http://brane-space.blogspot.com/2012/04/conductivity-in-solar-corona.html
We had the problem at the very end:
Find the temperature of the corona at the distance r = 2R_o. From this estimate the flux at this distance.
Solution:
Recall the temperature at distance r is given by:
T(r) = T_o (R_o / r) ^2/7
Then at a distance, r = 2R_o:
T(r) = (2 x 106 K) (Ro /2 Ro ) 2/7
T(r) = (2 x 106 K) (1 /2 ) 2/7 = 1.6 x 106 K
To get the flux, we assume: F = 2/7 [4 π R k T ]
Where: k = 1.8 x 10-10 (T ^5/2 / ln Z) W m^-1 K^-1
k = 1.8 x 10^-10 [(1.6 x 10^6 K) ^5/2 / 20] W m^-1 K^-1
k = 3. 1 x 10^4 W m^-1 K^-1
Therefore: the flux F =
2/7 [4 π (14 x 10^8 m) (1. 1 x 10 ^3 W m^-1 K^-1) (1.6 x 10^6 K)]
F = 2.5 x 10 ^20 W
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