) will balance G -- as a result, v will be constant and the ball will, in the cross-sectional view, come "right at you" (shown as point of arrow).
In the figure below, note that I have drawn what is shown above in a 3-D image, that may be more clear than the cross-sectional and plan views above.
It is easy to extend this example to a fluid flowing down a sloping sea surface being moved and deflected by the same two forces described above. At the moment that v becomes constant (with CE and G in balance and vectorally opposite in direction) we have, by definition, a Geostrophic Current.
Note: The slope of the sea surface does not have to be large to support a major Geostrophic Current. The Florida Current, after it has curved around the southern tip of Florida has a velocity as high as 8 km/hr (about 5 mph) and yet the slope of the sea surface is only about 1 m in 100 km.
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Some thoughts on determining sea surface slope |
How is the steepness of such a small slope determined?
We must contend with a curved sea surface (the ocean geoid) covered by chaotic ocean waves. We can get quite a good idea of the slope using precision radar satellites, but traditionally the sea surface was estimated indirectly by measuring the internal distribution of vertical density at two or more hydrographic stations (locations on the ocean surface occupied by an oceanographic research vessel) separated horizontally by a few tens to hundreds of kms. The relative slopes of the surface between these stations represents the dynamic topography of the upper mass of the ocean and it is from these estimated slopes that the velocity of the Geostrophic Current can be calculated.
Let me propose a problem that illustrates how the relative slope between two stations is estimated. Only inequalities will be used because, to understand the principles, we do not need to actually calculate densities at the stations -- only to compare them.
Suppose you had two hydrographic stations (A and B) on the ocean surface separated by 50 km (an arbitrary distance) over a flat bottom, as shown in the cross-section below. I chose a flat bottom so that the average pressures of the water columns at the two stations can be assumed to be constant (Pa = Pb).
Suppose you determine that the average salinities of the water columns at both stations were the same (Sa = Sb), but that the temperature of the water column at Station A (Ta) was greater than the average temperature of the water column at Station B (Tb). Therefore, you would have the following conditions:
and
Ta > Tb
which, because s is inversely proportional to T, will result in
Less dense water at Station A, therefore, has a larger volume of water in its column than Station B (remember that s = mass/unit volume, and we have not changed the mass). If we further assume that the columns of water at both stations have the same diameter, then the increased volume at Station A will result in a greater height of the column at Station A (so Ha > Hb), and the sea surface will slope down from Station A to Station B, as shown below:
Using the same cross-sectional view (and its plan view) of the slope of the sea surface given above (as shown below), we see that the down-slope component of gravity (G) will point from Station A to Station B. If we assume that the slope was caused by a Geostrophic current, we also know that CE will balance G (and be vectorally opposite to G), and will, therefore, point directly up-slope from Station B to Station A. Note that these two conditions will be true regardless of hemisphere.
In what direction, relative to this cross-section, will the Geostrophic current actually flow?
Remember that we are assuming that the geostrophic balance has been achieved, so we are not looking for a 'curving path'. To determine the geostrophic current, we need to know the hemisphere. If we assume we are in the Northern Hemisphere, we see in the plan view below that the current will flow parallel to the lines of constant height such that CE will be 90 deg. to the right of its velocity (v). On the cross-sectional view, this means the current will flow "out of the screen" (shown by the "tip of an arrow") -- think about 'tipping' the plan view up so you are looking at the end, and the arrow coming directly at you. Note on both views that v is perpendicular to the cross-sectional plane between Station A and Station B (in fact, that is the only direction that the current can flow -- either directly at you out of the screen or away from you into the screen).
Do you see that the current would flow "into the screen" (with the "tail of an arrow" being shown instead) if the stations and slope shown above were in the Southern Hemisphere?
Do you also see that if the slope was up from Station A to Station B (the opposite slope to that shown), the direction of the current for each hemisphere would be just opposite to that discussed above?
I suggest you look at the balance of G and CE (and the resulting geostrophic current) for the following set of conditions for both hemispheres (over and above that used in the illustration above) - these will give you ALL OF THE SETS OF CONDITIONS THAT YOU WILL ENCOUNTER ON THE HOMEWORK AND/OR YOUR EXAM:
Ta = Tb; Sa < Sb
Ta = Tb; Sa > Sb
For each set of conditions above (including those where the average salinities are not equal), determine the relationship of sa to sb and find the slope (assume always that if sa < sb, then Station A would have the least dense water). Though it is not as clear how changes in density (due to changes in salinity) will result in different heights of the water at the two stations, for this class (and for estimating the slope of the sea surface), you may assume that the column with the lowest salinity (and density) will be higher than the column with the higher average salinity (and density).
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Some final thoughts about geostrophic currents |
Note that the example given only estimates the relative slope between two stations (and, therefore, only the relative Geostrophic current perpendicular to the cross-section between these two station). To find the "true current", we would need to use three or more stations and then vectorally average the relative Geostrophic currents from each of the cross sectional planes between those stations.
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