Changes between Version 1 and Version 2 of u/adebrech/Matlab


Ignore:
Timestamp:
07/20/16 11:56:26 (9 years ago)
Author:
adebrech
Comment:

Legend:

Unmodified
Added
Removed
Modified
  • u/adebrech/Matlab

    v1 v2  
    11= Parker Wind =
    2 Solving
    32
    4 psi − ln(psi) = 4 ln(xi) + 4/xi − 3, where psi = (v/v,,s,,)^2^ and xi = r/r,,s,,.
     3== Differential Equation ==
     4
     5The Parker solution to the solar wind assumes a spherically symmetric radial outflow (with velocity u(r)). Then by conservation of momentum,
     6
     7{{{
     8#!latex
     9$\rho u \frac{d u}{d r} = -\frac{d p}{d r} -\rho \frac{G M}{r^2}$
     10}}}
     11
     12Continuity (conservation of particle #) gives
     13
     14{{{
     15#!latex
     16$\frac{1}{r^2} \frac{d (r^2 \rho u)}{d r} = 0$
     17}}}
     18
     19If we assume an isothermal corona, the ideal gas law is
     20
     21{{{
     22#!latex
     23$p = \frac{2 \rho T}{m}$
     24}}}
     25
     26Continuity shows that $r^2 \rho u = C$, where C is a constant. If we substitute for p and rho into the momentum equation, we get a differential equation in u and r:
     27
     28{{{
     29#!latex
     30$\frac{1}{u} \frac{d u}{d r}(u^2 - \frac{2 T}{m_p}) = \frac{4 T}{m_p r} - \frac{G M}{r^2}$
     31}}}
     32
     33Define $r_c = \frac{G M m_p}{4 T}$ so that $u_c^2 = \frac{2 T}{m_p}$ (i.e., sonic radius); substitute and rearrange to find
     34
     35{{{
     36#!latex
     37$\frac{d u^2}{u_c^2} - \frac{d u^2}{u^2} = \frac{4 d r}{r} - \frac{4 r_c d r}{r^2}$
     38}}}
     39
     40Integrate both sides to find the Parker wind equation,
     41
     42{{{
     43#!latex
     44$\psi - \log(\psi) = 4 \log(\xi) + \frac{4}{\xi} + C$, where $\psi = (\frac{u}{u_c})^2$ and $\xi = \frac{r}{r_c}$
     45}}}
     46
     47== Plotting ==
     48Solving the Parker wind equation:
     49
     50{{{
     51#!latex
     52$\psi - \log(\psi) = 4 \log(\xi) + \frac{4}{\xi} - 3$
     53}}}
    554
    655There 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.