J = 1/4p ò 4p I dw = 1/4p (1.4 x 10 7 Jm-2 s-1) 4p
So: J » 1.4 x 10 7 Jm-2 s-1
2) The net flux (H) passing through the Sun’s surface is estimated using: sTo 4/ p = 2H.
Therefore: H = sTo 4/2 p
Where: sTo 4 = (5.67 x 10-8 W m-2 K-4)(4800K)4
sTo 4 = 3.0 x 10 7 W m-2
H = (3.0 x 10 7 W m-2) /2 p = 4.8 x 106 W m-2
3) Consider the solar half-sphere and the energy going into it each second. We know on average photons are absorbed after traveling a distance with optical depth t =1 in the propagation direction. Averaged over all directions this corresponds to a vertical optical depth of t = 2/3. Based on this find:
a) The energy going into the half sphere each second.
Solution:
This is just: s(Teff)4 = p F = D 2p S =D 2p I
b) The change in (a) over each absorption and re-emission over vertical optical depth.
Solution:
Over a vertical optical depth one has t = 2/3, then:
D 2p S/D t = p F/ (2/3) = 3p F/ 2
or:
DS/D t = 3p F/ 2 (2p) = 3F/4
c) The total absorption and total emission and the relationship between them, defined over all space.
Solution:
Over all space, the total emission = 4p S and the total absorption = 4p J, and by radiative equilibrium:
4p S = 4p J so that S = J.
No comments:
Post a Comment