5
$\begingroup$

Assume that by some means, the moon could be given an atmosphere, of the same density and pressure at the surface as the earth's. Obviously in a stable atmosphere there are temperature variations from pole to equator, day to night, and a lot of interesting dynamics follow from that, but for argument's sake assume an average of something like 0 to 5 C.

How long would it last?

Clearly there's no air there now (unless you count traces of potassium, or whatever it has), so such at artificial atmosphere would not be permanent, but presumably the moon could hang onto some air, for some time. With the moon's lower gravity, the density or pressure vs altitude curve will be different, and I'd guess that a significantly larger fraction of the mass would be at higher altitudes than on earth, and there's no protection from the solar wind. Someone with the right equations could probably work out how long it would take the pressure at the surface to drop by half.

$\endgroup$
1
  • 1
    $\begingroup$ just wondering whether it'd be days or millenia. $\endgroup$
    – JustJeff
    Commented May 8, 2011 at 11:57

5 Answers 5

2
$\begingroup$

Not sure about the solar wind but when it comes to Jeans escape: The escape velocity from the Moon is 2.4 km/s:

http://www.wolframalpha.com/input/?i=escape+velocity+from+moon

This is not much higher than the velocity of the molecules. The mass of Nitrogen molecule, the prevailing one, is 28 u,

http://www.wolframalpha.com/input/?i=mass+of+nitrogen+molecule

which is $4.65 \times 10^{-26}$ kg. In the $\exp(-mv^2/2kT)$ Maxwell factor for $T=275$ Kelvin, we may see that $$\langle v^2\rangle = kT/m = 1.38 \times 10^{-23} \times 275 / 4.65 \times 10^{-26} =81,600$$ squared meters over squared seconds. We need $v^2=2,400^2 = 5,760,000$ meters per second which is 70 times more than $81,600$, so the relevant exponential is $\exp(-35)$ (don't forget the factor of $1/2$) which is $6\times 10^{-16}$ or so. Only this fraction of molecules going up escape. It takes about 100 seconds for a molecule to try another value of its velocity, after a collision and a return from the upper atmosphere, so one needs something like $10^{16}\times 100 = 10^{18}$ seconds for the Nitrogen molecules to reach the escape velocity. That's something like 30 billion years.

Nitrogen in the lunar atmosphere (and heavier molecules) would be safe against Jeans escape, I think. Of course, it would be much easier for Hydrogen to escape. 28 drops to 2, 35 drops to something like 2 as well, so much of the Hydrogen escape during its 10th attempt or so, within minutes. Modifying the prefactors in the exponent makes a huge impact on the result.

$\endgroup$
4
  • 3
    $\begingroup$ As the models of the moon's creation give a common origin with earth, and as the universe is 14 billion years old, one would expect to have an N2 O2 H2O atmosphere on the moon, with these numbers. Maybe the solar wind and the solar magnetic field play a big role then? $\endgroup$
    – anna v
    Commented May 8, 2011 at 19:42
  • 1
    $\begingroup$ @anna v - i've heard it said that the atmosphere of Mars is so thin, b/c that body doesn't have enough magnetosphere to keep the solar wind from interacting with the topmost reaches of its atmosphere. Presumably this effect would be more pronounced on the moon, w/weaker gravity still. I have no idea how one calculates the effect. $\endgroup$
    – JustJeff
    Commented May 8, 2011 at 22:43
  • 4
    $\begingroup$ High altitudess, are normally much hotter than the lower altitudes. I think the major heatsource is u is solar ultraviolet, and this is perhaps 1 to 2% of the total solar energy. But the infrared opacity of this highest atmospheric level is very low, so the temperature that balances absorbed energy and radiated energy is several times higher than the surface temperature. This means the effective temperature for Jeans escape is much higher. Then add in the effect of the solar wind, and the atmospheric lifetime becomes even lower. $\endgroup$ Commented May 8, 2011 at 23:20
  • 3
    $\begingroup$ Dr Motl: how do you get the "100 seconds for a molecule to try another value of its velocity" figure? $\endgroup$ Commented May 10, 2011 at 18:57
2
$\begingroup$

Calculating Jeans escape time for temperature of 275 would be correct if you were talking about temperature near the surface, on the average, but (1) the escape time is dominated by the hottest part of the atmosphere, which for the moon would be more like 400K, and (2) the surface temperature isn't what's important for Jeans escape; it is the exosphere temperature, which due to ultraviolet heating is much hotter.

$\endgroup$
2
$\begingroup$

Geoffrey Landis studied this. He says "thousands of years".

For the heavier gasses, though--like oxygen and nitrogen--the gravitational escape lifetime in the atmosphere is thousands of years. While the moon will lose atmosphere over geological time spans, it could hold onto gas for a very long time by human scales.

For these gasses a different mechanism removes them from the lunar atmosphere. The unfiltered light of the sun ionizes the gas molecules, and the ionized molecules are then quickly swept away by electric fields associated with the solar wind. This occurs in a time span of approximately 100 days. When the atmosphere gets thick enough this mechanism stops happening--but the gas generation needed to make it "thick enough" is something like 10,000 tons/day--considerably higher than anything produced in our lunar industrial facility--at least in the next century or two.

So, once you are able to contribute 10,000 tons per day to the atmosphere it will then continue to thicken naturally, he says. Because the atomsphere protects the lower layers from ionization. So the answer is certainly millennia rather than days. I've been trying to find a more exact figure and got here. So - I see from other comments here that the problem is the temperature of the upper atmosphere of the Moon so I can see why he didn't give an exact number of years now.

It might not be that wise though, as the Moon's surface would react readily and possibly violently with water in the atmosphere according to N. M. Hoekzema in his An Atmosphere for the Moon

A lunar oxygen atmosphere can exist for many centuries, but only if it is very thin and in the absence of water. What if this atmosphere were moist and oxygen rich?

The upper few kilometers of the lunar surface contain several times 1018 kg of iron(II) which in the presence of water would readily react with oxygen to form iron(III). Such an amount of iron(II) could easily absorb all of the oxygen in the Earth atmosphere.

A large fraction of the Moons crust consists of oxides of calcium, magnesium, and iron(II), which in the presence of water would react to form hydroxides that would (partly) dissolve in the forming seas to create a poisonously alkaline fluid, with pH 10--11. If enough oxygen were available to oxidize the dissolved iron(II)hydroxides, insoluble iron(III)hydroxides would precipitate on the sea floors and shores, creating vast quantities of slightly poisonous, orange mud. Such reactions would be violent and fast in the upper part of the crust, but their rate would decrease with increasing depth. The oxidizing, hydration, and other processes would continue for ages. In the meantime oxygen and other pressures would not be stable. Most of important all: the absorption of such enormous amounts of oxygen, water, by the upper part of the crust of the Moon would make the rocks expand by perhaps as much as ten percent or more. One can wonder if such expansion would be a tranquil process. It could create strong quakes for possibly many thousands of years.

I go into this some more in my Terraforming the Moon in my Case for Moon First with more information.

$\endgroup$
1
$\begingroup$

The solar wind tends to disassociate molecules. So without the magnetic field, you should test nitrogen as single atoms with atomic mass of 14 instead of the molecular mass of 28.

1.38×10−23×275/4.65×10−26 = 81600 = ~1/70
1.38×10−23×275/2.325×10−26 = 163200 = ~1/35

1/EXP(70/2*-1)100=158601345231343072.81296446257747 = ~5.026 billion years
1/EXP(35/2
-1)*100=3982478439.7576225021870676349852 = ~126 years

$\endgroup$
0
$\begingroup$

From another similar post: "A commonly quoted rule of thumb says that the escape velocity needs to be 6 times the gas's mean velocity in order for that gas to remain captive to gravity [...]"

$V_{rms} = \sqrt{\frac{3RT}{M}}$

to divide by 6 and solve for T,

$T = \left(\frac{V_{escape}}{6}\right)^2 * \frac{M}{3R}$

Substituting values for the moon:

$T = \left(\frac{2380 \frac{m}{s}} {6}\right)^2 *\left(\frac{.032 \frac{kg}{mol}}{3*8.314 \frac{J}{mol°K}}\right) = 201.9°K$

Therefore, to hold an atmosphere containing molecular $O_2$ on the moon, it would have to be less than -72°C. If we use atomic O, then the temp is -172°C, barely 10°C above the boiling point. Brrr.

Examples of other planets/moons:

Titan:

$V_{rms}(CH_4) \ (@ \ 93.7°K) = 0.382 \frac{km}{s}$

$V_{escape} = 2.639 \frac{km}{s}$

$2.639\frac{km}{s} \ / \ 0.382\frac{km}{s} = 6.91 > 6$

Because Titan is so far from the sun, and significantly shielded by Saturn's magnetosphere (see https://upload.wikimedia.org/wikipedia/commons/5/54/Plasma_magnet_saturn.jpg), and thus the solar wind and ionizing radiation environment is comparatively benign (Saturn's radiation belts notwithstanding), it has probably lost very little of its original atmosphere despite being relatively close to the figure of merit.

Venus:

$V_{rms}(CO_2) \ (@ \ 737°K) = 0.646 \frac{km}{s}$

$Vescape = 10.36 \frac{km}{s}$

$10.36 \frac{km}{s} \ / \ 0.646 \frac{km}{s} = 16.04 > 6$

Although Venus is $\mathit dang \ hot$, and getting fried by the solar wind and ionizing solar radiation, its gravity is sufficiently strong that those comparatively big heavy $CO_2$ molecules aren't going to be leaving very fast.

One more just for laughs:

Triton (chosen because it is the smallest object in the solar system where this could remotely work, Titania and Ceres are just way too small), with a hypothetical atmosphere of $SF_6$, would be stable up to a temperature of 343°K. Given that Triton's average surface temp is 38°K and $SF_6$ is gaseous only above 223°K I will leave as an exercise to the reader to make that one work $\mathit Hint$: $SF_6$ is a greenhouse gas with a warming potential ~24,000 times that of $CO_2$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.