1) Find the effective temperature of the Sun and the boundary temperature (To) and account for any difference.
(Take the Stefan -Boltzmann constant s = 5.67 x 10-8 W m-2 K-4 )
Solution:
The effective temperature is obtained using:
p( Fo ) = s(Teff)4
So: Teff = [p( Fo ) / s] 1/4
Where s = 5.67 x 10-8 W m-2 K-4
Is the Stefan-Boltzmann constant. Then:
Teff = [6.3 x 10 7 Jm-2 s-1/ 5.67 x 10-8 W m-2 K-4] 1/4
Teff » 5800 K
The boundary temperature is found from:
Teff = (2)1/4 To = 1.189 To
Or:
To = Teff /1.189
= 5800K/ 1.189 » 4800 K
The boundary temperature differs because of being referenced to a different optical depth. The boundary temperature (To) approaches the value of the effective (or surface) temperature when t = 0, but this still exhibits a difference in layers so will not be exactly the same!
2) Prove J = ½(I1 + I2)
By integrating I in the forward and backward directions.
Solution: We define:
J = 1/4p ò 4p I (q) dw
In forward direction:
J(+) = 1/4p ò2po òp/2 o I1 sin q dq df
In backward direction:
J(-) = 1/4p ò2po ò-p/2 o I2 sin q dq df
(Note change in upper limit of 2nd integral)
J = J(+) + J(-) =
1/4p ò2po òp/2 o I2 sin q dq df + 1/4p ò2po ò-p/2 o I2 sin q dq df
= 1/4p [ 2p (I1 + I2) ] = ½ (I1 + I2)
J = ½(I1 + I2)
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