# GG 140: The Atmosphere, the Ocean, and Environmental Change

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# The Atmosphere, the Ocean, and Environmental Change

## GG 140 - Lecture 12 - Circulation of the Atmosphere (Exam I review)

Chapter 1: Review of Exam Solutions [00:00:00]

Professor Ron Smith: You did a lab this week, and it was a little bit–the weather discombobulated us a little bit. There was fog hanging around. And some of you got a good launch. Some of you didn’t. For those that didn’t, depending on what your TA wants to do, there may be another chance to do a launch when you come to the second meeting for that lab next week. But remember, there are two parts to that lab. You have to do the pilot balloon launch, but you also have to learn about how to handle National Weather Service balloon launches. So you have to be sure you get both of those parts done in the two periods when you meet for this lab.

So the exam,10 questions. You’ve got two more exams like this coming in the course. They’ll be a similar structure, a mixture of short problems to work out and short written answers. They are a little bit long, as you noticed. I’ll try not to make them any longer. But it is what it is.

The average grade on this I believe was 81. If you got anywhere in that vicinity, plus or minus 10, I wouldn’t worry about it. If you fell way below that, you might want to talk to me about how you’re handling your work in the course, whether you’re putting your emphasis into the right areas of study and so on. I’d be happy to talk to anybody about that.

I’m going to go over each question today, and if you have any additional questions while I go through these, please stop me, and we’ll spend some time on them.

Question one was about computing the total mass of Venus’ atmosphere. And we’re going to use the same principle we used to do the Earth’s atmosphere in class. We use the hydrostatic relation for the full column, which says that the pressure at the bottom of the column is balancing the weight of the air above it. So by looking at the ratio of pressure to gravity, you can get the mass per unit area in a column of atmosphere. And then just multiply that times the surface area of the sphere.

So the formula is this, but you’ve got to be sure you use Venus parameters everywhere–Venus radius, Venus surface pressure, Venus surface gravity–and then you get that number. Can someone check in their notes, when we did this for Earth, what did we find for the total mass of the Earth’s atmosphere? Could someone just remind me of that? I wanted to do that comparison, but I don’t have my notes here. A similar calculation, but we did it for Earth in class.

Student: 5.4x1017.

Professor Ron Smith: 5.4x1017. Yeah, so that’s quite a bit less, then. And that’s remarkable that Venus, one of our nearest neighbors, has an atmosphere that is so much more massive than ours. So it’s not possible to assume that all these planets have a similar atmosphere when you see that kind of a remarkable difference between just Venus and Earth.

Any questions on this?

Question two was about water vapor. It had two parts, first to compute the partial pressure of water vapor. And I gave you the density–the mass density of water vapor, and I gave you the temperature. So you should have used the perfect gas law for that. I’ve written it here. The partial pressure for water is the density for water times the gas constant for water times the temperature.

And you’re given 0.01 kilograms per cubic meter for the density, but now you’ve got to be assured you use the right gas constant. So the gas constant for water is going to be the universal gas constant, 8,314, divided by the molecular weight for water. Water is H2O. The molecular weight of oxygen is about 16. The molecular weight of each hydrogen atom is 1, so that adds up to 18. That’s why I have it there.

So I’ve brought that value down into here, 462 for the gas constant. And then of course I had to convert the temperature, which was 20 Celsius, to Kelvin, and that’s what you get. And be sure to put units. Always have units on there. That’ll be Pascals if you’re following the normal SI system of units.

The relative humidity is the ratio of the partial pressure of water vapor to the saturation value at the temperature that you have. And there’s a table here that gives you a few values for saturation vapor pressure for water at given temperatures. And for 20 degrees, it’s 23.4 millibars. Got to keep consistent units. We’re working in Pascals, so I divided by 2,340, and I get 0.58, which is 58%.

A couple of you put not PSat down here but PTotal, the total pressure, air plus water vapor. Well, that’s going to be much, much larger than this value. The total pressure, for example, in this room, PTotal, would be something like 1,013 millibars, two more zeroes to get it in Pascals. And if I put that number into there, I’m going to get a very much smaller relative humidity.

But that’d be missing the point. Relative humidity is not the ratio of water to air. It’s the ratio of how much water you have to how much water vapor you can have, the maximum that you can have at the temperature that you’re given. Is that clear?

Question three, explain why clouds form in rising air. Well, the point is that clouds form when air rises, because when air rises, it moves to lower pressure. Pressure decreases as you go up in the atmosphere. When you put air into a lower pressure, it’ll expand. When air expands, it does work on its environment, and its temperature drops. That’s called adiabatic cooling.

When you cool air, its saturation vapor pressure drops–that is, the amount that can be held in the vapor state decreases–and if that continues, then eventually, the relative humidity will reach 100% and then go beyond. And the excess, then, must condense out as liquid or ice, some form of condensed water. So the idea for the answer to that has to do with adiabatic cooling and the decrease in the saturation vapor pressure because of that cooling.

Questions on that?

The greenhouse effect, why does it warm rather than cool the surface of the planet? The point is that the greenhouse effect has to do with atmospheres absorbing some fraction of the radiation that’s trying to penetrate through it. Now, that’s a good starting statement, but we’ve got to get more specific than that. It turns out that it’s easier for the Sun’s rays to get in than it is for the Earth radiation to get out.

If it were the opposite, then I suppose the greenhouse effect would be a cooling mechanism, making it difficult for the Sun’s rays to get in. But if it’s easy for the Earth’s radiation to get out, that probably would cool. But it doesn’t. So easy to get in, difficult to get out, therefore, the Earth’s temperature must warm in order to reach that radiative balance.

So what makes it easier to get in than to get out? It’s not the direction. It’s the wavelength. The wavelength of the radiation coming from the Sun is centered around–in the visible range, about 0.4 to 0.7 microns. And because of the gases we have in the atmosphere, the particular gases, the atmosphere is mostly transparent to those wavelengths. On the other hand, the Earth’s surface being cooler, when it radiates, it radiates at a longer wavelength. And for those wavelengths, the atmosphere is mostly opaque. So that’s why the greenhouse effect is there. And quite a few of you had a garbled explanation of that.

What gases contribute and why? So it turns out that none of the molecules we define as being air molecules–that is, N2, O2, and argon–none of them are active greenhouse gases. Their structure is too simple. They don’t have a dipole moment, and they cannot absorb or emit the infrared radiation.

But slightly more complex molecules, like CO2, water vapor, ozone, N2O, NO, they have enough of a complex structure–mostly asymmetry. They break symmetry in a way that allows them to have a positive-charged end and a negative-charged end. And that means that they’ve got a dipole, and they can–just like your cell phone, they can emit and they can absorb radiation using that oscillating dipole.

Questions on that?

Question five, a small rocky asteroid. OK, I give you the solar constant–that’s how much radiation is reaching that–and I give you its albedo. So what are you going to do? Here’s the thing here. So if you’ve got this little asteroid, it’s got a certain value of the solar constant. It’s going to be different for Earth, because it’s closer to the Sun, so the solar constant’s going to be a little bit larger.

Really, you’ve got to make the same assumption we made in class. You’re going to have to assume that you’ve reached some kind of a steady-state temperature, where it’s receiving an amount of radiation per unit of time and emitting at the same rate to space. And then you’ve got to find the temperature of the object that allows it to be in that steady-state balance. If the temperature’s too cold, it won’t be radiating enough to be in steady-state. If it’s too hot, it’ll be radiating too much to be in steady-state. So there’s only one temperature that’ll allow it to balance its heat, and that’s the one that you can derive from this formula.

So solar constant, 1 minus the albedo, over 4 times the Stefan-Boltzmann constant. So 2,000 watts per square meter, 1 minus the albedo, 4–there’s the Stefan-Boltzmann constant–take the fourth root, 257 K.

Any questions on that? I think people did pretty well on that one.

Question six, explain why a planet like Earth will have lost its light gases over geologic time but retained its heavier gases. So the point is the Earth is in this middle category of planets where it’s large enough to retain the heavier gases but not large enough to retain the light gases. And the way one reasons that out is to realize that the Earth has a certain escape velocity–that’s the velocity needed for particles to escape–and that is independent of the particle itself. That’s just a fixed value.

Then the light molecules are moving rapidly. The heavier molecules are moving more slowly. And it turns out that enough of the light molecules are exceeding the escape velocity, so we will have lost that over geologic time. But the heavier molecules are moving slowly enough, so they do not exceed the escape velocity, and they can be retained gravitationally by the planet. So most of you got that, while a few of you buggered up those descriptions a little bit.

Question seven was just a–here it is–it’s just a cloud. The earth is radiating, and the cloud top is radiating. The temperature of the Earth I gave you as plus 30, and this was minus 60. And I just asked you to compare how these two surfaces are radiating to space. The two things you had to do, you had to compute the wavelength that was being most strongly emitted by each surface. Each surface emits a broad range of wavelengths, but there’s a peak where it emits more than at other wavelengths. And that’s given by Wien’s Law, which is here.

So for the higher temperature–this is the constant divided by the temperature in Kelvin–it comes out to be 9.56 microns. And for the cooler temperature, it’s at about 13.6 microns. But both of those fall into the thermal infrared range of radiation. And in terms of the flux, how much radiation is being emitted, it’s the Stefan-Boltzmann Law. Again, put in the temperatures in Kelvin, and you get 483 and 118. A big difference arises because of that powerful power of four in the exponent there.

Questions on that? Yes.

Student: I just have a question about significant figures. So I know when you emailed us that one time, you said to give us back one more significant figure than you gave us. So I think I got some points off for that, but if you only gave us one significant figure, would I give an answer back with two?

Professor Ron Smith: One or two would be my recommendation. So here, I’m not sure I’ve been completely consistent with that. For example, when I gave you the temperature on that problem, I just gave it to you as kind of a round number, 30 and 60. So that’s not many significant figures. On the other hand, the constants are known to, in fact, many more than this. I’ve truncated that a bit too.

So you do have a bit of a dilemma there. Do you go with–here, there’s four significant figures. Here, I gave it to you in Celsius. There was only maybe one or two. By the time I convert that to Kelvin, it looks like maybe there’s now three. So, yeah, this is tricky. And I hope we weren’t unfair.

But if you were going to give me five or six decimal points there, I would have taken off credit. Or if you’d truncated that off to be, you know, 480 or something, I probably would have taken off some credit. But if you’re anywhere in this ballpark, I hope we didn’t. But show that to me if there’s a problem with that.

I think we can go on to question eight, then. So this was one of those release of a pollutant in an unconfined atmosphere with no wind. It says there’s no wind blowing. So I imagine a source that’s diffusing out in all three directions–north, south, and upwards–forming a growing hemisphere of polluted air. And the radius of that hemisphere is growing with time according to the square root of the diffusivity times time.

And so putting the numbers in we had for that–be sure you work in seconds–I get 1,039 meters. And then the way I did the problem was to compute the mass of air in that hemisphere. So that’s the density of air times the volume of the hemisphere. And I got 2.8 times 10 to the 9 kilograms. Sorry, kilograms.

And then this concentration we’re looking for is the mass of the pollutant divided by the mass of the air into which it has been mixed. And I said you put in 10 tons, I think, so that’s 10 to the fourth kilograms divided by 2.8x109 kilograms, just this number brought down, and I got 3.57x10-6. In mass units, kilograms per kilogram. So noticing the minus six, I could write that as 3.57 parts per million–there’s the million–by mass. By mass because it’s kilograms per kilogram. So that’s what I got for that.

Questions there?

People did well on nine. I just gave the–I reminded you of the lapse rate for Earth, and I gave you the height of 30 kilometers. You had to put it into this approximate formula for an isothermal atmosphere, and you got that for a mass density at that elevation in units of kilograms per cubic meter.

All right, the last one, then. Supercooled water is liquid water at a temperature below the normal freezing point of water–that is, below zero degrees Celsius. It typically has the property that it wants to freeze, and if something comes along that will give it that chance, it’ll freeze instantaneously. But until you do that, it will stay for a while as liquid water.

Unstable lapse rate. An unstable lapse rate is an atmospheric lapse rate–which, as you know, is the rate at which temperature changes with height, that’s a definition of a lapse rate–that is more negative than the adiabatic value. The dry adiabatic value is minus 9.8 degrees per kilometer. So if it’s more negative than that, that would mean, for example, minus 10, minus 11, minus 12 degrees Celsius per kilometer would be an unstable lapse rate. If it’s less negative or positive, like minus five degrees per kilometer, minus two, minus one, zero, plus two, plus five, those would all be stable lapse rates.

And the characteristic of that is that when a parcel rises, then, it cools less rapidly than the environment in which it’s penetrating. And therefore, it ends up warmer than its new environment. And that’ll allow it to continue to rise, and it’ll lead to convection if you try to set up an unstable lapse rate.

Archimedes’ Law, few of you got this. I guess it’s because it’s not in the book, and I only mentioned it once in class. But still I was a little bit surprised, because I think it’s well known. But the Archimedes’ Law I’m referring to here is that the buoyancy force that acts on an object immersed in a fluid is equal to the weight of the fluid displaced. So if I take a basketball and I push it down into a swimming pool, I will feel a force trying to push it back up that’s equal to the weight of the water that would be there if the ball were not.

Now, that turns out to be pretty simple to understand, because the pressure increases as you go down in that pool, and therefore, the bottom of the basketball feels higher pressure than the top. And that’s what’s causing the buoyancy force. And the rate at which that pressure increases as you go down is related to the density of the surrounding fluid. That’s why the surrounding fluid comes in, because the surrounding fluid determines the rate at which pressure increases with depth. And it’s that increase that gives you the buoyancy force pushing up on the ball. So that’s Archimedes’ Law.

The tropopause is the boundary between the troposphere and the stratosphere. The troposphere is that first layer where the temperature normally decreases with height. The stratosphere is the next one up where the temperature increases with height.

Advection fog is fog caused by moving warm air–warm, moist air over a cold surface. Advection means horizontal movement. So that’s a clue. And if you take warm, moist air and move it over a cold surface, it’s going to lose heat to that surface. As it cools down, the saturation vapor pressure’s going to drop. And the familiar story, you’re going to form some condensed water droplets in that way, and that’s the fog.

Questions on any of that?

Sorry to take so long on that, but I want to just be clear that you should spend some time going over those. If you want to do better on the next one, part of the key is to really go over the last one and understand everything that you did and how you could have done better.

Chapter 2: Differential Heating and Earth’s Energy Balance [00:23:06]

Let’s see here. It’ll be just a minute for that light to come on. I want to start a new subject today, which is the general circulation of the atmosphere. There it is. So we just finished a big section on water in the atmosphere, so this is a new topic now. We’re going to look at the whole globe and try to understand the large-scale air motions on the globe.

I always have a debate with myself at this point of the course. I wonder whether I should start with the smaller-scale air motions like storms and then talk about the general circulation, or whether I should do the general circulation first. And I never quite know the answer. This year, we’re going to start with general circulation first, then do storms. But then as we get into seasons, we’ll have to come back to review this subject of the general circulation of the atmosphere.

So the basic principles that are going to control how we think about this problem is that differential heating generates the circulation. Now, you know this. If I have a bonfire over here or a hot stove or something, it’s going to heat the air near it. That warm air wants to rise because of its buoyancy, and you’re going to set up a circulation in this room driven by the fact that you’re heating the air here but not there. Differential heating means you’re applying different heating at different locations.

But that circulation, then, is going to move heat–because the air carries heat with it, it’s going to move heat from hot to cold regions trying to balance out the local heat budgets. So we’re going to see this principle applied now to the atmosphere as a whole. And you can imagine starting with just this simple idea. Here’s the globe, and the Sun hits the equatorial regions most directly. In the equatorial regions in the middle of the day, the Sun is directly overhead. So this cosine theta factor is about one. You’re getting the full intensity of the Sun, 1,380 watts per square meter, directly hitting regions near the Equator.

And it turns out that there’s less thermal radiation being emitted in the equatorial regions than is being received from the Sun. So while we assumed a week or so ago that the Earth as a whole is in radiative balance, that is not true for each latitude belt. Each latitude belt is out of a state of radiative balance. As I’ve shown here near the Equator, getting more from the Sun than it’s emitting to space, and at the poles, where the radiation is coming in tangentially, you’re really getting very little sunlight.

Of course, with the tilt of the Earth, you do get a little bit, but let’s assume that ray is just parallel here. It’s just going to hit the North Pole tangentially. It’s not going to deliver any solar energy that way. But the temperature of the pole is not absolute zero, so it will be radiating to space.

So there’s an imbalance here. There’s an excess of heat here and a heat deficit near the poles. How can that be sustained? Well, the way that is sustained is to have some other mechanism for transporting the heat from the Equator to the pole. And that’s where the general circulation comes in. So not only is the difference in heating going to produce a circulation by buoyancy forces, but that circulation then is going to solve this problem we have of how to balance out the heat budget at each latitude.

So there’s two ways to look at this problem. Differential heating causes the circulation, and the circulation helps us to balance the heat budget at each latitude. I’m going to go through this now in some detail, because this is the key concept for understanding the Earth’s general circulation.

So on the x-axis is latitude from the North Pole down to the Equator to the South Pole. And there are two curves here. The red curve is radiation received from the Sun. That could be over a day or over a year, averaged out. But anyway, there’s a lot of radiation being received from the Sun in the equatorial region and much, much less at the poles.

I brought the globe over here just to remind you of this. If I forget the tilt for the moment and the Sun’s radiation is coming in like this and the Earth is spinning–let me get it spinning in the right direction, it spins towards the east–you’re going to get a lot of heat distributed around the globe near the Equator. But the Sun’s rays are just coming tangentially near the poles. So on a per unit area basis, you get very little radiation near the poles.

Now, that’s probably an annual average. And during Northern Hemisphere summertime with this tilt, OK, there is a little bit of radiation received at the pole. In the Northern Hemisphere winter, there’s absolutely none. So you don’t get a zero value. You get something that’s an average of the seasons. But it’s definitely much less than what is received at the Equator, just because of the geometry of the sphere. It’s nothing more than just the geometry of the sphere and the fact that the Sun’s rays are coming in parallel to each other.

Now, the equatorial regions are warmer than the other parts of the planet, so they radiate more by the Stefan-Boltzmann Law. I have it right here. The hotter the surface, the more it radiates, but not enough to make up that difference. So there is a heat surplus near the Equator. Near the poles, it radiates less than at the equator, but not so much less that it doesn’t leave a deficit there.

Chapter 3: Poleward Heat Transport [00:29:57]

So we’ve got this problem of how to balance the budget in the Equator and in the high-latitude regions. And that is done, as I’ve said before now, by a heat transfer mechanism that has to do with the ocean and the atmosphere transporting heat to the poles to balance out that heat budget.

So here’s what we estimate from what we know about the atmosphere using balloon soundings and so on to work this out. What’s plotted here on the x-axis is the heat transport in watts–it’s a power unit–as a function of latitude. North Pole, Equator, South Pole. And the total is shown in the black curve, the atmospheric component in red, and the ocean in blue.

Now, something about the units. It says capital P, capital P capital W up there. The capital of the P means “peta.” You know the prefixes “kilo” and “mega” and “giga” and “tera.” Well, you keep going up till you get to peta, which is 10 to the 15th. So a petawatt is 10 to the 15th watts. So for example, at a latitude of about 30 or 40 north latitude–and New Haven is right here at 40 north, just about in the peak of this curve–there is about six petawatts of energy being transported towards the North Pole by ocean and air currents in the atmosphere. Most of it’s by air currents.

Now, do we know this? When we walk outside today, do we know this? Well, I would argue that, yes, you probably do know this. Because from your experience now, you know that on days when it’s especially warm, the air is blowing from the south, typically. On days when it’s especially cold, the air is blowing from the north. So there’s this transport of air back and forth.

You have roughly the same amount of air moving north and south, but when it’s moving northward, it is–you can put that down on the floor, if you like. Sorry it’s in your way. When it’s blowing northward, it’s carrying more heat, because it’s hotter. And when it’s flying southward, it’s carrying less heat, because it’s colder. So just from that asymmetry that you would notice on a day-by-day basis, you can understand what’s causing this north-south heat transport. It’s just air moving back and forth, but carrying more heat on its way north than it does on its way south.

I’m going to illustrate that first for the ocean. This is a crude diagram of the gyres in each of the world’s ocean basins. For example, in the North Atlantic gyre, there is a warm current going northward. That’s the Gulf Stream, shown in red to indicate that it’s a warm current. There’s a cold current running southwards.

So look. You’re taking the same amount of water north and south. You’re not piling up water at the pole or at the Equator, but you are transporting a net amount of heat–not a net amount of water, but yes, a net amount of heat. Because the water going northwards is warm. The water moving southwards is cold. So they don’t balance each other out in terms of the heat that they’re carrying.

And the same thing goes on in the Southern Hemisphere. For example, there are warm currents and cold currents, so the net heat transport is towards the south. That’s why they plotted this negative on the last plot. So if we call this plot the northward heat transport, it’s positive in the Northern Hemisphere, because you’re transporting heat towards the North Pole. It’s negative in the Southern Hemisphere, because you’re transporting heat towards the South Pole, going in the opposite direction, so we give it a negative sign like that.

So the oceans are doing some of it, but the atmosphere does most of it. Typically, if you have warm air and cold air and the jet stream is running around the Earth kind of in a zonally symmetric fashion, well then, you’re not transporting any heat. But very frequently, it’ll break down into a wave-like motion, where cold air will come streaming down–in fact, here’s New Haven right about here–cold air will come streaming down from the pole, heading to the Equator. That has to be replaced by an equal amount of warm air moving poleward. And in so doing, you’ll have a net amount of heat transported towards the pole, and that’s what’ll balance out the heat budget at each latitude.

Chapter 4: Heat Flux by moving air [00:35:08]

Just to remind you about this basic physical principle, if you imagine a pipe carrying a fluid, what do you know about the pipe? You probably know the cross-sectional area of the pipe, you know the speed of the fluid in the pipe, and you probably know the density of the fluid in the pipe. Then if you wanted to know the mass flux–how much water or how much air is being carried by that pipe–it’s simply the product rho U A, and that’ll have units of kilograms per second. That’ll be a mass flow rate.

Well, it’s only a small step–once you understand that, it’s only a small step to go beyond that to understand how much heat is being transported in the pipe. Because we know the heat capacity, or the heat stored in a fluid, is proportional to the mass that you have times the specific heat capacity times the temperature. So this quantity C sub p–I’ve given the value here for air–it’s called the specific heat capacity at constant pressure. You can look it up. You can Google it. It has this value for air. The units here are joules per kilogram per degree Kelvin.

So the more mass you have, the more heat is stored. And the hotter the temperature, the more energy is stored in a particular kilogram of air. So to go from the mass flux to the heat flux, you just have to multiply by these additional two factors, the heat capacity, which is the amount of heat you have per degree of temperature, and then the temperature itself.

Student: And so what are the units on heat flux?

Professor Ron Smith: The units on heat flux would be joules per second. Or you may know that that is a watt. A joule per second is a watt. So the point is that when air moves, heat moves too, because heat is stored in air.

Now, this doesn’t quite convey the full picture I want to convey. If I went back to this picture, what I’d like you to envision is two of these pipes. One’s carrying air southward, and the other is carrying an equal amount of air mass northwards. The amount of mass flow in each pipe is the same, but the amount of heat flow is not, because the temperature is different for the two pipes.

So no net transport of mass, but yes, a net transport of heat. Because one stream of air is cold, and the other stream of air is hot. So I don’t think this is a particularly difficult concept, but it lies at the base of understanding how the atmospheric general circulation works.

Are there questions on this? I’d love to take a minute or two on questions. Yes.

Student: Why does that happen? Why does it not stay constant?

Professor Ron Smith: I’ll get to the question of why that happens in about a week. This happens because of an instability that occurs along the jet stream that generates what’s called a mid-latitude frontal cyclone. It’s the main weather producer around the mid latitudes.

And over the next six months, your life will be largely controlled by these events. Storms that come through New Haven are mid-latitude frontal cyclones. They are basically breakouts. The prison boundary has been breached, and the cold air is breaking out and going south, and the warm air is heading northwards when the jet stream develops a little instability in it and breaks down in that way.

So your book is good at talking about these mid-latitude frontal cyclones, and we’ll be talking about it in this course as well. It’s a spontaneous event. It happens throughout the winter, just day after day after day, just one after the other of these kind of breakout events.

Chapter 5: Effect of Earth’s Rotation: Coriolis Force [00:39:19]

Now, the subject gets a little more complicated, though, because of the effect of the Earth’s rotation. Remember, I’m talking about the Earth spinning on its axis. And what does that do? Well, there’s one obvious thing it does and one perhaps that’s not so obvious. The obvious one is that it distributes the Sun’s heat zonally. So if I just had this globe sitting fixed without spinning and the Sun was over there, it’s hitting this side of the Earth, but never this side. And so we’d have a hot spot forming here, and this would be icy cold on the other side.

Instead, because the Earth is spinning on its axis, it’s not a hot spot that forms, but a hot region near the Equator that forms. The spinning of it distributes the Sun’s heat zonally. We use the word “zonally” when we’re going this way around the Earth. And you end up with a latitudinal–or sometimes we say “meridional”–temperature gradient, or heating gradient. But zonally, the heat gets well distributed because of the spinning of the Earth. So that’s the first point.

The other one that we’ll get to next week is the Coriolis force. Whenever a planet spins, it generates a somewhat mysterious new force called the Coriolis force. Whenever a parcel tries to move in a straight line on a spinning planet, it gets deflected. And we’re going to talk about that next time. And that really has a major influence on the nature of the general circulation, as we will see.

And I’ll give you a hint of that here, and we’ll do it more quantitatively next week. So for example, let’s look at the case when the planet is not rotating. So the point that’s right under the Sun, if the Sun’s down there, would heat up. And every other place, you’d get less and less sunlight. Whether you moved this way or that way, you’d get less and less sunlight, and you’d get no sunlight on the backside.

So you’d have a subsolar hot spot, just one very hot area right here. The low-level air would move in towards it. Those are the red arrows. And then, because the air gets heated, it would rise, and then aloft, it would spread out in all directions. And those are the thin blue arrows. So that’s just like I built a bonfire here. You’d have air moving in towards it from all directions rising. And then when it hit the ceiling, it would go out in all directions.

Now, this is not at all what we have on Earth. This is the idealization or kind of a thought experiment of what the circulation would be if the Earth was not rotating on its axis. So don’t take this as what happens. Take it as a building block to understand what actually happens in the Earth’s atmosphere.

So let me bring in now the effect of distributing the heat zonally. So this would happen if I had weak rotation. I learned a few weeks ago in a lecture I went to on the planet Titan that actually the planet Titan fits into this kind of a description. It rotates rapidly enough to distribute the heat zonally but not so rapidly that the Coriolis force is a major player.

So here’s what happens. You get a hot equator, a fairly uniform hot equator. Air moves towards it at low levels, gets heated, rises, and then moves away from it, poleward, at high elevations. It would sink somewhere out here and fall to the surface, and then move towards the hot line again, and then rise. You’d have a kind of planetary, simple bonfire kind of circulation towards and away from this heated line caused by the solar heat.

Yeah.

Student: Would the surface at the equator be really hot and the surface at the poles be really cold?

Professor Ron Smith: It’d be cold. That’s right. And the circulation would be a rather simple one, rising near the equator, sinking near the poles, and then the air moves right back to the equator.

Chapter 6: Effect of Earth’s Rotation: Circulation Cells [00:43:46]

But now, that’s for weak rotation. For the Earth, we rotate somewhat faster than that. You know it takes 24 hours for us to have a day. That’s a pretty fast rotation. By the way, this is sometimes called the single-cell circulation. Well, I guess there’s two of them, but on one side of the equator, there’s just one, rising, sinking, rising, sinking. So the single cell refers to what happens on either the northern or the southern hemisphere.

On Earth, you get something that’s closer to a three-cell circulation, because the rotation rate is larger. So the Coriolis force is playing a larger role. You do get rising motion near the Equator, but it only goes slightly north and south and then sinks. And north of that, you get westerlies and then easterlies again, laid out like I have very crudely on this line diagram. I’m going to show you a better version of this in just a moment, but the main point is to show how much more complicated this is than that. And that all has to do with the stronger action of the Coriolis force.

So this is a more typical artist conception of the three-circulation system. It’s a little bit more complicated, of course. The rising motion along the Equator then gives you poleward-moving air aloft that descends at about 30 degrees north and south latitude. This is a very strong overturning cell. It’s called the Hadley cell. And you must know about this one. This is really one of the dominant features of the climatology on Planet Earth is the Hadley cell.

Rising motion near the Equator, air moving poleward till it gets to about 30 degrees and then sinks, and then returns towards the Equator at the surface of the Earth. Because of the deflecting force from the Coriolis force, that air doesn’t move directly towards the Equator. It comes in from the easterly direction. So it forms what we call the northeast trades, wind blowing from the northeast in the Northern Hemisphere, and the southeast trade winds in the Southern Hemisphere.

North and south of that, there is the Ferrel cell, which is a rather weak overturning in the opposite direction. But that’s not the primary feature. I don’t like this diagram, because they show the Ferrel cell so clearly, but they don’t show the most obvious and dramatic feature of the westerlies, and that’s the existence of these mid-latitude frontal cyclones that I was talking about just a moment ago. And when I show you a movie, you won’t see the Ferrel cell, but you’ll see those mid-latitude frontal cyclones in that region.

And then in the polar regions, you get also a fairly weak cell giving you some easterly flow at the very high latitudes.

So this is the three-cell pattern of general circulation for Planet Earth. And I want to show you a set of satellite loops to try to illustrate a few of these features. This is a global view pieced together from a number of geostationary satellites posted at various longitudes around the globe. And it’s looking at the emitted long-wave radiation. So we won’t see day and night here. We’re not counting on the Sun to illuminate these clouds. If we did that, it would go black, light, black, light as the Sun illuminated or not.

But instead, we’re looking at the emitted radiation–Stefan-Boltzmann-type emitted radiation in the thermal infrared–and that’ll tell us about where the clouds are and where the clear air is. For example, these light areas are clouds. They’re higher, so they are colder. They emit less. Just like the problem you had on the exam, they emit less because they’re higher. And that reduction in emission is shown here as a white region.

The lower parts and the surface of the ocean and the surface of the planet are hotter. They emit more. And that’s illustrated here by a dark pixel. So you’re basically looking at a moving temperature map of the Earth, where the temperature’s mostly controlled by the height of the object that you’re seeing. The clouds are high, so they’re cold. The ocean and the land are low, so they’re warm. I don’t think I’ll have much time to talk about this, but I just want to leave you with this image of how this moves.

This is three or four days. And some of the things you see here are the Intertropical Convergence Zone, where the trade winds are converging and the air is rising. And then your eye is really caught by these mid-latitude frontal cyclones. I’m going to play that one more time, and look at all these little comma-shaped cloud features here that represent this rapid flux of heat northwards and southwards, trying to balance the Earth’s heat budget latitude by latitude. I’ll get this to show again. I’ll show it just one more time, and then we’ll quit.

We’ll quit there and continue this discussion next time.

[end of transcript]