Answer to Question #24943 in Astronomy | Astrophysics for leah rabuku

Question

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?

Answer 1

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.