## Monday, June 24, 2013

1) If I is taken over all space (i.e. 4 p steradians), the mean intensity is:

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.