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2 answers

Good question. The brightness temperature of a surface, body, or whatever is found, as you no doubt are aware, by inverting the Planck function which relates the emissive power of a black body over a range of wavelengths to the temperature and works for all wavelengths (infrared, vis, microwave, radio, etc.). This temperature is related to the actual temperature of the body by its emissivity, its variance from a black body. If I've done my arithmetic properly you may find the actual temp is equal to the brightness temp times the emissivity to the negative 1/4 power, using the Stefan-Boltzmann law.

2007-02-16 07:36:49 · answer #1 · answered by 1ofSelby's 6 · 0 0

Assuming we're talking infrared here, you need to know the emissivity of the surface because the brightness temp would be the equivalent blackbody temperature. Once you knew your emissivity, I'd use brightness temp (in Kelvin) / emissivity ~= real temperature (in Kelvin). You can look up emissivities of surfaces here on any of these sites.
http://www.google.com/search?hl=en&q=emissivity+table&btnG=Google+Search

My satellite meteorology textbook also mentions the Rayleigh-Jeans approximation to get at brightness temperature and temperature for the microwave portion of the EM spectrum. If you don't have "Satellite Meteorology: An Introduction" by Kidder and Vonder Haar, you should be able to borrow it from a college or university library for the equation and constants.
http://www.amazon.com/Satellite-Meteorology-Introduction-Stanley-Kidder/dp/0124064302/sr=1-1/qid=1171653307/ref=sr_1_1/102-2353553-6566558?ie=UTF8&s=books

2007-02-16 06:18:34 · answer #2 · answered by tbom_01 4 · 0 0

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