The coronal heat density is generally given by:

**q = - k grad T**where k denotes the conductivity and grad T is the gradient of the temperature.

The simplest physical model is a static corona for which heat inputs cancel outputs, so:

div q = 0

Assuming a spherical symmetry for the corona one can write:

1/r^2 [d/dr (r^2 k_o T^ 5/2 dT/dr) = 0

Obviously the preceding assumptions mean there must be some distance where the coronal temperature becomes zero. From the above equation. one should be able to show:

d(T^ 7/2) = 7/2 (F T_o 5/2)/ 4π k_o d(1/r) = C d(1/r)

where C is a constant.

The integral is:

To ^7/2 - T^ 7/2 = C[ 1/R_o - 1/r]

Now, set the temperature at infinity (T) to zero and obtain:

C = R_o T_o ^7/2

which fixes the total flux at:

F = 2/7 [4 π R_o k_o To ]

After another step, one finds:

T(r) = T_o (R_o / r) ^2/7

which gives the temperature T(r) as a function of a distance r based on taking the reference 'boundary' of the corona as having T_o = 2 x 10^6 K. For example, at the Earth's distance, r = 1.5 x 10^11 one would find:

T(r) = (2 x 10^ 6 K) [7 x 10^8 m/ 1.5 x 10^11 m] ^ 2/7

where R_o is the solar radius, hence T_o is referenced to

*the base of the corona*. Then:

T(r) = 4.3 x 10^5 K

__Problem:__

Find the value of the constant C at the solar corona boundary.

*Solution*:

We use: C = R_o T_o ^7/2

So: C = (7 x 10^8 m) (2 x 10^6 K) 7/2 = 7.9 x 10 ^30 m K^7/2

There is also an "empirical" thermal conductivity given by the formula:

k = 1.8 x 10^ -10 (T ^5/2 / ln Z) W m^-1 K^-1

here ln Z = 20.

If we separate out the T^ 5/2 factor, then it follows that the 'baseline' thermal conductivity is:

k_o = [(1.8 x 10^-10 )/ ln Z ] W m^-1

Or: k_o = (1.8 x 10^-10 )/ 20 W m^-1 = 9 x 10^-12 W m^-1

__Problem__:

Find the temperature of the corona at the distance r = 2R_o. From this estimate the flux at this distance.

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