Setting Atmosphere Density

Constants:

Assume every photon entering a tube of dimensions (R0-RP, dx, dx) is absorbed by the time it reaches RP, and all the hydrogen in the tube starts ionized (X=1). Using an adiabatic atmosphere, the density profile is given by

,

with , , and the number density at the surface of the planet (to be solved for).

Volumetric recombination rate is

,

(at the wind temperature of 104 K).

Total recombinations (per unit time) are then given by

Total photons entering the tube per unit time are given by

,

where is the photon number flux.

Solving the equality​ of these two quantities for np, find

.

For the currently-running case, our flux of 2x1013 gives a mass density of , much larger than our current value of 1.32x10-16 (but we're not recombination-limited in the current case). For the high-flux case, Jphot = 2x1017,

.

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Comments

1. mccann -- 7 years ago

Hey Alex diff ​my corrections and see if you agree with them. I get , I've also got that magnitude from my notebook. The biggest being a square root on the sound speed, reversing bounds of integrations and consistently working in planet radius units (as that is how R0 and Rcrit are defined in your file). The integration also has analytic solutions in terms of incomplete beta functions, or simple polynomials when those would have blown up.

2. adebrech -- 7 years ago

Yep, that's what happens when I work late. All looks good to me.