Fig. 1. This NASA Earth Energy Budget is based on data over 10 years of observations as of 2009. As of the time of this writing, it is still the Earth Energy Budget on NASA websites.
NASA says that the Earth emits 239.9 W/m2 of longwave infra-red radiation into space. This implies an effective Earth system radiative temperature of
P = 239.9 W/m2 = σ T4 = (5.6697 x 10-8 W/m2K4) T4,
so T = 255.0K, by application of the Stefan-Boltzmann equation.
Now we can go a step further, since NASA says that 40.1 W/m2 of longwave infra-red radiation emitted from the surface alone passes through the atmospheric window without absorption by the atmosphere directly into space. This allows us to calculate the effective radiative temperature of the atmosphere alone. Consequently,
P = (239.9 - 40.1) W/m2 = σ T4 = (5.6697 x 10-8 W/m2K4) T4,
so T = 243.7 K, the effective radiative temperature of the atmosphere alone as seen from space.
According to the U.S. Standard Atmosphere Table of 1976, this is the temperature at mid-latitudes at an altitude of 6846 meters by interpolation of table data. This is very close to 7000 meters, where gas molecule parameters are given in the U.S. Standard Atmosphere Table of 1976.
What the NASA Earth Energy Budget critically fails to note is that the Earth's gravity causes a linear temperature gradient in the troposphere, the Earth's lower atmosphere, which was well-known to the American scientists who devised the U.S. Standard Atmosphere tables from the 1950s to the final table of 1976. The U.S. Standard Atmosphere Table of 1976 is found in many editions of the CRC Handbook of Chemistry and Physics, such as the 71st and 96th Editions. One can still obtain it from the government for a fee. This gravitational field induced linear temperature gradient was also understood by Prof. Richard Feynmann who discusses it in his Feynmann Lectures late in Vol. 1. Let me provide a simple explanation of this linear temperature gradient due to gravity and the heat capacity of air molecules.
EK = (3/2) kT, where EK is the kinetic energy for a perfect monatomic gas molecule, where k is the Boltzmann constant. However, the lower atmosphere is made up almost entirely of diatomic molecules, with N2 and O2 more than 99% of the atmosphere. EK = (5/2) kT for a diatomic perfect or ideal gas molecule and (6/2) kT for a polyatomic molecule with more than two atoms. This is because a diatomic molecule has rotational kinetic energy around each axis perpendicular to the bond between the two atoms in the molecule. There are equal amounts of energy in each of the 5 degrees of freedom of the diatomic molecule. Molecules such as CO2 and CH4 with more than two atoms have 6 degrees of kinetic energy freedom. This allows us to tie the total kinetic energy at an altitude to the translational velocities of molecules given in the U.S. Standard Atmosphere table of 1976 for dry air. The total kinetic energy of the diatomic molecules making up more than 99% of the lower atmosphere is then 5/3 times the translational kinetic energy.
Conservation of energy for a diatomic gas molecule requires that, where 7000 meters altitude is chosen as a reference altitude, since it is the altitude in effective radiative equilibrium with space:
Where EK0 is the kinetic and the total energy of the gas molecule at sea level, v0 is its translational velocity there, E7000 is the total energy at 7000 meters altitude, v7000 is the translational velocity of the gas molecule at 7000 meters altitude, m is the mass of the molecule, g is the gravitational constant at 7000 meters altitude, and h is the altitude, here 7000 m. From the U.S. Standard Atmosphere table of 1976, the mean gas molecule in the atmosphere has a mass of 28.964 amu or 4.8080 x 10-26 kg, which is greater than the mass of the most common N2 molecules and less than the mass of the second most common O2 molecules. The gravitational constant at 7000 meters altitude is slightly less than that at sea level and is found in the table to be 9.7851 m/s2. The temperature is 242.7K. The translational velocity of the mean molecule at 7000 meters altitude from the table is 421.20 m/s. The value of EK0 is calculated to be approximately 1.040 x 10-20 Joules per mean molecular weight air molecule at sea level based on the translational velocity at 7000 meters and assuming that we are only calculating the static gravitational effect.
We can now set the gravitational effect EK0 kinetic energy into the EK = (5/2) kT equation and calculate what T should be if there were no other cooling effects, such as the evaporation of water. Note that air convection is not a net changer of the energy here, except for the effect of volume expansion cooling as the warm air rises and the pressure drops. This temperature gradient exists in the static air, yet there is no flow of heat.
The surface temperature due to the action of gravity alone on the atmosphere is found from:
EK0 = 1.040 x 10-20 J = (5/2) kT = (5/2) ( 1.381 x 10-23 J/K )T
T = 301.2 K
This temperature is actually warmer than the 289.5 K temperature implied by NASA in their Fig. 1. Energy Budget, since they always assume the surface emits as a black body. With a 398.2 W/m2 emission, the implied black body temperature is 289.5 K. More commonly, the average surface temperature is taken to be about 288 K.
If the Earth's surface were 301.2 K due to the gravity temperature gradient alone, the equivalent black body power density to allow us to compare its effect to other effects in the NASA energy budget would be:
P = σ T4 = (5.6697 x 10-8 W/m2K4)(301.2 K)4 = 466.6 W/m2
Note that this is not an actual flow of power to the surface. The atmospheric temperature gradient due to gravity exists under conditions of energy conservation, with the energy of gas molecules always constant. There is no heat flow in this effect.
Let us now consider other inputs and outputs of energy from the NASA Earth Energy Budget of Figure 1. We will use this effective gravitational power, the power absorbed by the surface from the sun directly, the power loss due to thermals, the power loss due to water evaporation, and the power loss due to surface longwave infra-red emission through the atmospheric window directly into space.
P = (466.6 + 163.3 -18.4 - 86.4 - 40.1 + R) W/m2 = σ T4 = (5.6697 x 10-8 W/m2K4) (288 K)4,
Solving for R, we find that
R = - 94.9 W/m2
According to NASA in Figure 1, R is equal to the - (358.2 - 340.3) W/m2 = - 17.9 W/m2 , the difference between the longwave surface emission absorbed by the atmosphere and the back radiation from the atmosphere. This is a modest cooling mechanism for the surface. What we see from our calculations though is that there is really much more cooling needed than 17.9 W/m2 .
Now there certainly is surface radiation which is absorbed by the atmosphere. I have maintained that there is very little back radiation and what there is is due to those situations in which the air temperature near the surface is higher than that of the surface. This is not the common daily condition, but it does happen at times. I have also maintained that the electric dipoles that oscillate to emit radiation characteristic of a body at 288K cannot provide kinetic energy to the evaporation of water and to transfer to air molecules in collisions. Energy must be conserved. Some of the molecules in the surface do emit longwave radiation as though they were part of a surface at 288K, but some do not, because they gave up their energy to the evaporation of water or in collisions with air molecules. The sum of the radiation energy emitted from those that emit radiation is less than that of a surface at 288K which was not also being cooled by water evaporation and air gas molecule collisions. For radiation purposes, only a portion of the surface area emits longwave infra-red and this portion is distributed nanoscopically over the entire surface area. Other nanoscopic areas are dumping their energy instead into evaporating water. Still others are exchanging it with cooler air molecules which are impinging upon a water surface. At sea level, the average air molecule has 6.92 x 109/s collisions with other air molecules. Due to the more than 100 times as many water molecules at the surface of a square meter of water compared to the two-thirds power of the number density of air molecules in a cubic meter, the collision rate for air molecules with water molecules at the surface is more than 100 times greater than that for air molecules with other air molecules. This gives surface water molecules many opportunities to transfer energy to cooler air molecules. The story is similar for the other materials of the Earth surface.
In fact, we need more cooling effects to explain the surface temperature being as low as it is. A big part of that is surface longwave radiation absorbed by the atmosphere, but I believe another part of that is other cooling mechanisms due to infra-red active gases (greenhouse gases) and underestimation of the cooling by water evaporation and thermals.
Infra-red active gases (water vapor, carbon dioxide) have a higher heat capacity than do nitrogen and oxygen molecules. Consequently, they can remove more energy from the surface upon collisions with it and add to the heat loss of the surface due to rising thermals. In addition, they make minor contributions to the transfer of heat upward through the atmosphere by radiating infra-red from warmer air layers to the usually cooler air layers just above them. Since that transport of energy is at the speed of light, this effect transfers energy faster than a rising thermal and acts as a cooling mechanism.
The advocates of catastrophic man-made global warming pose the problem of our average surface temperature as one of explaining why it is as high as it is. The real problem is explaining why it is as low as it is. This is the sad state of NASA and generally of the entire set of alarmist man-made global warming advocates understanding of the climate. For this poor service, NASA now spends more than half of its budget on climate change, with the emphasis heavily on catastrophic man-made global warming.