I didn’t think that a Part Two would be needed after the initial installment – The Sun and Max Planck Agree
Anyway, I always appreciate commenters explaining why the article hasn’t done its job, and so following various comments, hopefully, this article can address the deficiencies of the first.
I recommend reading the First Article before diving into this one. The main thrust of the article was to explain that solar radiation and terrestrial radiation have quite different “signatures”, or properties, which enable us to easily tell them apart.
For reader unfamiliar with how radiation varies with wavelength, here are two “blackbody” curves:
What you can notice is that the curve for 10°C radiation is – at all wavelengths – greater than the curve at -10°C radiation. The peak radiation for both of these curves occurs around 10μm. This is Wien’s law where the peak occurs at:
λpeak = 2898/T
where λpeak is the wavelength of peak radiation. So for 283K (10°C) the peak wavelength = 10.6μm. But radiation – as you can see – occurs at all wavelengths.
The total radiation from a body varies in proportion to T4, and here is the comparison of solar (at 5780K) and terrestrial radiation (at 260-300K):
Where is the terrestrial radiation? It can’t be seen on a linear plot, it is so small.
Let’s see it on a log plot:
For those not used to seeing log plots, check out the left hand side axis – the peak of the 300K radiation is around 1000x lower than the solar radiation at that wavelength (each major division corresponds to a factor of 10). And the spectral intensity of the higher temperature radiation exceeds the value of the lower temperature radiation at every wavelength.
This is the radiation measurement you would get from a spaceship parked just off the surface of the sun (for the solar radiation at 5780K) and just off the earth (for the terrestrial radiation of 260 – 300K).
The earth is some distance away from the sun and so the radiation at the top of the earth’s atmosphere will be reduced quite a lot.
If we consider the radiation from the sun expressed as per m2 then the solar radiation incident on the earth’s atmosphere will be reduced by a factor of (rsun / distancesun-earth)2 – this is known as the inverse square law – and there is a nice explanation at Wikipedia.
The amount the solar radiation is reduced for top of atmosphere = (696×106/150×109)2 = 1 / 2152 = 1 / 46,225.
(If we calculated it the other way, we would work out the total solar radiation from the whole surface area and conclude that this value must be divided by 2×109).
So if we look at the incident solar radiation at top of atmosphere (TOA) compared with terrestrial radiation:
We can see that the cross-over is quite small.
And this is not the whole story. Some solar radiation is reflected from the earth (on average around 30%). And depending on the angle of the sun from the zenith, the amount of solar radiation per m2 will be reduced accordingly.
So the above graph is the best case.
If we take the average of 30% of solar reflected, and at an angle from the zenith of 45°, we will have these curves:
You can see the crossover is very low.
Just for completeness, here it is on a linear plot, with the crossover section clear:
Hopefully, it’s clear, hopefully, there is no possibility of confusion – solar radiation can be easily distinguished from terrestrial radiation within the earth’s climate system.
If we measure radiation > 4μm we can conclude it is terrestrial and if we measure radiation < 4μm we can conclude it is solar.
Of course, at night, it is even clearer.