# Kinetic Energy of the Earth's Atmosphere

Evening,

Being the studious student that I am, I’m doing the extra credit for my “Rocks for Jocks” lab science. (not that I need it. but why not, eh?)

Regardless, the only thing I’m stuck on is finding the kinetic energy of Earth’s Atmosphere. I have a funny feeling that’s variable, depending on season and other assorted variables, so I’d be more than happy with an average.

You want us to do your homework?

More like, I’m doubting that there IS an answer. And this is part b of a 7 part question, and I only need that information to actually solve the problem. If you don’t want to help, that’s your perogative.

I would start by getting a figure for the temperature of the atmosphere. It will vary along some curve from some average surface temperature to very low temperatures. There is a formula for the relationship of temperature to mean kinetic energy/atom that I can’t think of at the moment. I know it involves the Boltzman constant.

That’s as far as I can go in pointing out a possible direction to go.

The answer will depend on scale. On a large scale, you’d want to estimate some sort of typical wind speed, find the total mass of the atmosphere, and use good ol’ K = 1/2 mv[sup]2[/sup]. On a microscopic scale, you’d want to estimate a typical temperature for the atmsphere, which will give you the kinetic energy per particle (approximately k[sub]b[/sub]T, to within some small dimensionless factors), and multiply by the total number of particles in the atmosphere. The latter will be much larger, but may or may not be relevant to your problem.

Note that both of these would only be estimates (probably good to an order of magnitude or so), and neither of these is a constant. I presume that the question only calls for a rough estimate.

About 1000mph at the equator, slowing to zero at the poles, or is he ignoring rotation?
As far as mass goes, the atmosphere exerts a force of 14.7 pounds per square inch, at sea level. That corresponds to a mass of 1.11 kg/cm[sup]2[/sup]. Multiplying by the surface area of the earth (5.1 X 10 [sup]18[/sup]cm[sup]2[/sup]) gives a mass of 5.66 X 10 [sup]18[/sup]kg. (figures not checked)

pretty close, as it turns out. Wikipedia gives the mass of the atmosphere as 5.1 X 10^18kg, which is what I’d planned on using.

The real trick is finding an average velocity, or KE = .5mV^2 isn’t much help. I thought about taking the average windspeed of the globe, but I’m not sure at what rate the windspeed changes as the latitude changes.

Ke = 3/2 nRT where n is the number of moles of gas, T is the temperature in Kelvins, and R is the gas law constant usually expressed in units of 1.98 cal/deg mole , would also work, but then I’d need to know the average temperature of the atmosphere

Well. I think NASA publishes a temperature curve for the standard atmosphere. You could fit a mathematical curve to it and get an estimate of the average by integration.