The model is built on one of the great conservation laws of science -- conservation of energy. If you track all the ways that energy enters your 'system', and all the ways that it leaves, you can know quite a lot about it even without knowing very much about what's going on inside. That word 'system' may not mean what you're thinking. What I mean here is that we draw an imaginary box around, in this case, the earth, and see what passes through the box.

We have, in general, three ways to get energy through the box. We can carry some material from one side to the other (advection or convection). But the earth doesn't gain or lose material from space, so can't gain or lose energy this way. Another way is to conduct heat across the boundary. A frying pan does this in making the handle hot even though the flame is a good distance away. But with a vacuum on the other side of our box, there's nothing to conduct

*to*. The third way is as radiation. Everything emits some radiation. The hotter is is the more it emits. A traditional thermos tries to minimize all three -- the inside is glass, a poor conductor, the glass is almost totally surrounded by a vacuum to cut down conduction even further, it is sealed, to prevent convection, and the inner wall of the thermos is reflective, to prevent the wall from gaining or losing heat by radiation to your drink.

In looking at what can cross the boundary of the earth system we have only two things -- radiation from the sun, and radiation from the earth. We all know that many things can happen inside the box: storms, building and melting ice sheets, forest growth and decay, and so on. But they're all inside. For our simplest model of the temperature of the earth we don't need to know about those, just income versus outgo. (Well, more to it than this, but hang on for now.)

The sun's radiation all comes in to the box, passing through a disk whose area is pi*r^2, where r is the radius of the earth and pi is the usual 3.14159... Same as the sun and moon look like disks to us, we look like a disk to the sun. Some of that solar energy gets bounced right back out. The fraction of the incoming solar energy that gets bounced out is called the albedo. For the earth, it is about 0.30 averaged over the whole planet, through the whole year. Particular surfaces can be much higher (snow can be 0.8) or much lower (oceans can be 0.06). But since we're only concerned with what passes through the box, we don't need to worry about those details yet.

The earth also emits radiation, as does everything that's not at absolute zero. For an ideal black body, that emission is proportional to s*T^4. This is the Stefan-Boltzmann law, and s is the Stefan Boltzmann constant, equal to 5.67e-8 Watts per square meter per K^4. Every square meter of the earth radiates like this. Since the earth is (very nearly) a sphere, the total area is 4 * pi * r^2. Again, r is the radius of the earth.

Conservation laws are written in terms of a balance of income and outgo. If the earth is in energy balance, then the following equation is exact. If it isn't, we'll have to look at how far out we are.

Income = pi * r^2 * S

Outgo =

pi * r^2 * S * a (albedo reflecting energy immediately away from the earth)

+ 4 * pi * r^2 * s * T^4

For the conservation law being Income = Outgo, we get

pi * r^2 * S = pi * r^2 * S * a + 4 * pi * r^2 * s * T^4

We notice two things rapidly. First, pi * r^2 is in every term. We can divide the equation by that and simplify. Second, the solar term is on both sides. We can subtract the albedo term from both sides to get all the solar-related terms on one side, and the earth on the other. So now we have:

S*(1-a) = 4*s*T^4

If we solve for T, we're back to the equation I gave in the original note

It's remarkable that the size of the earth doesn't show up here. Actually, it does -- that 4 is the remaining part. It's 4 because the entire area of the (spherical) earth emits energy to space, while the sun only contributes energy through a circular disk, and the area of the sphere is 4 times larger than the area of that disk.

This leads to a point that I see commonly missed. When solar observers talk about how much the solar output changes, they're referring to the S -- seen by the disk of the earth. During a solar cycle, it varies by about 1 Watt per square meter. In looking at climate, though, we think in terms of the spherical earth we're standing on, rather than the disk in space a satellite is using. When climate people talk about Watts per square meter, we're talking about 4 times as many square meters as solar observers. To make the two comparable, climatologists often divide the equation by 4, meaning that when a solar observer says 1 Watt per square meter, climate folks will treat it as 0.25. Not because we don't think the sun's important, but because the 0.25 gives a description that we can compare to earthly climate processes we observe all over the sphere.

On Monday, there'll be a third post in this set -- analyzing this model. In the mean time, a project for the more mathematically skilled out there: How is this model affected by using a proper oblate spheroid for the earth? Is it? And, for everyone, try out the spreadsheet version and see how it behaves as you change the parameters.

Ah, I should note that the S above isn't exactly the S in the original. Here it's the solar constant divided by mean earth-sun distance (in AU), while I left the two separate in the original. As tamino showed, the two are awfully close to each other since the mean earth-sun distance (over a year) doesn't change much even as the orbit varies on tens of thousand year time scales.

A different and more important matter if you're not used to this is that the temperatures will come out in Kelvin -- the scientific absolute temperature scale. Celsius is used, but when we are concerned about energy, it's best to use Kelvin (K). Fahrenheit is probably never used in doing science (even if someone did, it never gets published that way). Kelvin is Celsius plus 273, so water freezes at about 273 K, boils at about 373 K. (Or Celsius is Kelvin minus 273.) The earth's black body temperature is about 255 K, or -18 C.

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