(ii) According to Ekman's theory, the surface current should be 45° cum sole of the wind direction. The situation under consideration is in the Southern Hemisphere, and so 45° cum sole is 45° to the left. The wind is westerly (towards the east) and so the surface current should be towards the north-east.

(c) According to the 'rule of thumb', the surface current is typically about 39r of the wind speed. In this case, the wind speed is 5 m s_l and so the surface current might be expected to be about (3/100) x 5 = 0.15 ms_l. This is about five times the value we calculated in (b); however, in that calculation, we used a value for A- towards the upper end of the likely range of values (10-5 to 10_1 m-s_l), and a smaller value for A- would have resulted in a larger value for u0.

Question 3.5 The period of an inertia current is given by Equation 3.7: 2k

Now. Q = 271/24 hrand at the poles <|) = 90° and sin <J> = 1.

2n 24

Question 3.6 (a) Pressure gradients act from areas of higher pressure to areas of lower pressure. In this case, the slope of the isobars tells us that at any given depth, the pressure is greater on the right than on the left. The horizontal pressure gradient force must therefore be represented by the arrow pointing to the left, and the Coriolis force by the arrow pointing towards the right (Figure A2).

horizontal pressure gradient force

Conolis

Figure A2 Answer to Question 3.6(a).

horizontal pressure gradient force

(b) We are told that the current is (lowing into the page, and we know that the Coriolis force acts cam sole of the direction of motion, i.e. to the right ¡n the Northern Hemisphere and to the left in the Southern Hemisphere. In tilts case, the Coriolis force is acting to the right of the direction of motion, and so the situation illustrated must be in the Northern Hemisphere.

Question 3.7 (a) The first step is to calculate/ At 30" of iatitude. stn $ = so/= 2Q x 0.5 = 2x7,29* 10^x0.5 = 7.29X 10" V1. In this case,-, is 2000 - 1000 = 1000 m, and we are assuming that is the same:

L= 100 x 101mand^ = 9.tims-:. Therefore, substituting into Equation 3.13:

fl- \ Pa _ 9.8 X 1000 ( i - 0.9997) _ 9800 x 0,0003 7.29xl0r5Xl05 ~ 7.29

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