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The following statement is from this article:

The behavior of a dense atmosphere is driven by collisions between its atoms and molecules. However, the moon's atmosphere is technically referred to as an exosphere because it’s so thin, its atoms rarely collide.

However, it doesn't explain what is meant by "rarely collide". I find this strange, especially since the word 'technically' is used. There doesn't seem to be anything technical about such a vague definition. I mean, someone might think that the atoms of the atmosphere rarely collide at an altitude of 400 km above the earth's surface. So he or she might think that that must be exosphere, but it's not. It's the thermosphere.

The following statement is from this article:

However the moon's atmosphere is so thin, atoms and molecules almost never collide. Instead, they are free to follow arcing paths determined by the energy they received from the processes described above and by the gravitational pull of the moon.

Much the same thing in other words. I would have thought that between collisions atoms and molecules follow arcing paths whenever there is gravity. The higher the pressure the shorter the arcs, but arcs nevertheless. So this doesn't clarify anything for me.

So my question is: what does it mean to say that "its atoms rarely collide"? Does it mean they normally bounce off the moon rather than each other?

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    $\begingroup$ "Technically" doesn't always imply precision. It can also be used as a way to say "this is factually correct but generally irrelevant for the topic at hand". Such as "Frank is the most Scottish person I have ever met, even though he is technically Irish". That doesn't answer your question, but it does counter the notion that "technically" somehow implies great precision. $\endgroup$ – Flater Apr 27 at 12:53
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    $\begingroup$ Either you edit the question to ask about specific data or this is some exercise at semantics. $\endgroup$ – Mithoron Apr 27 at 13:41
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    $\begingroup$ also: en.wikipedia.org/wiki/Exosphere#Lower_boundary "If we define the exobase as the height at which upward-traveling molecules experience one collision on average, then at this position the mean free path of a molecule is equal to one pressure scale height." $\endgroup$ – eps Apr 27 at 14:42
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    $\begingroup$ like all definitions for the boundary though, you are just going to get a rough estimate (which rapidly changes anyway depending on local conditions). $\endgroup$ – eps Apr 27 at 14:44
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    $\begingroup$ @eps "the height at which upward-traveling molecules experience one collision on average". One collision per what? $\endgroup$ – Matthew Christopher Bartsh Apr 27 at 18:14
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Any given gas molecule goes a long way on average without colliding with another gas molecule. The mean free path for a gas molecule is on the order

$$ \lambda\sim \frac{k_BT}{d^2p} $$

where $k_B$ is the Boltzmann constant, $T$ is the temperature, $d$ is the kinetic diameter of the molecule in question and $p$ is the pressure. Plugging in some rough values for the lunar environment gives

$$ \lambda\sim 10^5~\rm km $$

as a back of the envelope estimate. The moon's circumference is only $\sim10^4~\rm km$, so most gas molecules probably escape the moon's atmosphere or collide with the moon without ever colliding with another gas molecule.

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    $\begingroup$ Note that on Earth at STP, the mean free path is ~60nm (nanometers, not nautical miles...) $\endgroup$ – Jon Custer Apr 26 at 22:09
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    $\begingroup$ @honeste_vivere $10~\rm eV$ corresponds to a temperature of $10^5~\rm K$, about twice the temperature of the sun's surface. Fortunately for the Apollo astronauts, the moon is much cooler than that. ;) $\endgroup$ – Chris Apr 26 at 22:39
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    $\begingroup$ @honeste_vivere An energy of $10~\rm eV$ corresponds to a speed much higher than the escape velocity so presumably any such sputtering does not stay in the moon's atmosphere for long- if you even consider it to have ever been part of the atmosphere. NASA measurements of the moon's atmosphere have its temperature range from $95~\rm K$ to $390~\rm K$. $\endgroup$ – Chris Apr 27 at 20:32
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    $\begingroup$ That mean free path is crazy. I'm used to the idea that atmosphere holds itself up through molecular collisions, but those particles seem not to ever hit each other, so why doesn't it just... fall down? Is it just elastic collisions with the ground (like little bouncing balls) or orbital trajectories from the solar wind? $\endgroup$ – Mario Carneiro Apr 27 at 23:48
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    $\begingroup$ @MarioCarneiro: Indeed, that's the whole point of the question: NASA is saying that an exosphere is qualitatively different from an atmosphere, because the mean free path is on the same scale as orbital / escape distances. So as you say, it's more like a bunch of stray molecules individually bouncing around for the duration they're near the moon at all. (At least that's my understanding.) $\endgroup$ – Peter Cordes Apr 28 at 2:13
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The phrasing is a reference to some of the assumptions made in the physics of gasses. For gasses at a pressure level we are used to, like on the surface of earth, the mean free path length is very tiny (about 68nm at STP). This means that a gas molecule undergoes a great many random collisions as it travels a distance. By the central limit theorem, this makes the behavior of the gas very predictable in a statistical sense. Many of our laws are formulated on these assumptions.

In rarefied gasses, such as the atmosphere of the moon, mean path lengths can be much longer. At $2\cdot10^{-12}$ Torr, that atmosphere qualifies as "ultra high vacuum" with mean path lengths on the order of $10^5$ km!

This means many assumptions we make about how gasses work require more complex physics to model them. In our atmosphere, it is very reasonable to assume that the velocity of a particle is almost uniformly distributed, and they travel in almost straight lines. On the moon, we must account for the fact that particles have been under the influence of gravity for quite some time between collisions.

The practical differences show up in high vacuum chambers as well. Turbo-pumps look a lot like fans, but they aren't designed to create pressure gradients. The idea of a "pressure gradient" is not as meaningful in such vacuums. Instead, the fan blades act more like baseball bats, knocking the gas atoms away.

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  • $\begingroup$ @ruakh Thanks. I have too much use for the word "reified." Finger memory is strong! $\endgroup$ – Cort Ammon Apr 28 at 3:05
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    $\begingroup$ Is "almost uniformly distributed" meant to read "almost normally distributed"? $\endgroup$ – jacob1729 Apr 28 at 11:56
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"Rarely" doesn't have a black-and-white, sharply-delineated definition, because there is no non-arbitrary cutoff for when things "almost never" happen (unless you're a mathematician - then "almost never happen" means "happens with probability $0$"; that's not the case here, though!).

The key point is that, the thinner a gas gets, the less and less likely it is for its molecules (including single atoms as "molecules") to collide with each other, because they're farther apart and so it is easy for an atom moving in any given direction to miss a lot of atoms before it finally finds one to collide with - if it ever does at all.

The reason though why this is important is that the usual "thermal" distribution of molecular speeds in a gas - the Maxwell-Boltzmann distribution - comes about by repeated collisions of the gas molecules with each other. Note that if you think about the ideal gas model, there is no a priori reason that the molecules can't have any old speed distribution whatsoever: so that we model a gas as having this distribution must come from some sort of assumption. That assumption is collisions, and that assumption is what starts to fail here. In a sense, the gas starts to lack a well-defined "temperature" under generic circumstances.

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