Quote:
Originally Posted by Extravoice
Does your calculation result in an average temperture for the sphere?
|
It does. Or rather, an average infrared luminosity. (Luminosity varies with the fourth power of temperature.)
It assumes a sphere intercepts radiation over a cross-sectional area of πr
2 and reradiates a black body spectrum uniformly over its surface area, 4πr
2. The ratio between these two gives us the factor of 4 sitting on the bottom of the albedo section of the equation. Lumpy objects will generally have a higher ratio of radiating area to interception area, and so may equilibrate to lower temperatures.
Of course, in the real world, something the size of a moon or planet will be able to develop areas of its surface which are locally hotter and colder than the average, just so long as the complicated temperature distribution results in the right amount of infrared reradiation to maintain equilibrium.
[I
still can't get the edit function to work, and now I've noticed a misplaced decimal in my equation. If I can ever get back to it, I'll edit the original, but meanwhile here's the corrected version (the typo was in the exponent of the albedo section):
T = Ts x ([1-A]/4)^0.25 x (Rs/d)^0.5
Sorry!]
Grant Hutchison