Atmospheric retention

[granthutchison]

Where do you get those UV proportions from? Did you have to calculate the blackbody curves for the stars yourself, or has it already been done somewhere? And what would be the UV proportion for K and M stars (a lot lower than G, I'd imagine).

I did them myself - same little program I used to give you luminous efficacies a while back. It just numerically integrates the black body spectrum over the appropriate wavelength interval. Here's a little table of energy output in UV, visible and IR for the middle of the various main sequence spectral classes.

Code: Select all

     O5   B5   A5   F5   G5   K5     M5
UV   98%  75%  34%  18%  12%   3%   0.4%
vis   1%  18%  39%  39%  37%  26%   11%
IR   0.4%  7%  27%  43%  51%  71%   88%

(But note that much of the visible energy is going to be at blue wavelengths in F and above, and red wavelengths at K and below - the eye is less sensitive to these, so actually illumination will be less than you might guess from the above figures. Use the luminous efficacy numbers to get a real feel for the level of illumination provided by a given spectral class.)

Do I need to put the fourth root values of those UV ratios into the MU equation because it's the UV flux that's important, not so much the temperature?

Temperature is the important thing, since it determines the velocity of the atoms/molecules. UV absorption seems to be an important determinant of the high temperature of Earth's exosphere. I'm just taking the fourth root of the UV flux in order to convert energy flux to temperature.
(You'll see that this fourth-root relationship is also implicit in the equation for planetary temperature, which involves the fourth root of [one minus the albedo].)

Grant

[Evil Dr Ganymede]

Sweet. Thanks for that table, that's going to be handy .

Sooo... taking the 4th root of the UV ratios we have:

O: 1.7
B: 1.6
A: 1.3
F: 1.1
G: 1.0
K: 0.7
M: 0.4

So it seems that if it's down to EUV (extreme UV) flux then planets around K and M stars should suffer less atmosphere loss than an identical planet around and F or G star?

[granthutchison]

So it seems that if it's down to EUV (extreme UV) flux then planets around K and M stars should suffer less atmosphere loss than an identical planet around and F or G star?

For an Earth-like temperature and oxygen content, it seems like the exosphere should be cooler with an M-type than an F-type. And to this level of very shakey approximation, it seems like an Earth-like planet might just be able to hang on to helium if it were in orbit around a cool M dwarf - but in that setting you've got the likelihood of tidal locking, which would shove the general dayside temperature higher than we've figured so far, perhaps offsetting the reduction in UV flux.
At the other end of the scale, it doesn't look like the other atmospheric components are in much danger, since the cut-off molecular weight doesn't get much higher than 14, even around an O-type.

Grant

[Evil Dr Ganymede]

Indeed... hmm. You've reminded me that I'm gonna have to figure out a way to take tidal locking into account too.

Though is it really just a case of dividing the UV ratios?

Interpolating from your table, a G2 V star has a luminosity of 1 Sol, 13.8% of which is in the UV, 37.6% of it in the Visible, and 48.6% of it in the IR. So it has a UV luminosity of 0.138 Sols, Visible luminosity of 0.376 Sols, and an IR luminosity of 0.486 Sols.

An F5 V star has a luminosity of 7 Sols, and 18% of that in the UV, 39% in the visible, and 43% in the IR. So that means that it has a UV luminosity of 1.26 Sols, a visible luminosity of 2.73 Sols, and an IR luminosity of 3.01 Sols, right?

So the F5 V star is emitting about (1.26/0.138 =) 9.13 times as much UV as the G2 V star isn't it? Even if I take the fourth root of that, that comes to 1.74, not 1.1 as we figured earlier.

EDIT: Waitaminute, I think I grok this now.

The 9.13 value I calculated above is basically the 'UV Luminosity' of the star relative to Sol, right? Ordinarily, to find the blackbody temperature of a planet orbiting the star, you use the blackbody equation:

Tb = [L/(16.PI.sigma.D^2)]^0.25

But to get an 'Exosphere temperature' (which is really what I'm after here) you need to put in the UV luminosity (UVL) instead of the total luminosity, so:

Te = [(UVL)/(16.PI.sigma.D^2)]^0.25

Which is the same as multiplying Tb by (UVL^0.25).

So very roughly we can say that Te = Tb * (UVL^0.25) [* 4 if there's an O2 atmosphere], right?

[granthutchison]

But it's the proportional UV flux that's important here, not the absolute value. Your baseline temperature is calculated from the black body equilibrium. Then you've introduced a constant of proportionality (4) to account for UV absorption by an oxygenated exosphere. But you need to adjust that value for the increased UV flux associated with hot stars - if, for a given temperature, three times more UV is arriving, it's going to have three times greater an effect - which translates to a multiplier of fourth-root-of-three (~1.3) on the temperature. So you multiply by four and then by 1.3. The absolute luminosity has already been dealt with in the black body calculation, and we're only seeking proportional multipliers.
If you plug absolute UV luminosity straight into the blackbody equation, then you're neglecting the blackbody equilibrium temperature of the planet, which has got to be a major determinant of the exosphere temperature - the exosphere gas is just atmospheric gas that's reached high altitudes, after all.

I don't think we know what to do about non-oxygenated atmospheres, though - the UV-absorption scenario applied only to the Earth, as an explanation of its anomalously high exosphere temperature. We don't know if there are any components of other exospheres that are specifically absorbing UV, though I imagine that's written down somewhere.

Grant

[Evil Dr Ganymede]

I'm not assuming that the planet is at the same temperature and distance as Earth though - I'm trying to make a general case here.

Anyway, I'm still using a proportional UV flux here aren't I? The F5V star I calculated earlier has a UV luminosity that is 9.13 times larger than the G2V star that we orbit. Taking the fourth root of that is 1.74.

As I see it (which admittedly may be wrong ) I think I am actually doing what you're suggesting. I'm taking the relative UV luminosity, and modifying the BB temperature. I think you're saying that the exosphere temperature for an O2 atmosphere should be:

1) Te = Tb * [UV luminosity relative to Sol]^0.25 * 4

where Tb = [L/(16.PI.sigma.D^2)]^0.25

I'm saying that the exosphere temperature for an O2 atmosphere should be:

2) Te = [UVL/(16.PI.sigma.D^2)]^0.25 * 4

which is exactly the same thing as the first equation. So the Blackbody Temperature is still in there.

Plug in the numbers - let's say that we have Earth at 1 AU from an F5 V with the previously described parameters (UVL=9.13), and Earth at 1 AU from a G2V (UVL=1).

Using equation (1), I have:

Te = 278.3 * 1 * 4 = 1113 K for the G2V case, and
Te = 278.3 * (9.13^0.25) * 4 = 1935 K for the F5V case.

Using equation (2), I have:
Te = 278.3 * 1 * 4 = 1113 K for the G2V case, and
Te = 483.74 * 4 = 1935 K for the F5V case.

So I think we're saying equivalent things, just phrasing it differently. Perhaps your version is clearer though. If I'm still not following you though, can you illustrate what you're trying to say with an equation? (sorry to be such a pain ).

[granthutchison]

I'm just realizing that this is going nowhere. The problem is that we're using a multiplication factor, which doesn't make any physical sense to me, now I'm thinking about it properly. It implies that very low UV flux would effectively cool the exosphere below the blackbody temperature of the vast bulk of the planet. But what's going on is that the exosphere is populated by gas diffusing up from below, in equilibrium with the black body temperature of the planet. If it contains oxygen, it is able to absorb a significant proportion of the solar flux (because it absorbs UV), but can't reradiate it effectively (because this rarified gas can't behave like a black body). So the temperature seeks upwards to a new equilibrium at which reradiation does match absorption. If there's no UV, or no oxygen, the energy influx isn't there, so there's no rise in temperature (but no fall, either). So this is better modelled by an increment in temperature. Trouble is, the physical processes involved are going to be horrendously non-linear, since they depend on the precise radiative and absorptive behaviour of the specific exosphere gas mix.
Sorry, after a brief foray into modelling this, it seems I should have stuck with my original protestation - too few data.

Grant

[Evil Dr Ganymede]

Hmm. I see your point.

Though interestingly enough, the graphs I've seen comparing Mars, Venus, and Earth show that Venus' exosphere is actually cooler than Mars' (seems to be around 175K, according to this), and cooler than its blackbody temperature (about 332K).

Oh well. I'm starting to think we're going in circles too . Still, thanks for all the help, you've given me much to ponder! (hopefully this has been useful to everyone else too!)

[granthutchison]

Though interestingly enough, the graphs I've seen comparing Mars, Venus, and Earth show that Venus' exosphere is actually cooler than Mars' (seems to be around 175K, according to this), and cooler than its blackbody temperature (about 332K).

It gets hotter higher up. By 200km altitude, the dayside of Venus is at almost 300K, according to the graphs in The New Solar System. And if you factor in Venus's high albedo and slow rotation, the equilibrium blackbody temperature comes out around 270K. Interesting that Mars has a significantly higher increment, though: 215K for its albedo-corrected blackbody temperature, and a tad over 300K in the exosphere. Both presumably have CO2 as the dominant molecule - which I guess just goes to show that there are non-linearities we wot not of.

Grant