The following graph shows the absorption of long-wave radiation by the first three kilometers of the Earth's atmosphere. The graph is typical of a clear day in April, at latitude 30° North. Water vapor content is 10,000 ppm (that's 1%), CO2 is 330 ppm, air pressure is 1000 mbar, and air temperature is 280 K (that's 7°C). This line is the same as the 0-km line from our Earth's Atmosphere post.
The 5 μm to 30 μm range includes over 95% of the heat radiated by the Earth's surface (look at the brown power density line in this graph). We see that the first 3 km of the atmosphere absorbs almost all radiation from 13 μm to 30 μm, and from 5 μm to 8 μm. According to the Spectral Calculator, the absorption between 5 μm and 8 μm is due to water vapor, between 13 μm and 17 μm is due to CO2, and above 17 μm is due to the water vapor again.
Most radiation between 8 μm and 13 μm passes through the first 3 km of the atmosphere. The higher layers, being thinner and more sparsely supplied with water vapor, absorb even less of this radiation, as you can see in our compound graph. To the first approximation, the first 3 km layer of the atmosphere is transparent from 8 to 13 μm and opaque to other long-wave radiation. The 8 to 13 μm radiation emitted by the Earth's surface will pass right through the atmosphere and into space, provided there are no clouds. As we declared in our recent review, we are ignoring clouds for the time being.
In our up-coming posts, we will consider each 3-km layer of the atmosphere up to altitude 18 km in turn. We will approximate each layer as being transparent to some wavelengths and opaque to others. Once we have considered all the layers, we will write a computer program to calculate the radiation of heat from the various layers, and so obtain an estimate of the Earth's greenhouse effect without clouds.