# Blackbody radiation and thermal equilibrium?

I'm new to blackbody radiation and I have been struggling with the following question:

The dominant wavelength of light from the sun is 510nm. Estimate the equilibrium temperature of the earth.

I'm not looking for any sort of full solution, just a suggestion for a starting point, and possibly sources which would be useful in working through this question. I currently have notes in front of me on; modes in a cavity, the Rayleigh-Jeans formula, the ultraviolet catastrophe, Planck's solution to the blackbody problem, and wein's displacement law. I am having a hard time understanding the derivations for the equations involved in these topics and how the topics and equations are related to eachother. Any guidance on any of the above would be much appreciated as i dont have a large amount of time left to attempt this question.

Don't know what assumptions are behind this question or the amount of rigor expected in the answer, but for a quick and rough calculation you could try using Wien's displacement law to calculate the temperature of the Sun (assuming that it is acting as a blackbody radiator). Then from that and the distance of the Earth to the Sun, you could calculate the amount of radiant power absorbed by the Earth from the Sun. You'll have to assume some emissivity, maybe 1 as a first approximation. Then you'll need to calculate what the equilibrium temperature of the Earth is so that it is radiating away (by means of blackbody radiation) exactly as much power as it is receiving from the sun. Kind of curious what this simple calculation would yield for the equilibrium temperature of the Earth, so please let us know what you get for an answer.

• There are so many stars ,not only the sun,so equilibrium temp of the earth is impossible to calculate that way.I think there is a typo in the question.
– Paul
Oct 19, 2015 at 4:11
• @Paul: The radiated power received by the Earth from all the stars in the universe except for the Sun is practically nil compared to the radiated power received from the Sun. For the purposes of this exercise, I think that we can safely ignore all stars except for the Sun.
– user93237
Oct 19, 2015 at 4:41
• @SamuelWeir this was very useful, i messed around with it for a bit, used a given value of .92 emissivity for earth's atmosphere and ended up with a value of around 255K i think? which is obviously much lower than the correct value but as the source 'anna v' provided says the calculation doesn't take into account the greenhouse effect, and the answer i got was similar enough to the one contained in that source so im pretty content, thanks for your help.
– user95945
Oct 20, 2015 at 21:09
• @RhysGI - T=255K is a surprisingly good estimate considering how simple the model is. Thanks for the info.
– user93237
Oct 20, 2015 at 23:07