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Can Moon hold an atmosphere?

  1. Apr 15, 2012 #1
    Hi, I am trying to calculate if moon can possibly hold an atmosphere if we decide to terraform it.
    The assumption is that it is somehow miraculously completely shielded from Solar Wind,
    let's say by strong earth like magnetic field.

    I found somewhere on the web, that atmosphere of any planet will hold if the velocity of the gas molecules is less than 1/5 of the escape velocity of that planet (any link to why is it so is appreciated).

    1. Vair <= 1/5 Ve

    from this we get (substitute Vair with ideal gas velocity and Ve with Escape Velocity):

    2. √(3KT/Mair) <= 1/5√(2GMp/r)

    Now my questions are:
    a. What temperature T should i use in the equation, The maximum temperature in the upper atmosphere of the moon (380K)? or the average temperature of the upper atm (220K)? Maybe the average surface tmp that will be in the end?

    b. What do I use for r: is it top of the Thermosphere or Exosphere or maybe something else?
    (of course i could assume r to be Rmoon, but i just want to understand the physics here.)

  2. jcsd
  3. Apr 15, 2012 #2
    The idea behind escape velocity is that if you have an object traveling straight upwards (away from the surface) with a speed greater than that of the escape speed, then it will escape from the atmosphere and fly off into the solar system. Since in thermal motion all directions are uniformly preferred, some portion of the gas molecules will at some point be flying straight upwards and if they are near the top of the atmosphere then they can escape. If the speed of your particle is less than the escape speed, the gravitational pull will prevent it from completely escaping. Now, in thermal motion there are basically a range of speeds centered on the average speed, so if your average speed is just under the escape speed a substantial number of particles will actually be traveling faster than the escape speed, and therefore eventually escape. So people tend to pick a (somewhat arbitrary) factor below the escape speed to place the average speed at for holding an atmosphere, because you want few enough particles to be above the escape speed cutoff. I say it's somewhat arbitrary because, more or less, there will always be some particles above that cutoff. The fewer there are, the longer it takes for the atmosphere to disappear. So with the average thermal speed set at 1/5 of the escape speed, compared to, say, 1/10, you will get a lower lifetime for the atmosphere, as it were -- but they will both still be really long amounts of time.

    This is a slightly subtle question. Particles are more likely to escape from the upper atmosphere, because of the higher temperature (and therefore higher average speed) and lower density (and therefore longer mean free path between collisions of molecules). Note that this is not related to the fact that they are closer to outer space; unless the height of the atmosphere is significant compared to the radius of the moon, that difference won't account for much (see answer to part b). On the other hand, because of the lower density, you won't get a significant fraction of the atmosphere to disappear by getting rid of just the upper atmosphere part. So you want to use the average temperature of the atmosphere, or maybe even the surface temperature, to get a better estimate of whether the moon can hold the gas down.

    One way to think about escape speed is in terms of energy: if the kinetic energy of a particle is directed upwards (to abuse the meaning of a scalar quantity) and is greater in magnitude than the gravitational potential energy of the particle, then the particle will be able to convert enough of its kinetic energy into removing itself from the gravitational potential of the moon, and still have some left over to fly away. So the velocity comparison equation is essentially setting the kinetic energy multiplied by some arbitrary factor (5 or 10, as I discussed above) equal to the gravitational potential energy at some location. So if you're talking about the likelihood of particles in the upper atmosphere escaping, you would use the gravitational potential energy at the upper atmosphere, which is basically the radius of the moon plus the height of the atmosphere. On the other hand, if you want to get an estimate of the lifetime of the total atmosphere, you would use the radius of the moon, because as I mentioned in the answer to part a, you have to remove the significant amount of gas sitting at the surface to eliminate the atmosphere.
  4. Apr 15, 2012 #3
    Thanks for the answer Steely Dan,
    actually I do understand the basic physics of the process,
    What I actually need is formulas or ideas as to how to calculate the lifetime (lets call it Lt) of atmosphere as a function of Vair/Ve (which is now 1/5).

    Considering that time is important to me, i can't use T of the surface (which is only correct for lim(t→∞)), also i can't use Rmoon (because again it is for lim (t→∞))
    So that's why i need to know exactly which T and r to use there, they should be the same T and r that where used (or implied) in the calculation of lifetime.
    Btw, I do understand that to get a perfect answer here i will need to take an integral on Vair and Ve as a function of height h (where r<h<r+∞), and also as function of temperature.
    Well this would be an overshoot, so i just want to get a maximum close assumption here based on some facts (empirical or statistical).

    Last edited: Apr 15, 2012
  5. Apr 15, 2012 #4
    That calculation does sound non-trivial, and the reason may be the time dependence of the problem. It's not just an integral over the atmosphere's height but also over the time it takes for particles to escape. But to make an estimate, note that the particles that will escape will be the ones in your Maxwellian velocity distribution that are above the escape speed. Those particles will escape, and then the distribution will become Maxwellian again (the particles at lower speeds will be boosted up to the higher speeds, and then promptly escape). So if you estimate the time it takes to re-create the Maxwell distribution for the velocities, that timescale (divided by the fraction of particles above the escape speed in your distribution) should roughly correlate to the lifetime of the atmosphere. The thermal timescale should simply be the mean time between collisions of particles in the atmosphere. But the fraction of particles above the escape speed will often be the determining factor, since for mean speeds, say, 1/10 of the escape speed, only a tiny fraction will be above the escape speed.
  6. Apr 17, 2012 #5
  7. Apr 17, 2012 #6


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    Compared to your number of 1/5, I wouldn't worry about precise values for r and T: The moon has a radius of ~1700km, the atmosphere would be of the order of ~200km (larger than on earth to compensate the lower g). If you take the radius of the moon, the error is ~10%, and as you take the square root it gives an error of 5%, compared to the arbitrary 1/5.
    T can vary by a larger amount, but for terraforming, I think humans would want something close to 280K on the surface.

    I think you should try to work with numbers like that first. If you get a lifetime of years for the atmosphere, you don't have to consider the temperature values again. If you get millions of years: Similar. If you get something in between, it might be interesting to work more carefully.
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