How to calculate the earth's effective temperature

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).

The Greenhouse Effect

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.

1. An iterative approach

Assume that the surface of the planet emits all the solar radiation
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.

2. A short cut

We can get the above results directly by recognizing that the
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.

A multi-layer atmosphere

Suppose that we pile on another perfectly absorbing layer on top of
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.