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Home , Coriolis force

ATM S 211 Solutions Homework 4 Due Tuesday November 13 at 6 pm PDT

Problem #1: in a bathtub (11 pts)

In class, we defined the as

F 2Ω sin(φ)v (1) Co =

5 1 where Ω 7.27 10− s− is the ’s rate of , φ denotes the and v (units of m/s ) is = × relative to Earth of the object in question.

Consider a circular bathtub initially filled with draining through a hole at the bottom center of the bathtub. By unplugging the bathtub we create a low pressure area at the center of the bathtub.

For parts (a), (b), (c) and (d) assume that the bathtub is located in Seattle where φ 45◦. =

(a) In this situation, is the Coriolis force deflecting the draining water either toward the center of the bath- tub, toward the edges of the bathtub, to the right of the horizontal force or to the left of the horizontal pressure gradient force ? (1 pts)

In the , the Coriolis force deflects to the right. Here, motion is induced by an horizontal pressure gradient force pointing toward the center of the bathtub.

(b) Compute the magnitude of the Coriolis force using equation (1) assuming the water drains with a speed

of v 5 m/s . Useful information: sin(φ 45◦) 0.71. Make sure to include the correct units! (2 pts) = = =

Very simply, using equation (1)

5 1 4 F 2(7.25 10− s− )(0.71)(5 m/s ) 0.00052 m/s2 or 5.2 10− m/s2 . Co = × ' × Note that the units are of even though we called it a force.

(c) Using your answer to part (b), is the Coriolis force acting on the water either greater than, less than or equal to the force of acting on the water? (2 pts)

The acceleration caused by the force of gravity is g 9.8 m/s2 which is roughly 20,000 greater than = the Coriolis force computed in part (b).

Now, let’s investigate whether the water could have attained a speed of 5 m/s as a result of the Coriolis force only. The Coriolis force as written in equation (1) has units of acceleration (i.e. m/s2 ). So, dividing a speed by FCo gives us a scale. That is, an estimate of the amount of time it would take the Coriolis force to act and make the water move at that speed. ATM S 211 Homework 4 Solutions 2

(d) Compute the aforementioned time scale using F found in part (b) and v 5 m/s . Co = Report your answer in minutes. (2 pts)

Converting into minutes, this is

speed v 5 m/s 1 minute time scale 161.5 minutes . 4 m 2 = acceleration = FCo = 5.2 10− /s 60 s ' × | {z } 1 = Or, 2 hours and 41.5 minutes.

For parts (e) and (f), assume that the bathtub is located at the South Pole.

(e) Repeat part (a). (1 pts)

To the left of the horizontal pressure gradient force, the Coriolis force switches sign across the .

(f) First, find the magnitude of the Coriolis force as in part (b) with v 5 m/s and then compute the corre- = sponding time scale as in part (d). Report only the time scale in minutes. (1 pts)

At the South Pole φ 90◦ and sin(φ) 1. So, again using equation (1), = − = −

5 1 4 F 2(7.25 10− s− )( 1)(5 m/s ) 0.000727 m/s2 or 7.27 10− m/s2 , Co = × − ' − − ×

4 2 and its magnitude is 7.27 10− m/s . Now dividing v with F gives × Co

5 m/s time scale 114.6 minutes . = 7.27 10 4 m/s2 ' × − Or, a little under 2 hours.

(g) Using your results of parts (d) and (f), briefly comment on whether the Coriolis force is responsible for the swirling motion observed in draining bathtubs. Please keep your discussion to one or two sen- tences. (2 pts)

The Coriolis force is not responsible for the swirling motion observed in draining bathtubs anywhere on Earth. Its magnitude is too small to be responsible for the generation of the observed swirling .

Problem #2: A backward rotating Earth (9 pts)

Suppose that Earth was rotating in the opposite direction, but everything else was the same. That is, the rises in the west and sets in the east. In this situation,

(a) Would the Coriolis force deflect moving objects with respect to Earth either in the same direction as today, in the opposite direction as today or would the Coriolis force vanish ? (1 pts)

Changing the sense of rotation takes Ω Ω. Hence, the Coriolis force switches signs and its deflection → − is in the opposite direction. ATM S 211 Homework 4 Solutions 3

(b) Would there still be an equator-to-pole temperature gradient ? Yes or no? Briefly explain in one or two sentences. (2 pts)

Yes. Properties such as insolation, the latitudinal distribution of water vapor and albedo are not affected by rotation (in particular its sense).

(c) Would the hydrostatic balance relation have to be modified ? Yes or no ? Briefly explain in one or two sentences. (2 pts)

No. Hydrostatic balance describes the vertical pressure gradient force exactly balancing the force of gravity independently of rotation (in particular its sense).

(d) Would there still be jet streams? If so, would the jets streams blow either toward the East or toward the West ? (2 pts)

Combining hydrostatic and geostrophic balance gives us jet streams. The Coriolis force has the opposite on this backward rotating Earth; hence, the jets streams blow in the opposite direction. That is, toward the West.

(e) Considering the effects of surface , would the surface either blow into low pressure areas and out of high pressure areas or out of low pressure areas and into of high pressure areas. (2 pts)

The sense of rotation does not alter the phenomenon. The Coriolis force is proportional to the of the winds. Surface friction slows down the winds and correspondingly the Coriolis force. So, the horizontal PGF (pointing into lows pressure areas and out of high pressure areas) wins and there’s is a net flow into lows and out of highs.

Problem #3: General circulation in the Pangaea Era (10 pts)

About 300 million years ago all land were connected into a supercontinent called Pangaea. In addi- tion, land masses were evenly distributed on both sides of the equator; unlike today where the majority of land masses are located in the Northern Hemisphere.

Note also that in the present day,

The Arctic is isolated from warm currents, ¦ Antarctica is a continental ice sheet. ¦ In answering the questions below, assume that orbital parameters, the Earth’s spin, ocean heat transport and the greenhouse gas concentration, etc. are the same as present day. More specifically, consider only the impact of changing the Earth’s land-ocean distribution.

(a) Averaging over the entire Northern-Hemisphere, was the winter-time (i.e. months of December, Jan- uary, February) equator-to-pole temperature gradient either greater than, less than or the same as the present day value ? (2 pts)

The Arctic Sea being exposed to warm ocean currents is less likely to freeze into , so the North Pole is warmer than today. Also, the equatorial latitude band contains more continents than today, so,

on average, it cools more readily (lower Cp ) away from the equinoxes than today. Therefore, unequivo- cally the equator-to-pole temperature gradient is less than the present day value. ATM S 211 Homework 4 Solutions 4

(b) Now averaging over the entire Southern-Hemisphere, was the winter-time (i.e. months of June, July, August) equator-to-pole temperature gradient either greater than, less than or the same as the present day value ? (2 pts)

Similarly, sea ice is harder to form (and to maintain) than land ice, so the South Pole is also warmer than today. As in part (a), the equatorial latitude band is cooler. Again, the results follows unequivocally.

(c) Was the Northern-Hemisphere winter-time either stronger, weaker or the same strength as the present day ? (2 pts)

The strength of the winter-time jet stream is modulated by the magnitude of equator-to-pole tempera- ture gradient. From part (a), we infer that the winter-time jet stream is weaker.

(d) In the Northern Hemisphere, Were mid-latitude winter storms either more likely, less likely or equally likely than today? (2 pts)

The frequency of mid-latitude storms is modulated by the strength of the winter-time jet stream. So, a weaker jet implies that storms are less likely.

(e) Repeat part (c) for the . (1 pts)

Weaker. With the result of part (b) and the same analysis as in part (c)

(f) Repeat part (d) for the Southern Hemisphere. (1 pts)

Less likely. With the result of part (e) and the same analysis as in part (d).