Meeting Update

Thoughts on parker wind

The solution for the parker wind is found by solving the equations below

Note however, that the actual radius of the planet (and lambda) does not really matter…

All that a higher lambda implies, is a smaller planet radius (compared to the sonic radius) - and sampling more and more of a subsonic atmosphere before having a hard boundary.

Also - per our discussion the other day, it seems that

So this effect is more pronounced when the orbital sepration is small and when the planet radius and mass ratio is close to - ie Hot Jupiters

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Comments

1. Adam Frank -- 9 years ago

So smaller r_p would mean a higher \rho_p since we would be driving the wind from deeper in the atmosphere(?)

2. Jonathan -- 9 years ago

Yes - for a fixed and , a smaller would mean a larger . It would also mean that the velocity at the surface would be more and more subsonic.

The parker wind doesn't have a 0 velocity inner boundary… Even at the planet 'surface' the parker wind solution has a non zero velocity. Although for r_p << r_s, or equivalently, for lambda >> 1, this velocity is almost 0. See the dimensionless solution to the parker wind.

Presumably if lambda ~ 1 and if the velocity at r_p were zero, the solution would deviate from the parker solution. And then different lambda's would give different dimensionless curves.

3. Jonathan -- 9 years ago

Which I guess is what happens with the Stone and Proga type BC's