I am no specialist in astrophysics, and i am seeking some basic understanding about whether (and how exactly) Stefan-Boltzmann Law could be applicable to rocky planets with a significantly thick athmosphere.

Few days ago, i've found a link (provided by someone in this forum; thanks!) to a simple calculator:

http://www.endmemo.com/physics/radenergy.php . Entering values into the thing confirms one of my old amateurish conclusions: that considering amount of solar radiation reaching Earth and possible albedo and greenhouse effect changes, it is very unlikely to see some +10C warming (for the average annual surface temperature of Earth), and quite impossible to see some +20C or higher (both values are above-pre-industrial, of course).

For said conclusion to be valid, though, one has to assume that Earth's athmosphere is quite transparent - in both directions. Yet, we know that both water vapour and CO2 allow most of Sun's energy in (visible spectrum), but does not (as easily) allow most of Earth's own emission out - since it's in IR spectrum.

And then i had this idea: well, we have a practical example of an extremum of greenhouse athmpshere: Venus. 96,5% of its athmosphere is CO2 (with the rest being 3,5% nitrogen and some trace amounts of other gases), and it's darn thick and heavy athmosphere with that - some 9+ MPa pressure at the surface (Earth has ~0,1MPa surface pressure). What if i'd try to use the calculator to see how much different from Stefan-Boltzmann Law temperature such an athmosphere would produce at the planet's surface? Assuming that long-term, incoming energy = emitted energy in terms of the whole planet, of course.

Here are assumptions and numbers i've got when i tried:

Venus has (known from measurements):

(Bond) Albedo = 0,9 // which is drastically different from Earth's 0,306

ergo, emissivity = 0,1 (Kirchhoff's law)

surface area = 4,6E+14 meters

average surface temperature = 737 K

Using the calculator, we get:

**7,7E+17 W** //that's her "

**would be output**" roughly

We also know that for Venus:

solar constant = 2611 W/m^2

crosssection = 1,15E+14 m^2 //not counting extra size her athmosphere is

ergo,

**input =** 2611*1,15E+14*(1-0,9)

**= 3,004E+17 W**Obviously, Venus can not (long-term, at least) emit more than twice amount of energy than it absorbs. Any output from its core - i mean fission of uranium-like radioactive matherials, - can't be of this order of magnitude after some 4+ billions years of the planet's existance in solid form. Therefore, the only possible explanation - is that temperature of Venus' hard surface is higher because of lack of transparency of its athmosphere. If by any chance Venus would have no athmosphere whatsoever, yet its surface would still have the same 0,9 albedo (all wavelengths average) the planet has today with its athmosphere present, - then its surface temperature would be lower. How much lower? sqrt(sqrt(7,7/3,0)) lower, since it's 4th power of the temperature in the formula; i.e. 1,2657 times lower: 737 / 1,2657 = 582,3 K. Therefore, Venus' athmosphere greenhouse effect warms up the surface by 737 - 582,3 = 154,7 degrees Kelvin. No wonder, with so much higher than Earth's solar constant (Earth has it 1366,1W/m^2 iirc), and with such a "greenhouse hell" athmosphere.

The difference is significant, but not awfully dramatic. Back to Earth, our athmosphere has some ~90 times lower pressure at the planet's surface than Venus', and CO2 % content is ~250 times lower. I'd say, Stefan-Boltzmann Law is much more applicable to Earth directly, with greenhouse effect giving relatively minor effect.

Same calculations as above, but this time for Earth, albedo 0,306, surface temperature 288 K, surface area 5,1E+14 m, solar constant 1366,1 W/m^2, crosssection 1,275E+14 m^2:

**"would be output" for Earth = 1,38E+17 W**

actual input for Earth = 1,21E+17 WOnce again, the crosssection did not include the extra radius given by the athmosphere; this explains some of the difference. The rest (and the most) is the greenhouse effect.

Which is, finally, where i get to a question. Problem is, "official" strength of the greenhouse effect on Earth - is some ~33 K warmer than otherwise expected. This is supposed to be good science. According to the calculator, "would be output" of 33 K colder Earth (so it's 255 K for the temperature, emissivity = 1 - albedo = 0,694) - would be 8,5E+16 W, which is some 30% LESS than actual input for Earth calculated just above. Which seems quite impossible (for a body without an athmosphere and with a given bond albedo / emissivity). If to estimate Earth's greenhouse effect using the numbers above, then it seems the greenhouse effect is just 8 to 9 degrees kelvin - some ~4 times less than the official amount of +33 K. So the question is:

where's an error in my amateurish attempts to apply Stefan-Boltzman to Earth? What am i missing? ><

Thank you in advance for any help...