Thermal emissions by solid bodies can be analyzed to provide information on the temperature at a few wavelengths below the surface. Observations of the night hemisphere of Mercury longward of ∼10 cm yield temperatures close to the diurnal equilibrium temperature. As the radiation is obviously able to escape directly from the regions being observed, why does it not get much colder than this during the long mercurian night?

Question

Geography

Ruben Fry

21 June, 03:53

Thermal emissions by solid bodies can be analyzed to provide information on the temperature at a few wavelengths below the surface. Observations of the night hemisphere of Mercury longward of ∼10 cm yield temperatures close to the diurnal equilibrium temperature. As the radiation is obviously able to escape directly from the regions being observed, why does it not get much colder than this during the long mercurian night?

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Answers (1)

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Alma Keith · Answer 1

Answer:T = 2898*10^-5K

Explanation:

wavelength = 10cm

Using Wien's Law

maximum wavelength = 2898*10^-6/temp (K)

thus T=2898*10^-6/10*10^-2

T = 2898*10^-5K

this is the maximum observed temperature of 10 cm wavelength black body

since this temperature is close to the diurnal equilibrium temperature, that's why it not get much colder.