Interstellar Planets/Hydrogen vapour pressure questions

[Evil Dr Ganymede]

I've been thinking about David Stevenson's paper on Interstellar Planets (see this link for the PDF), and am puzzled by something he says on page 6. Rather than bug him about what may (for all I know) be a very silly question, I thought I'd try asking here first

He's talking about how some worlds can retain hydrogen in their atmospheres in interstellar space after they've been ejected from the systems in which they form. Most of the paper is about the largest worlds, that retain very thick atmospheres, but for less massive worlds he says:

For sufficiently low masses, an alternative (collapsed atmosphere) solution exists with a molecular hydrogen ocean overlain by a thin vapor pressure-equilibrium atmosphere.

...and then conveniently doesn't say what the pressure of this atmosphere would be

Is there a way to calculate the pressure of a thin vapor pressure-equilibrium atmosphere made of hydrogen (over oceans of liquid hydrogen?) at a temperature of about 18 K? I've tried looking on the web, but I find a range of values from 0.213 atms, up to 1570 torr (about 2.1 atmospheres), which isn't exactly helpful. Or can it actually be a range of values?

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And as another question, if you were in orbit around a planet drifting in interstellar space (say, 1 or 2 lightyears from the nearest star at least), would you be able to see it in visible light?! Would starlight alone make it visible at all, or would it just be visible as a dark circle in the sky that blots out stars behind it??

[revent]

I've been thinking about David Stevenson's paper on Interstellar Planets (see this link for the PDF), and am puzzled by something he says on page 6. Rather than bug him about what may (for all I know) be a very silly question, I thought I'd try asking here first

He's talking about how some worlds can retain hydrogen in their atmospheres in interstellar space after they've been ejected from the systems in which they form. Most of the paper is about the largest worlds, that retain very thick atmospheres, but for less massive worlds he says:

For sufficiently low masses, an alternative (collapsed atmosphere) solution exists with a molecular hydrogen ocean overlain by a thin vapor pressure-equilibrium atmosphere.

...and then conveniently doesn't say what the pressure of this atmosphere would be

Is there a way to calculate the pressure of a thin vapor pressure-equilibrium atmosphere made of hydrogen (over oceans of liquid hydrogen?) at a temperature of about 18 K? I've tried looking on the web, but I find a range of values from 0.213 atms, up to 1570 torr (about 2.1 atmospheres), which isn't exactly helpful. Or can it actually be a range of values?

It won't be a range, but I don't remember how to calcuate it (My thermodynamics prof would yell at me ). What you really need is a CRC handbook, or just call your local university physics dept and ask the thermodynamics prof.

Seriously, vapor pressures are normally experimentally determined. AFAIK, the calculation would be a REAL major pain, and probaby wrong because you omitted some supposedly minor effect.

And as another question, if you were in orbit around a planet drifting in interstellar space (say, 1 or 2 lightyears from the nearest star at least), would you be able to see it in visible light?! Would starlight alone make it visible at all, or would it just be visible as a dark circle in the sky that blots out stars behind it??

It would definitely be visible if you looked only at the surface, since it would be reflecting light from the stars, and a sufficently dark-adapted eye can see a single photon (this has been experimentally determined). The illumination would almost certainly be so dim you could only see in black and white, tho, and the stars probaby so bright in comparison they would dazzle you and make you blind for a bit.

[granthutchison]

If you make the assumption that the liquid phase doesn't change in volume with changes in pressure, then the boiling point curve for a given substance follows the equation:

ln(P2/P1) = -deltaH/R*(1/T2 - 1/T1)

where deltaH is the molar latent heat of vaporization, R is the molar gas constant, and temperatures are in kelvins. So for liquid hydrogen the above reduces to:

ln(P2/P1) = -108.7*(1/T2 - 1/T1)

Unfortunately, the assumption of constant volume is a pretty poor one for liquid hydrogen over the range of vapour pressures between its triple point and its boiling point at one atmosphere. However, the International Practical Temperature Scale of 1990 gives us two combinations of temperature and vapour pressure we could plug into the above equation, and the triple point values gives us a third pairing. The list looks like this:

13.8033K 7040 Pa (triple point)
17.035K 33321 Pa
20.27K 101292 Pa (= 1 atmosphere)

If we select appropriately from the above, and use them as T1/P1 values, we can get a reasonable estimate of the vapour pressure of hydrogen across its liquid range at 1 atmosphere. For instance, if I plug in the 20.27K values, I get a vapour pressure at 18K of 51500 Pa (0.51 atm); if I plug in the 17.035K values, I get a vapour pressure at 18K of 47000 Pa (0.46 atm). Taking a weighted average of these two gives a final best estimate of 48300 Pa (0.48 atm).

Grant

[Evil Dr Ganymede]

You reckon it'll be around 0.5 atms, eh... So let me get this straight - what basically happens is the atmosphere collapses and liquifies, so the liquid hydrogen starts evaporating into the vacuum, but when the H2 vapour above it accumulates so much that the pressure at the base is about 0.5 atms, it stops evaporating and it's stable in the long term, so you can have the oceans remaining liquid under a permanent H2 gas atmosphere at 0.5 atms, yes?

[granthutchison]

The luminance of a perfect white diffuser illuminated by starlight alone at the Earth's surface is 6e-5 cd/m?. (You never see this in practice, because airglow and zodiacal light effectively increase the illumination by a factor of 10.) You can multiply this by around 1.3 one you're outside the atmosphere => 8e-5 cd/m?. Multiply by the albedo of your chosen object to get the actual luminance - so I guess this could be anything between 8e-7 and 8e-5 cd/m? in practice, for albedos between 0.01 and 1.
That's dim. The Milky Way has a luminance around 3e-4 cd/m?; the black sky seen from a back-country location at midnight is 3e-5 cd/m?; the estimate for the luminance of the interstellar black background is 6e-6 cd/m?, as random photons come in from all directions after being scattered by dust and gas.
Visual threshold for a dark-adapted eye is somewhere between 1e-7 and 1e-6 cd/m?, so your planet is going to be reflecting detectable light levels across the plausible albedo range: for the highest albedos, it could be up to 13 times brighter than the dark sky, and so might be visible as a dim grey loom, but it would still look black seen against the Milky Way. But if it were made of very dark material, it could actually be darker than the night sky - I think that would probably make it difficult to see ...

Grant

[granthutchison]

but when the H2 vapour above it accumulates so much that the pressure at the base is about 0.5 atms, it stops evaporating and it's stable in the long term, so you can have the oceans remaining liquid under a permanent H2 gas atmosphere at 0.5 atms, yes?

Exactly. If the atmospheric pressure is higher than 0.5 atmospheres, the net movement of molecules is from gaseous to liquid phase; if atmospheric pressure is lower than 0.5, the net movement is from liquid to gas. So the whole thing equilibrates at 0.5.

Grant

[Evil Dr Ganymede]

Cool. Thanks again, Grant (and Revent).