Answer to Question #24943 in Astronomy | Astrophysics for leah rabuku
energy at a rate
Eb = σT
where σ is the Stefan-Boltzmann constant, given by σ = 5.67 × 10−8 W m−2 K−4
If we assume that the Earth acts as a black body of radius R with eﬀective(radiative)
temperature Te, and that it is in radiative equilibrium, then
e = πR2
(1 − a)Q,
Te =!(1 − a)Q4σ
Computing this value for the Earth using the parameters above yields Te ≈ 255 K. A bit chilly, but not in fact all that bad! Actually, if the average eﬀective temperature is measured(Tm) via the black body law from direct measurements of emitted radiation, one ﬁnds Tm ≈ 250 K, which compares well with Te. On the other hand, the Earth’s (average) surface temperature is Ts ≈ 288 K. The fact that Ts > Te is due to the greenhouse eﬀect,to which we will return later. First we must deal in some more detail with the basic mechanisms of radioactive heat transfer.
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