Use the idea of an energy balance: What comes in = what goes out

**How much energy is received by the earth?**

Solar radiation incident in the earth's disk (1368 Watts per square

meter) --comparable to energy incident a flat, horizontal surface

when the sun is directly overhead on a clear day.

We need to multiply the incoming solar energy by the factor 1/4--the

ratio of the area of the earth's disk (pi r squared) to the earth's

surface area (4 pi r squared)-- You can think of this as spreading

out the incident solar radiation uniformly over the earth's surface

(the night side of the earth as well as the day side) 1368 / 4 = 342

watts per square meter.

Then we need to multiply by the factor 0.70, which takes into account

the fact that 30% of the incident solar radiation is reflected back

to space by clouds, snow and ice, the light colored sands of the

deserts, and even just a bit from the daisies. 342 x 0.7 = 239.4

watts per square meter. The 0.70 is equivalent to the factor
(1-A)

in the formula in the text.

**How much energy does the earth have to get rid of?**

239.4 watts per square meter.

**How does the earth get rid of that energy?**

By emitting radiation. That's the only way an isolated planet like
the

earth can get rid of energy. Most of the energy emitted by the earth

and the other planets is in the infrared part of the electromagnetic

spectrum. (They're not warm enough to emit an appreciable amount

of visible radiation).

**What does the emitted radiation tell us about the earth's temperature?**

We can calculate the temperature a black body would have

to have in order to emit the same amount of radiation as the earth.

This is called the effective radiating temperature. It's the

temperature that would be measured by an infrared thermometer

(radiometer)in outer space, pointed at the earth.

**How do we calculate the effective temperature?**

From the Stefan-Boltzmann Law which says that the emitted radiation

is equal to a constant times the effective radiating temperature raised
to the

fourth power.

To solve this equation, all we need to do is divide the emitted

radiation (239.4 watts per square meter) by the constant (5.67 x 10

to the minus 8) and take the fourth root of the result. Dividing we

obtain 42.2 x 10 to the eighth power. (We could also express this

result as 4.22 x 10 to the 9th power, but since we're going to take

the fourth root of this number, it's convenient to use a power of 10

that's divisible by 4.) We'll take the fourth root on a calculator,

but to check it's a good idea to estimate the result by taking the

square root of 50, which should be just about 7 and taking the square
root of 7 which

should be around 2.5. The fourth root of 10 to the eighth power is

100. Hence, the answer should be a number around 2.5 x 100 or 250.

The calculated result is 255. Remember that all results obtained from

the Stefan_Boltzmann Law and other radiation laws are expressed in

degrees Kelvin, so this is 255 K (-18 C, 0 F).

**An analogy:**

Consider yourself sleeping comfortably under one blanket, your

metabolism operating at a fixed rate, putting out heat to keep your

body warm. The temperature of the top surface of the blanket is

measured with an infrared thermometer. It's a little warmer than the

room. Someone comes along and throws another blanket on you. Will the

temperature of the top surface of the top blanket be any different

from what it was before (after it has had time to adjust)? Not unless

your metabolism changes. But the top surface of the bottom blanket

will get warmer due to the influence of the top blanket, and you, the

sleeper will be warmer. Adding greenhouse gases to the atmosphere of

a planet is like adding another blanket. It doesn't change the

effective radiating temperature of the planet, but it raises the

temperature of the surface of the planet.

The treatment in the text The box on p.43 illustrates the greenhouse

effect by assuming an isothermal atmosphere-- (an atmosphere that is

all at the same temperature) that is perfectly transparent to solar

radiation, but acts like a black body in the infrared part of the

electromagnetic spectrum, where the planet emits radiation. It

absorbs all the radiation emitted from the surface of the planet, and

re-emits it: half in the upward direction to space, and half in the

downward direction, back to the surface of the planet. The problem
is

solved by means of simultaneous equations: one is the radiation (or

energy) balance for the surface of the planet and one is for the

atmosphere. Here are two alternative approaches, neither of which

requires solving simultaneous equations.

that it receives-- 239.4 watts per square meter. For convenience,

let's just call this E (for emitted). All the emitted radiation will

be absorbed by the atmosphere. Half will be reemitted to space and

the other half back to the surface of the planet. On order to get rid

of this returned radiation, the surface of the planet will have to

warm up so it can emit an extra E/2 units of infrared radiation. But

half of this amount will be returned from the atmosphere. So it will

need to warm up enough to emit another E/4 units. This bouncing back

and forth goes on and on, but fortunately the amounts of radiation

involved are getting smaller each time. It is clear that the total

radiation emitted from the surface of the planet (E + E/2 + E/4 + E/8

+ E/16......) is going to add up to 2E, and the total emitted to

space from the top of the atmosphere will add up to (E/2 + E/4 + E/8

+ E/16......) = E. Hence, the surface of the planet will have to be

hot enough to emit twice as much radiation as a planet wothout a

greenhouse effect. From the text it follows that the absolute

temperature of the surface (in deg K) will have to be warmer by a

factor of 1.19 (the fourth root of 2). The temperature of the

atmosphere will be the same as the effective temperature of the

planet.

atmosphere must emit E units of infrared radiation to space and that

it will emit an equal amount downward to the surface of the planet.

Hence, for thermal equilibrium, the surface of the planet must emit

enough radiation to balance not only the amount it receives from the

sun (E), but also what it receives in the form of downward infrared

radiation from the atmosphere (E). Hence, its emission must be

E + E = 2E.

the one that we've already considered. Using the short cut we infer

that the radiation emitted from the top layer of the atmosphere must

still be E; the radiation emitted upward from the bottom layer of the

atmosphere must be equal to 2E and the radiation emitted from the

surface must be 3E. It's not difficult to extend this simple analysis

to additional layers.

*Limitations of this calculation*

1) It's assumed that the atmosphere is isothermal. The layer of the

real atmosphere that's most important in terms of the greenhouse

effect is the troposphere, where temperature decreases with height.

Because of this height dependence, the real atmosphere emits more

radiation in the downward direction than in the upward direction (88

units vs. 70 units in Fig. 3-19).

2) It's assumed that the atmosphere absorbs all the outgoing

radiation at all wavelengths in the infrared part of the

electromagnetic spectrum. In reality, the absorption of radiation by

the atmosphere is highly wavelength dependent. At some wavelengths

there's very little absorption and the radiation emitted by the

earth's surface escapes to space, while at other wavelengths it gets

absorbed, reemitted, absorbed and reemitted many times before it

finally escapes. To carry out this calculation accurately it has to

be done wavelength-by wavelength... to capture the fine scale detail

in the spectrum requires literally thousands of calculations

analogous to the one we did in class.

3) Radiative transfer isn't the only process by which energy escapes

from the earth's surface. Conduction of heat and evaporation of water

transfer about twice as much energy from the earth's surface to the

atmosphere as the net upward flux of infrared radiation from the

radiation does. If the temperature distribution on earth were

determined only by radiative transfer (as in this example) the earth

would be so hot as to be uninhabitable. In this sense the true

'greenhouse effect' on earth is much larger than the 33 K difference

between the observed surface temperature (288 K) and the effective

radiating temperature (255 K) ascribed to it in your text.