From the Top of the Atmosphere

Here is a different kind of error; I'm not at all sure where the error lies. Maybe there is none, but in that case I'm wrong to put it this way. Bottom line: the molecules in the air around you are moving about as fast as if they had just fallen from the "effective height of the atmosphere", and I think that this is true anywhere you go, on any planet. But I'm extremely unsure of this.

Definition: the "effective height of the atmosphere" is pressure/density, which at sea level is about one ton per square foot divided by about one ounce per cubic foot, which gives you somewhere near 30,000 feet. It's the height which the atmosphere would have if its density were constant. (Actually, of course, if you go up 300 feet then you've risen above 1% of the atmosphere, so the pressure goes down by 1% and [unless the temperature has dropped already] the air expands to match, so that the "effective height" will still be some 30,000 feet over your head.)

The pressure and density I mentioned may be unfamiliar, but it's easy to check them. You probably remember sea-level pressure as being 14 pounds per square inch, and one square foot is 144 square inches -- so that's about a ton. Density is not so easy; perhaps you remember that air is nearly 80% N2 with a gram-molecular weight around 28, with the residue almost all O2 of about 32, so one mole (22.4 liters, a bit more than a cubic foot) will weigh 29 grams, just about an ounce. Well, it's really about an ounce and a quarter per cubic foot. Or you can look it up somewhere -- Galileo wrote somewhere in his Dialogues Concerning Two New Sciences (that's the less famous, more geeky of his main books) that air was about 1/800th the density of water, and if you know that a cubic foot of water is 60-some pounds or about 1000 ounces, then you've got it. The effective height of the atmosphere, at least when you start from sea level, is somewhere near 30K feet. Okay? (Is it a coincidence that we're roughly locating the tropopause, the troposphere-stratosphere boundary, here? Is there a tropopause for other planets? I have no idea.) Now, what about that speed?

Thirty-five years ago, as a sophomore physics student at Nacional de Buenos Aires, I was really quite surprised to learn that air molecules have an average velocity that's higher than the speed of sound; it took a while for my intuitions to accept that this is necessarily true because, loosely speaking, when air molecules (going in all directions) carry a wave (in one direction) some of their "effort" is literally misdirected. If the speed of sound is 1000 feet per second, the average speed of the molecules is 1500. As I recall, the 3/2 is exact and drops out of some volume integral whose details I no longer recall at all, but you may have noticed that exactness doesn't worry me here anyway.

Okay, so I've claimed that the speed of the molecules, which you now know is about 1500 fps, is about what they'd get from falling 30K feet. What was that formula again? Well, constant acceleration just means V=A*T which integrates to D=0.5*A*T^2 which we solve for T to get T=SqRt(2*D/A), and now we plug that T back in to the first equation to say V=A*SqRt(2*D/A)=SqRt(2*A*D) and if A is 32 fps^2 while D is 30K, V has got to be SqRt(2,000,000) or 1,414 which is definitely close enough. So yes, it works at sea level. But is it really a coincidence, or is it a Law that the Galactic Patrol's Lensmen will come after you for violating?

Imagine a tube that goes all the way up, to the top of the atmosphere, but is only wide enough for one molecule. Imagine that it's 100% nitrogen, which shouldn't make much difference at all, except that it helps the visualization. Imagine as usual that it's an ideal gas: as each molecule bounces off the molecules above and below, energy and momentum are conserved, so that if you blink as an actual collision happens then you can't tell that the two molecules didn't simply go through each other. Okay, that should mean that the bottom molecule is moving just as fast as if it were the top molecule of a a minute ago. (The actual time T of the last paragraph is about 50 seconds.)

Problem with that tube-to-the-top-of-the-atmosphere model: it had nothing at all about 30K feet being the top of the atmosphere. The tube went to the actual "top" of the atmosphere. Another problem is that it has no way to talk about averages; everything should be at a fixed velocity relating to height. I remember this puzzle every now and then -- probably at least once a year. Sometimes I convince myself, for a minute or ten, that these two problems actually cancel out: the 30K is an effective average. Then I think I really ought to make a computational model. Then I think I really ought to do some work instead. But at least I have written my puzzlement down.