Blackbody radiation represents cases where all incident radiation upon a body is absorbed and then re-radiated at some characteristic black body temperature. We have previously derived this temperature for the earth but will quickly repeat
the procedure here.
F = σT4
For the Earth: (Albedo = 0.3 assumed)
Problem: A blackbody satellite (A=0) of mass = 1000 kg and radius = 1 meter orbits the Earth. It is made of material that has specific heat c = 1000 J (kg K)-1. What is the cooling rate of the satellite when it passes behind the Earth's shadow? For this case A=0 which yields Flux = 342 W per square meter and equivalent blackbody temperature of 279K. When the satellite passes into the shadow, it is no longer in radiative equilibrium as F goes to 0 and then the satellite cools via blackbody radiation at temperature of TE  Atmospheric gases, in contrast, are not blackbody radiators. However, if we consider specific emissivity and absorptivity we can approximate the atmosphere as a blackbody. Emissivity of a body is defined by the ratio of Intensity of emitted radiation to the corresponding blackbody radiation: 
Similarly, specific absorptivity, reflectivity and transmissivity are defined as: 
A guy name Kirchoff also came up with a very useful law: 
This law holds when an object is placed in a field of radation that is not isotropic or necessarily in radiative equilibrium. For gases, the law holds only when the frequency of molecular collisions is much larger than the frequency of absorption and re-emission of radiation. In essence, this is a density affect and when this condition is satisfied the system is in a state of Local Thermal Equilibrium (LTE). For our atmosphere, LTE prevailes below altitudes of about 50km. The emissivity of the earth's surface is quite complicated and varies as a function of time (due to seasonal and systematic vegetation and land use changes). Emissivity is also a function of wavelength. The figure below shows emissivity for a variery of common materials.  Examples of Radiative equilibrium: Suppose we have an object with absorptivity of 0.1 to solar radiation and emissivitiy of 0.8 to thermal infrared radiation at the top of the atmosphere. Solar constant = So = 1370 Watts per square meter (at the top of our atmosphere) The equilibrium temperature is found as follows: Energy absorbed = energy emitted = 1370 x 0.1 = 137 Energy emitted is 0.8*σ *T4 = 137; Solve for T to get 235 K. Radiative Equilbrium of the Earth: Integrating aλ and eλ produces the effective albedo and effective emissivity respectively. The energy absorbed by the Earth system is: Ea = So π Re2*(1-a) The earth heats up and re-radiates away this energy as a blackbody radiator so that the Energy emitted is Ee = 4 π Re2σTe 4 * effective emissivity (e) So that the equilibrium temperature of the earth can be specified as T4 = (1-a)*S0/4eσ or T = 255*e-.25 The observed average temperature of the Earth is 288K or 1.129 times higher than 255. To reproduce that temperature requires the effective emissivity of the earth system to be e-.25 = 1.129 or e = .62 Now consider a planetary atmosphere consisting of multiple isothermal layers, each of which is transparent to shortwave radiation and completely opaque to longwave radiation. The layers and the surface of the planet are in radiative equilibrium. How is the surface temperature of the planet affected by the presence of this atmosphere? Begin by considering an atmosphere composed of a single isothermal layer. Because the layer is opaque in the longwave part of the spectrum, the equivalent blackbody temperature of the planet corresponds to the temperature of the atmosphere. Hence, the atmosphere must emit F units radiation to space as a blackbody to balance the F units of incoming solar radiation transmitted downward through the top of the atmosphere. Because the layer is isothermal, it also emits F units of radiation in the downward direction. Hence, the downward radiation at the surface of the planet is F units of incident solar radiation plus F units of longwave radiation emitted from the atmosphere, a total of 2F units, which must be balanced by an upward emission of 2F units of longwave radiation from the surface. Since F has increased to 2F, this corresponds to an increase of 48K in equilibrium temperature for the earth. The earth does not do this because our atmosphere is relatively thin. Changes in albedo are also quite influential. For instance a change of +/- 10 % changes the equilibrium temperature of the planet to 257 and 252 K respectively. Later we will look at the rate of loss of ice coverage of the planet. For now, snow is 85% reflective and ocean water is about 7% reflective. What effect on the global albedoe would occur if all the arctic sea ice were gone?
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