for the lectures on Monday October 7 + Tuesday October 8
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?
We need to multiply the incoming solar
energy by the factor 1/4--the
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?
3) How does the Earth
get rid of that energy?
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?
Radiation emitted by the Earth = constant x T4
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:
absorbed by the Earth = Radiation emitted by the Earth
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.
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 treatment in the textbook (box on page 43) illustrates the greenhouse
A simple approach
We can get the above results directly by recognizing that the top
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
2) It's assumed that the atmosphere absorbs all the outgoing
3) Radiative transfer isn't the only process by which energy escapes
(a) Inverse square law
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:
so we can write:
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
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