Answer to Question #24943 in Astronomy | Astrophysics for leah rabuku
due to the sun-earth distance, the average temperature of the earth is maintained at 255K. Derive this temerature of earth and state any assumptions made and state the variables used?
We denote the incoming solar radiation by Q; it has a value Q = 1370 W m−2 (watts per square metre). A fraction a of this (the albedo) is reﬂected back into space, while the rest is absorbed by the Earth; for the Earth, a ≈ 0.3. In physics we learn that a perfect radiative emitter (a black body) at absolute surface temperature T emits 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 4πR2σT4 e = πR2 (1 − a)Q, whence 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.