# Moon's negative greenhouse effect?

Using the standard formula for effective temperature of a planet, Venus has T(eff)=220K, Earth has T(eff)=255K and Mars has T(eff)=212K (using albedo for Venus a=0.7, Earth a=0.367, Mars a=0.25). Average surface temperatures of 730K, 288K and 218K respectively are easily explained by differences in greenhouse effect. The moon however with an albedo of a=0.1 should have a T(eff) of about 270K, but the average surface temperature is about 250K. Here is where I got the T(eff) formula and here is where i got the average temperature of the moon. (the latter is a new citing and I have edited the mean temperature value) I have revised my planet temperatures using this reference. What explains this negative difference?

• How did you calculate the "average" temperature? It varies massively between the day and night sides. So I suspect that is your solution. Doesn't the formula you have used assume the surface has a uniform temperature? You probably need the area-weighted average of $T^4$. – Rob Jeffries Oct 13 '15 at 15:59
• Please edit your question and specify the standard formula for effective temperature of a planet, at least by making it a link. – Jan Doggen Oct 14 '15 at 7:21
• It would be interesting to do Mercury too, with it's slow rotation. That should be similar to the moon-effect. It's elongated orbit around the sun might be a little complicated to work in but average temp of 332 degrees - here: space.com/18645-mercury-temperature.html – userLTK Oct 14 '15 at 10:21
• Same answer to your edited question, but the closing of the gap between the two just means that the (area-weighted) differences between min and max temperatures must be smaller. (i.e. I assumed one hemisphere is very hot and the other very cold, but in reality it will be smoother than this). – Rob Jeffries Oct 15 '15 at 11:37

However, in terms of blackbody radiation, if we treat both hemispheres as separate blackbodies, then together they radiate as much energy as a hotter blackbody because of the $T^4$ factor in Stefan's law. The "effective temperature" will be $$T \simeq (0.5\times 100^4 + 0.5\times 370^4)^{0.25}= 311\ K$$ So this is more like your equilibrium temperature and a more appropriate area weighting and/or tweaks to the exact range of temperatures could bring it into close agreement.