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GG 140: The Atmosphere, the Ocean, and Environmental Change
Lecture 6
 Greenhouse Effect, Habitability
Overview
A simple model of the overall Earth’s heat budget is derived. The Earth is assumed to be in equilibrium with the input of solar radiation balanced by the output of infrared radiation emitted by the Earth’s surface. Using this model, the Earth’s surface temperature is calculated to be cooler than in reality due to the lack of an atmosphere and the greenhouse effect in the model.
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htmlThe Atmosphere, the Ocean, and Environmental ChangeGG 140  Lecture 6  Greenhouse Effect, HabitabilityChapter 1: Earth Energy Balance [00:00:00] Professor Ron Smith: So last time, upstairs, we did this systems experiment where we had a tank of water. We tried to analyze it and its three components–the reservoir, the input, and the output. We talked about things like steady state balance, transient response to changes in the input and the output. We derived a little mathematical relationship describing how the outflow depends on the depth of the water. We put those equations together, we came up with a prediction of what the equilibrium water level would be for various values of flow inputs. It worked out pretty well. We got fairly good quantitative agreement on that. Are there any questions on any of that? So that’s basic system analysis. We’re going to apply that idea today to the energy budget of the Earth. And we’ll be applying it later in the course to other simple systems relating to the Earth. In this one, the input is going to be the solar radiation. Sometimes it’s called insolation, with an o, not insulation, but insolation. That’s the rate at which solar energy is approaching and then finally being absorbed by the Earth. The loss is going to be the infrared radiation to space. We’ll talk about the laws that govern that today. And the heat storage will be not the whole Earth, because it takes too long for heat to conduct in and out of a whole Earth–millions, if not hundreds of millions of years. So really the system we’re talking about is just the skin of the Earth, the first few meters. In the ocean, the first five kilometers or so. The part that is responsive to the heat that’s being put in and taken out. So we will have a reservoir. We’ll have inputs, we’ll have outputs. Now, there’s one input we’re going to neglect. There is some heat in the deep interior of the Earth that comes from the decay of uranium that makes the interior of the Earth quite hot. And that heat can leak out to the surface of the Earth. That’s called geothermal heat. That can be important for certain subjects, but for climate studies that is negligible. That’s only a few–a tiny fraction of a watt per square meter that reaches the Earth’s surface from the interior. And we’ll see, these fluxes are orders of magnitude larger than that. So we’re going to neglect geothermal radiation–geothermal heat from the interior. We’re also going to assume that the Earth has a uniform temperature. That’s not a great assumption, but in the beginning we’re going to just try to find a formula that describes the average temperature of the Earth. Later on we’ll be dealing with the pole to equator temperature gradients. But for now, the Earth will have a single temperature today. And of course, we’re going to assume steady state. So we’re going to use the kind of formulation we did with the water tank where we understand the inputs and outputs. Sometimes they’ll be out of balance, but we’re going to make the assumption they’ll be in balance and use that to solve for certain properties of the system. Chapter 2: Black Body Radiation – Wien’s Law and StephanBoltzmann Law [00:03:34] So that’s the goal today, but before I do that I’ve got to talk to you about the fundamental laws of radiation. Now, I mentioned this the other day when we were talking about the stratosphere and the thermosphere. If you have an object like anything in this room, or the Sun, or any object at all, and you plot the emitted radiation as a function of wavelengths–I’ll use the Greek letter lambda (λ) to represent wavelength. That’s the distance from the crest to crest of the wave. It looks something like this. I’m going to draw a bunch of curves for different temperatures. These are called black body curves. It’s also referred to sometimes as the Planck Function. So if you’ve got a book on physics, look up black body or look up the Planck Function and read what it says about this. Now, this would be for a cooler temperature. This would be for some intermediate temperature. And this would be for a high temperature. So for example, this might be the surface of the Earth with the temperature at 288 Kelvin. This might be for the surface of the Sun with the temperature of 6,000 degrees Kelvin. Notice two things about this. When you make the temperature greater, of course, you’re increasing the amount of radiation that’s coming out at every wavelength. Every wavelength the intensity of the emitted radiation increases with temperature. As does the integral under the curve. The total sum of all the radiation is the area under each of these curves. That total increases as well. Furthermore, notice that the peak of the curve shifts with temperature. It’s there for the cooler temperature, there for the intermediate temperature, and there for the higher temperature. So the color, if you like, the color of the radiation changes as the temperature of the object changes. Now, there are names for these particular aspects of the radiation law. The shift in the peak is referred to as Wien’s Law. And it’s given as by the following formula. Lambda max is equal to a constant divided by the temperature expressed in Kelvins. And this lambda is going to be in units of microns. A micron is 10^{6} meters. Before I compute an example of that, let me just remind you of the electromagnetic spectrum for a second. Wavelength increasing to the right. There’s a range we call the visible part of the spectrum that runs from 0.4 to 0.7 microns. The human eye is sensitive to those waves. Then, as we spoke about last time, the ultraviolet is over here, and the infrared is over there. Microwave is further over, and then radio waves are further over to the right, and Xrays are here. So just to remind you, you might want to remember 0.4 and 0.7 as the range of the visible part of the spectrum. Now, let’s do a calculation here. For the Sun, the surface of the Sun is approximately 6,000 Kelvin. If I plug into that formula to get lambda max, it’s going to be 2897 over 6,000. That’s approximately 0.48 microns. I’ll write it here as μm, micrometers. And that is going to lie right about there, roughly in the middle of the visible range of the spectrum. That’s kind of interesting, and you might want to think is that just an accident? I’ll say more about that later. But make a little question mark there. Why does it happen–yeah? Student: What is the 6,000 again? Professor Ron Smith: The surface temperature of the Sun. Is it just an accident? That’s my question to you for thought. Is it just an accident that the Sun’s peak radiation happens to fall in the human range of high sensitivity? Interesting question. Let’s do the same thing for Earth. Typical temperature for Earth is about 288 Kelvin. Lambda max then is 2897 over 288. That’s an easy one to do. That’s approximately 10 microns. That is way out–no, that’s in the infrared. Let me not exaggerate this. That is somewhere–this is kind of a log scale I’ve got here. That’s in the infrared part of the spectrum, the socalled thermal infrared. Question? Student: Is that the surface temperature? Professor Ron Smith: This is the surface temperature of the Earth, 288 Kelvin. We’ve used it before. It’s 15 degrees Celsius–it’s like the temperature in this room. Maybe we’re a few degrees warmer than that, but this is a typical Earthtype temperature. So this makes a big difference. This means that the radiation coming from the Sun is in one wavelength. And the radiation emitted by the Earth is quite a different wavelength altogether. So remember that. That’s going to be a key for understanding the greenhouse effect. The other law that arises from this is called the StefanBoltzmann Law. And it relates to the total power, the area under the curve, power per unit area of an object–an object emitting–is given by a constant–I’ll use a Greek letter sigma for that–times the temperature to the fourth power. This constant is the StefanBoltzmann constant. It’s derived in physics classes from first principles. And the value is 5.735 times 10^{8} watts per meter squared per Kelvins to the fourth power. You could look it up. You can Google it. If you ever forget it, you can find it very easily. It’s a very famous physical constant, the StefanBoltzmann constant. It will never be any problem with knowing what that value is. Now before we go any further, let’s see if these two laws make any sense. The StefanBoltzmann law that says that the total power goes like the fourth power of the temperature. If I were to double the temperature, the power would go up by a factor of not two, not four, not eight, but 16. So you’d increase the power by a factor of 16 by doubling the temperature. Very powerful sensitivity to temperature. And then Wein’s law that says that the wavelength goes inversely. The wavelength of maximum radiation goes inversely with the temperature. The higher the temperature, the shorter is the wavelength that’s being strongly emitted. Now, I’ve got this little light bulb over here. And I’m just going to hit some lights. I’ve got a variable transformer, and I’ve got a light bulb that has no frost on the glass, so you can see the filament in there. I can change the temperature of the filament by how much electrical current I pass through it. So when, of course, it’s at room temperature like that, it’s radiating a little bit, but it’s radiating at wavelengths in the infrared for which the eye is not sensitive. Then as I gradually increase the temperature, you can begin to see it there. There’s a little bit now in the visible part of the spectrum, but it’s red. Remember that diagram–I should have put–let me just back up and get that up here for a second. On this end of the spectrum is the red, and this is the blue. So originally the wavelength being emitted is here, now it begins to move over. As I increase the temperature you begin to see some red. Then as I continue to move it over, at some point you’ll be seeing equal amounts in all of the—in all the parts of the visible spectrum. So there it’s becoming white. Because that peak has moved over, centered on the visible part of the spectrum. Notice it’s also very much brighter. So with this one little device, we are looking at both the StefanBoltzmann law and Wien’s law. They’re both illustrated by this little experiment. Any questions on that? So all I’m doing is changing the temperature of the filament and what I’m seeing is an increase in power and a shift in the wavelength from the red into the center of the visible part of the spectrum. Yes, question. Student: Why does it appear to turn white when it does, directly after going to yellow? Professor Ron Smith: Well remember, think of this peak sliding over. So first you see only the red, then you begin to see some red and some green is in the middle. That’s going to give you yellow when you get the red and the green together. Then finally, when you get an equal set of all red, green and blue, that looks white. So white light, as Newton figured out, is the sum, an equal sum of all the visible parts of the spectrum. So we’re just seeing that happen as we shift that peak over and center it on the visible part of the spectrum. Yes. Student: Does the StefanBoltzmann Law give brightness or power? What is power?? Professor Ron Smith: Power is the emitted power, it’s the power in that radiation. It’s the rate at which energy is being sent from the object out into space with electromagnetic radiation. You can think of as brightness if you like, but brightness usually refers to how something is perceived by the human eye, and that’s not our context here. Our context here is an energy budget. How much energy is being lost. So I’m going to use the word power there instead of brightness. Anything else on this? Chapter 3: Infrared Emission [00:16:05] Well now I think we’re ready to do–oh, well there is one other thing maybe we should do while we remember. Some of you may not even believe me when I tell you that all the objects in this room are emitting. They’re just not emitting at a wavelength that the human eye can see. What do you see? You see light reflecting off various objects. That’s not what I’m talking about here. Objects yes, objects can reflect light coming from light bulbs and so on. But I’m talking about the emission of radiation. That’s a different process than reflection. So this is just an infrared thermometer. It’s not much different than the one you use to take your own temperature or the temperature of a child. It measures the temperature of an object by measuring the intensity of the radiation that’s being emitted. So it measures emitted radiation. Then from this curve, it figures out what the temperature must have been in order to radiate that intensely. So when I look at this table–by the way, I have to warn you of something. There is a little laser in there. See the red dot? But that’s just to help you aim it. It doesn’t play any role in the measurement whatsoever. It just aims so you know what object it is that you’re aiming at. But please don’t aim this in your eye. So anyway, the temperature of this table is 24 Celsius. Temperature of my hand is 30 Celsius. Try it yourself. I’ll pass that around. Anyway, that little gadget is a perfect illustration of the physical process I’m talking about. It’s an object emitting radiation in proportion to its temperature–in proportion to its temperature. Higher temperature, more radiation is being emitted. Chapter 4: Simple Model of Earth’s Energy Balance [00:18:08] Now I think we’re ready to do the main part of the derivation. I’m going to redraw the Earth here a little smaller. We’re going to assume that the Sun’s radiation–radiant beams are coming in parallel to each other. Because the Sun is so far away from us, we can assume that those rays are actually parallel by the time they reach the Earth. Therefore, they’re going to hit the Earth and cast a shadow out into outer space behind the Earth. The radiation that’s missing there is going to be the radiation that hits the Earth. Now, the reason I’m saying it that way is that you might be a little bit confused because this Earth’s surface is a sphere, and some of the radiation is hitting it in a normal direction, some of it is hitting it in a oblique direction. For the moment, I don’t care about that. I just want to know the total amount of radiation that hits the object independent of where it hits and what angle it hits. The easiest way to do that is just to imagine that the Earth is casting a shadow and has removed that amount of radiation from the beam. So the solar constant, the intensity of the radiation from the Sun at the orbital position of the Earth is about 1,380 watts per square meter. In other words, if you go outside the atmosphere facing the Sun and draw a square meter, there’s 1,380 watts of radiant energy passing through that square meter. If you had two square meters, it would be twice that value. So that’s on a per unit area basis. The area of this shadow then is just a circle. It’s πr^{2} where r is the radius of the planet. So the intercepted radiation is given by the product of Sπr^{2}. S is a solar constant, πr^{2} is the intercepted area. I want to illustrate that for a second. I’m going to hit the lights again here. I just have a source of light that’s nearly parallel and I have my globe here. Now the Earth is spinning on its axis of course, and its tilt may change with the seasons, but basically it’s casting a shadow as you see there. That shadow is just related to the projected area of this sphere, which is πr^{2}, the area of a circle. So that’s all I’ve done here is just to compute the total radiation hitting that globe by using the idea of casting a shadow. Questions on that? Pretty simple idea. Now, some of that radiation, however, is going to be reflected. And we define here something called the albedo. Albedo is the average reflectivity of a planet. It can be expressed as a decimal or as a percent. I’ll give you a number for Earth. For planet Earth the albedo is approximately 0.33, or you could call that 33%. Approximately 33% of the radiation that hits the Earth reflects off. And because that heat then doesn’t really enter the Earth, doesn’t really add energy to the Earth, we’re going to drop that off. So that was the intercepted radiation. The absorbed radiation then is Sπr^{2} times 1 minus the albedo. If 33% is reflected, well then, 67% is absorbed. That’s the 1 minus the albedo. It’s the other part. The amount reflected is Sπr^{2}A, and the part absorbed is Sπr^{2}(1A). So we’ve just partitioned that radiation, the part that is absorbed and the part that is reflected. Now, let’s try to compute what the emitted radiation is. We’re going to assume that the surface of the earth is a black body. It emits according to the StefanBoltzmann law. So we have to get the area–remember, this is per unit area–we need to find the area that is emitting. The area that is emitting is not the projected area of the sphere. It’s the actual area of a sphere. What is the expression for the surface area of a sphere? Anybody remember? Student: 4πr^{2}?. Professor Ron Smith: 4πr^{2}, right. So the area we’re talking about here is 4πr^{2}, and so the emitted radiation then is going to be 4πr^{2}σT^{4} using the StefanBoltzmann law. Questions so far? Yes. Student: For the surface area thing, we’re including the entire Earth, even though the Sun’s only shining on half of it? Professor Ron Smith: That’s right. That’s right. Remember, I’ve assumed here that the Earth has a uniform temperature, and since the emission depends only on the temperature, we’re going to use the entire surface of the Earth. After all, even in the real world, when expressed in Kelvins, that temperature is not so different. This might be 310 Kelvin, this might be 250 Kelvin, but out of a range of 300, so that’s not a huge difference. So this is not a huge problem. It’s not great, but it’s not a huge problem, the assumption of uniform temperature. Chapter 5: Equilibrium Calculations of Earth’s Energy Budget [00:25:51] Well, I think you know what I’m going to do now because of what I did in the tank experiment. I have laws that govern the input and the output, and I’m going to assume steady state and see what kind of a–how does a system come into equilibrium? So I’m making the steady state assumption. Simply, all I have to do is balance the emitted radiation, which has units, by the way, of watts with the intercepted or the absorbed radiation, which has units of watts. So I will simply equate those two. I’m going to write Sπr^{2}(1A) is now equal to, this is the statement of steady state, 4πr^{2}σT^{4} . That’s the profound and very useful step. And I could then go the next step and solve mathematically for the temperature of the planet. So let me do that. First of all, notice that the πr^{2} are going to cancel out from both sides of the equation. In other words, the result is not going to depend on the radius of the planet. Why is that? Well, the larger the planet is, the more radiation it receives from the Sun, but the more effectively it radiates to space. And those two things exactly cancel out. So there’s no direct impact of the size of the planet. Notice that four remains, however. That important difference between the projected area of a sphere and the surface area of the sphere, that stays in our calculation. Now let me bring T^{4} over here on the lefthand side. That’s going to be S(1A)/4σ. Have I done that right? Divide through by 4σ. Then the final step would be to take the fourth root of that equation. So I’m going to write it as T=[(1A)/4σ]^{1/4}. That is a prediction based on some simple laws of physics for how hot each temperature in the solar–each planet in the solar system will be. You need to know the solar constant for that planet, and of course, the further away from the Sun you are, the smaller will be that value. You need to know the albedo of the planet, how much radiation does it reflect. And then you need to know the StefanBoltzmann constant, but that’s a universal constant so that doesn’t change from planet to planet. Student: And the sigma is the StefanBoltzmann– Professor Ron Smith: So let’s put some numbers in here for Earth. What am I using for solar constant? 1,380. The albedo for Earth is 0.33. This is a formula for which you’re going to want to check the units. I haven’t put the units up here, but you should do that to be sure all the units are going to work out. That should be just in Kelvins. So when you do all the unit crossings out, you should get something just in units of Kelvin. Leave an inch in your notes to check on that. I worked this out and that’s about 252 Kelvins, predicted temperature for Earth. Questions on that calculation? Remember all the assumptions that went into it. These, as well as assuming that the Earth emits like a black body, because we used that. Now, if you convert that to Celsius, that’s minus 21 degrees Celsius. That’s a little too cold, right? Remember, the actual temperature for Earth is something around 288 Kelvin, which is about 15 degrees Celsius. Now if you look at this in the Celsius–in the Kelvin scale, 252 versus 288, that seems like a pretty good estimate. It’s only off by 10% or 15%, something like that. But if you look at it in terms of Celsius, it looks like a bad approximation. Why is that? Because remember, the Celsius scale–zero on the Celsius scale is based at the freezing point for water. Whether water is frozen or not is a big deal for us on Earth. So this prediction with the assumptions I made indicates the earth would be in a permanently frozen state. All the water on it would be frozen, whereas this would say most of the water’s going to be in the liquid state. So it’s a big deal to think of it in terms of Celsius or in terms of water, but it’s not such a bad approximation if you think of it in terms of absolute temperature. Now what do you think is the reason for the error? Where have I gone wrong with this calculation? Anybody in the back? Yeah. Student: You’re assuming no geothermal heat?. Professor Ron Smith: No. I did make that assumption, but it turns out that that’s a very good assumption. So that’s not the problem. Yes. Student: Are you assuming that it’s like a black body?. Professor Ron Smith: Yeah. What’s the catch word there? So yeah, the greenhouse effect. So the planet has an atmosphere. I’ve neglected the atmosphere here. What the atmosphere does on our planet and others is that when the radiation is emitted from the Earth’s surface, instead of letting all of that escape, it stops some of it, it absorbs it, and sends some of it back to Earth again. It’s as if that object is not able to radiate with its full black body potential. Something’s holding that heat in, and it’s the greenhouse gases in the atmosphere. So obviously, we’re going to be coming back to this time and time again in the course. But this important discrepancy is largely due to our neglect of the atmosphere, that is to say our neglect of the greenhouse effect. Chapter 6: Greenhouse Effect in Earth’s Atmosphere [00:33:29] Let me give a very brief description of how the greenhouse effect works on planet Earth. And then we’ll take a quick look at some other planets to see if other planets have a greenhouse effect as well. So once again, this is wavelength versus intensity of radiation emitted. For Earth’s temperature, the curve looks something like that. For the Sun’s temperature it looks something like that. We’ve already talked about this shift in the wavelength of maximum emission. But let me put underneath this some of the absorptive properties of the Earth’s atmosphere. On the same wavelength scale, I’m going to draw here the percent absorbed as radiation tries to pass through the atmosphere. In other words, I’m looking at a segment of the atmosphere and imagining radiation is either coming in and trying to penetrate to the Earth’s surface, or it’s trying to get out and escape. Either one, it doesn’t matter what direction it’s moving, but some fraction of the radiation’s going to be absorbed as it tries to pass through the atmosphere. That’s the fraction I’m going to plot on the yaxis here. Now, the visible part of the spectrum is about here. Let’s call that the visible part. I’m going to need these guidelines to help me be accurate. Here’s 100%. Basically, this absorption curve looks something like this. I’m just making some random wiggles here. There are a few areas that are nearly transparent in the wavelength spectrum. And then you get to the radio waves and it comes way down. So this is very schematic. I haven’t meant to be quantitative on this. The point is that most of the Sun’s radiation falls in a part of the spectrum for which the Earth’s atmosphere is transparent. When this percent absorption is low, that means the atmosphere is transparent. On the other hand, when the Earth tries to radiate back out to space, because it’s cooler, it radiates at a longer wavelength. That wavelength tends to fall in a region where there’s a lot of atmospheric absorption. The atmosphere is almost opaque at those longer wavelengths. So it’s a bit like a oneway valve. I don’t like that analogy, but let me say it and then correct it. It’s really like a oneway valve. The radiation coming from the Sun can pass through the atmosphere and heat the Earth. Now I’m neglecting the little bit that’s ultraviolet that heats the stratosphere, a little bit of the Xrays that heats the thermosphere. The bulk of the radiation comes right on through and hits the surface of the Earth. But when the Earth tries to radiate out to space, most of that radiation is absorbed and it has to be reemitted before it can finally escape from the planet. Now, I don’t like the term oneway valve because that implies that it makes a difference what direction the radiation is moving in, and that’s not the point at all. The point is that these are at two different wavelengths. Absorption doesn’t depend on what direction the photons are moving, but it does depend on their wavelength. So the short wavelength can penetrate, the longer wavelengths cannot. And that’s the origin, or that’s the physics behind the greenhouse effect. Yes. Student: Does it enter the atmosphere and then–? Professor Ron Smith: Enter and penetrate through. When I say penetrate– Student: And hit the surface?. Professor Ron Smith: That’s right. So this radiation will come through all the way to the Earth’s surface. There it will be absorbed. It’ll heat the Earth’s surface and then that heat will come back up and heat the atmosphere and so on. But the point is it gets all the way through the atmosphere without being absorbed. It isn’t until it hits the Earth’s surface that it gets absorbed. Thanks for that clarification. This is the atmosphere only I’m talking about. All those radiations, of course, cannot penetrate into the Earth. They’re absorbed immediately when they hit the Earth’s surface. Yes. Student: So you’re saying that the longer wavelengths–? Professor Ron Smith: Cannot penetrate through the atmosphere. Student: So they don’t go through–do they get into the atmosphere? Professor Ron Smith: They do. They get part way–for example, this happens to be radiation trying to get out, because it’s the longer stuff. So let’s just imagine for a moment, you’ve got the Earth’s surface radiating at, say, 10 microns, wavelength of 10 microns. It’ll penetrate up a few hundred meters or maybe even a kilometer into the atmosphere, but it’ll get absorbed. It’ll heat the air there, the air will then reradiate some up and some down. And some of the down will come all the way back and add a source of heat to the surface. So the fact that–I’m not saying it can’t penetrate a little bit. It’ll penetrate on average tens, hundreds, maybe even a few thousand meters, but it won’t get all the way out for the most part. That’s not 100%. Some will get out, but most will not. There are few special parts of the spectrum that are transparent, even to these longer waves, and we refer to those as windows. For example, the most famous one, the socalled infrared window is from 8 to 12 microns. It’s a fairly narrow window. It’s only 4 microns in width. And out there that’s not very much. But there, that particular wavelength can penetrate through. And this gadget, by the way, you might ask why does this work? If there’s such a thing as atmospheric absorption, then how come the radiation emitted from the surface hasn’t been absorbed before it gets to the instrument? This instrument is designed to work in that particular window. But it’s a narrow window, and I had to use special optics to get it just to work in that window. So be careful, the atmosphere is not opaque to every wavelength, but to most wavelengths out in the infrared part of the spectrum. Any questions there? Chapter 7: Energy Budgets for Other Planets [00:40:39] I think I have time to put up a small table. I’m going to keep this final formula here because that’s the one that I used to construct the table. Planet, albedo, predicted temperature, and the actual temperature, and something about the atmosphere of the planet. Venus, Earth, and Mars. We’ll just do those three Earthlike planets. The albedo for these planets are 0.71, 0.33, and 0.17. Venus is a very bright planet, if you look at in the sky. It’s very often the brightest object in the sky because it reflects radiation so strongly. We see that represented by that high albedo number there. The Earth we’ve already spoken about. Some of that reflectivity comes from the clouds in our atmosphere, which are quite bright. Mars is a rather dark planet, Mars with an s, because there are very few clouds in its atmosphere. So it’s only the surface itself that we’re dealing with when we’re computing the albedo. When I put the–I haven’t written the solar constant. Remember, S is different for each of these planets as well, but when I put the appropriate S in there and use that albedo, here’s what I get from this formula. I get 244, 252–we’ve done that one–and 216. And here’s the actual. The actual observed surface temperature for these planets. Mars is pretty good. We predict 216, the actual is 230. There is a little bit of a greenhouse effect on Mars, but it’s not terribly strong. The Earth we’ve already spoken about. That’s a pretty big difference when you express it in Celsius and think about the properties of water. But look at Venus. We are not even close. Not even in a Kelvin scale does that look like a reasonable estimate. That’s because Venus has the granddaddy of all greenhouse effects. That huge difference is due to the effect of Venus’s atmosphere trapping infrared radiation as it tries to leave the surface. Let me just mention here, Venus has a massive CO_{2} atmosphere. Earth has a moderate intensity, mostly air, but with some greenhouse gases like H_{2}0, CO_{2}, et cetera. I have to erase this now. And Mars has a thin CO_{2} atmosphere. CO_{2} is a greenhouse gas, but because the atmosphere of Mars is so thin, the greenhouse effect is not particularly strong on Mars. So there we have it. We have the fact that the greenhouse effect is–occurs on all planets that have atmospheres. It tends to make the planet’s surface temperature warmer than it would otherwise be. The physics of it has to do with radiation being able to penetrate through the atmosphere. The Sun’s radiation versus the radiation that’s emitted by the planet. Then it depends on the nature of the atmosphere how much of a greenhouse gas you’re going to have. CO_{2}’s a greenhouse gas. The surface pressure on Venus is something like 60 times that of Earth. Has a very massive atmosphere, and it’s almost all CO_{2}. You can just imagine you’re going to–kind of like the worst possible scenario for having a very strong greenhouse gas. Not only is there a lot of it, but almost all of it is an absorbing gas. Chapter 8: What is a Greenhouse Gas? [00:45:24] I have to say a few words–I think I can squeeze it in–about what is a greenhouse gas because I use that term loosely now. It needs to be defined. It’s basically a gas that can absorb and emit infrared radiation. What determines whether a gas can do that is its molecular structure. So for example, an Argon atom just by itself, because it’s only a single point mass, it cannot rotate, it can’t vibrate, it can just move around. And by moving around, it doesn’t radiate any energy. Nitrogen, N_{2}, two point masses held by a chemical bond. It can rotate, it can vibrate, but because it’s a symmetric molecule, it’s electrical charge distribution is symmetric and it has no dipole moment. A dipole moment is a separation of charges. If I’ve got a positive and a negative charge and they’ve been moved apart, then we say that object has a dipole moment. But you need to break symmetry in the molecule before you can have a dipole moment, and nitrogen is a perfectly symmetric molecule. Nitrogen, nitrogen, symmetry, no dipole moment. Oxygen, same problem, right? Perfectly symmetric molecule, no dipole moment. Look at this. The primary constituents of our atmosphere, nitrogen first, oxygen second, Argon third, none of them have a dipole moment, none of them can absorb or emit infrared radiation. These are not greenhouse gases. It doesn’t matter much how much of these you have. What are the greenhouse gases then? Well CO_{2}, which has a structure like this. That’s a greenhouse gas. Why, do you say? It looks symmetric, doesn’t it? But it can vibrate. If that carbon vibrates to the left and that oxygen vibrates to the right, suddenly I’ve broken symmetry. And I can have a dipole moment, and that will absorb and emit infrared radiation. What about water vapor? H_{2}O. Oxygen, two hydrogens. Well, that’s symmetric, isn’t it? Well no, not about that axis. It’s not symmetric about that axis. So it’ll have a dipole moment. And it will absorb and emit radiation. So see, the distinction I want to make here is between gases that are so simple in their structure that they can never, no matter what they do, they can’t generate a dipole moment, they can’t get a charge separation. Remember, when you’re talking on your cell phone, you’re transmitting radiation. There’s an antenna in there that’s creating a dipole moment–positive, negative, positive, negative, positive, negative, and that is what is emitting the radiation. When you’re listening, the same thing is happening. That radiation is coming to you. It’s being captured by the antenna in the cell phone with a dipole moment. So it’s exactly the same physics. In all cases, you have to have an oscillating dipole moment if you want to absorb or emit radiation. Air cannot do it, but these greenhouse gases can. We’re out of time, but we’ll pick up on the theme next meeting. [end of transcript] Back to Top 
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