ATM S 211: Winter Quarter 2002
 Climate and Climate Change
Notes for the lecture on Monday Jan. 14 & Tuesday Jan. 15 

Planetary Energy Balance

How do we calculate the Earth's effective temperature?

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

1) How much energy is received by the earth? 
Solar radiation incident on the Earth's disk (1370 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 R2) to the Earth's surface area (4 pi R2)-- 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) 1370 / 4 = 342.5 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. 342.5 x 0.7 = 239.7 watts per square meter.   The 0.70 is equivalent to the factor (1-A) in the formula in the text. 

    Energy flux absorbed by the Earth = 1370 x (1-0.3) / 4 = 239.7 W/m2

2) How much energy does the Earth have to get rid of? 
239.7 watts per square meter. 

3) 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. 

4) 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. 

       Radiation emitted by the Earth = constant x T4
(with the constant = 5.67 x 10-8 W /m2/K4)
(Remember that the effective radiating temperature is expressed in Kelvins!)

We now have all the elements to calculated the effective temperature.  We know how much energy the Earth receives and how much radiation is has to emit to get rid of that energy.  Using the idea of an energy balance: What comes in = what goes out:

Energy flux absorbed by the Earth = Radiation emitted by the Earth
                                    239.7 W/m2 = constant x T4

To solve this equation, all we need to do is divide the emitted radiation (239.7 watts per square meter) by the constant (5.67 x 10-8) and take the fourth root of the result.  Dividing we obtain 42.3 x 10-8. 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): 

                        T = 255 K

The figure below illustrates how we derived this energy balance.


This effective temperature of 255 K is the temperature the Earth's Surface would have if it didn't have an atmosphere.  It would be awfully cold! In reality, the Earth's surface temperature is closer to 288 K (15 °C, 59 °F).  This difference of 33 K is the magnitude of the greenhouse effect.  Before we go into more details about what this greenhouse effect is, let's look at Venus and Mars, our closest neighbours and calculate their effective temperatures.

5) Effective temperatures of Mars and Venus

We express the distance from the Sun in Astronomical units.  Venus is 28% closer to the Sun than the Earth, while Mars is 52% further away.  The solar flux arriving at Mars and Venus is calculated using the inverse square law (p. 38 in textbook). In the case of Venus:

Flux(Venus)/Flux(Earth) = (Rearth/Rvenus)2

such that Flux(Venus) = 1370 x (1/0.72)2 = 1370 x 1.9 = 2643 W/m2

Venus thus receives almost twice as much solar flux as the Earth.  Mars, which is further way, receives 43% less solar flux than the Earth.
Venus Earth Mars
Distance from Sun (A.U.) 0.72 1 1.52
Flux, W/m2 2643 1370 593
Albedo 0.8 0.3 0.22
Effective Temperature, K 220 255 212
Actual observed Temperature, K 730 288 218
Greenhouse Effect, K 510 33 5
1 Astronomical Unit = distance between Earth and Sun

You will notice in the Table above that even though Venus receives more solar energy than the Earth is, its effective temperature is colder.  This is due to the high albedo on Venus (0.8): 80% of solar radiation is reflected to space and only 20% is absorbed by the surface (see Figure 3-1 in textbook for a comparison of albedoes of the three planets).  The actual temperature at the suface of Venus is much warmer: 730 K ! The enormous greenhouse effect of 510 K (=730-220) on Venus is due to high levels of CO2 in its thick atmosphere.  Mars has a much weaker greenhouse effect because it's atmosphere is much thinner than that of the Earth.

The Greenhouse Effect

The treatment in the textbook (box on page 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 blackbody 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 radiation 
balance of the atmosphere. Here is one alternative approach, which doesn't require solving simultaneous equations.

A simple approach

We can get the above results directly by recognizing that the top
layer of the atmosphere must emit 239.7 W/m2 of infrared radiation 
to space (same amount of solar radiation that enters the atmosphere:
what goes in must go out).  The bottom layer of the atmosphere
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 (239.7 W/m2), but also what it receives in the form of downward 
infrared radiation from the atmosphere 239.7 W/m2). Hence, its emission
must match 239.7+239.7 = 479.4 W/m2.  Applying the Stefan-Boltzmann
law: constant x T 4 = 479.4 W/m2.  We thus calculate T = 303 K. 
The figure below illustrates this calculation.   Contrast it to the figure
above where we assumed no atmosphere, and you will see where
the greenhouse effect comes in.

The effective temperature we calculate in this manner is much warmer than the actual temperature of the Earth (288 K), because we made a number of simplifying assumptions.

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. 

Sample exercises:

(a) Inverse square law
Take two planets, planet A and planet B.  Planet B is at 10 times the
distance from the Sun as planet A.  We know the solar flux at A is
2000 W/m2. What is the flux reaching planet B?
Based on the inverse square law, we have the following relationship between the fluxes arriving at both planets:

Fb/Fa = (Da/Db)2

Where Fa and Fb are the solar fluxes arriving at planets A and B, and Da and Db are the respective distances of these planets from their sun.  We are given the following information:
Fa = 2000 W/m2 and Db =10 x Da

so we can write:
Fb = Fa * (Da/Db)2 = 2000 * (Da/(10 x Da))2 = 2000 x (1/10)2 = 2000 x 1/100 = 20 W/m2

We thus find that the flux arriving at planet B is 120 W/m2 (100 times less than the flux arriving at planet A).

(b) Planetary energy balance and effective radiating temperature
We calculated the effective radiating temperature for Earth in class, as summarized in the above notes. Let's do the same for Mars.  Here is the information we have: the albedo on Mars is A = 0.22, and the solar flux reaching Mars is equal to 593 W/m2.
Based on the planetary energy balance applied to Mars, the solar radiation reaching the planet's surface must be equal to the infrared energy emitted by the planet:
F(1-A)/4 = constant x T4
with F = 593 W/m2, A = 0.22, constant = 5.67 x 10-8 W /m2/K4
so we find T4 = F(1-A)/4/constant = 593 x (1-0.22)/4/(5.67 x 10-8) = 20.4 x 10+8 K4
Now let's take the fourth root of 20.4 x 10+8 K4, we find T = 212 K. The effective radiating temperature of Mars is thus 212 K.  The observed temperature on Mars is 218 K, so it's atmosphere has a small greenhouse effect of 218-212 = 6 K.


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