Parker Wind

Solving

psi − ln(psi) = 4 ln(xi) + 4/xi − 3, where psi = (v/v_s)² and xi = r/r_s.

There are 4 types of solutions - combinations of sub/supersonic at the surface and v → 0/v_f as xi → infinity. The only physical solution is subsonic at the surface and has a nonzero velocity infinitely far away, and passes through xi = 1, psi = 1.

First attempted to solve equation in Mathematica - solution appears correct up to the sonic radius, but the incorrect solution (v → 0) appears for r > r_s. Next attempted to plot approximations for r << r_s and r >> r_s in Matlab, but as would be expected, the approximations fail near r = r_s. Numerically solving the equation in Matlab gives the opposite case of Mathematica - correct solution for r > r_s, but not for r < r_s.

Looking at Jonathan's code for calculating the Parker wind to find the correct approximation, try to implement it in a different way to test understanding. Works for r > r_s, but not r < r_s.

Pertinent line is: yy(i)=vpasolve(y - log(y) == -3 + 4*log(xx(i))+4/xx(i),y,(xx(i) > 1) * xx(i) + (xx(i) ⇐ 1)*exp(3-4*log(xx(i))-4/xx(i)))

Interpreted as:

if xx(i) ≥ 1

yy(i)=vpasolve(y - log(y) == -3 + 4*log(xx(i))+4/xx(i));

else

yy(i)=vpasolve(y - log(y) == exp(3-4*log(xx(i))-4/xx(i)));

end

Correct & incorrect plots:

Change in Bow Shock with Magnetic Field

If sigma_* and sigma_p are equal, the bow shock radius is unchanged with or without magnetic field - ratio of radius to orbital separation, chi_bow = 0.240468. With sigma_* = 1, sigma_p = 0.1, chi_bow = 0.148204; sigma_* = 0.5, sigma_p = 0.1, chi_bow = 0.187300; sigma_* = 0.1, sigma_p = 0.5, chi_bow = 0.302483; sigma_* = 0.1, sigma_p = 1, chi_bow = 0.363674; and with sigma_* = 0.5 and sigma_p = 1, chi_bow = 0.297793 ≈ chi_Coriolis.