## The Size Of The Atmosphere

I have had several discussions where people have said that the atmosphere is so enormous than man could not affect it. When you look up at the sky and think about the size of the Earth it is not difficult to see why they think this. I began to wonder what volume it would take up if it was a liquid such as water – will it be as big as an ocean, a sea or a lake.

Air is a gas and is easily compressed – a fact that is obvious if you climb a high mountain or fly in a plane – the air becomes a lot ‘thinner’ i.e. less dense. Another indication is that the atmosphere is not that big is that we live beneath it. We have the weight of the entire atmosphere above pressing down on us – something we call *atmospheric pressure.*

So this is my quick ‘back of the envelope’ calculation on the mass of the atmosphere which was done with the help of my friend Ethan.

The way we have done this is to take the average atmospheric pressure at sea level (101325 Pa). One Pa (Pascal) is a unit of pressure equal to 1 Newton per meter squared (N/m^{2}). This is acting on the entire surface of the Earth (radius 6371Km = 6.371 x 10^{6}m). The surface area of a sphere is 4Ï€r^{2} which gives a area of 5.1 x 10^{14}m^{2}. If we multiply this area by the pressure (force per unit area) we get the force of the atmosphere on the Earth which turns out to be 5.17×10^{19}N.

We now use Newtons law F=ma – force equals mass times acceleration which means that mass is force divided by acceleration m=F/a. The acceleration due to the Earth’s gravitational field is 9.81m/s. This gives a mass for the atmosphere of 5.26 x 10^{18}Kg. Looking on wikipedia we find that they give 5.1480 Ã— 10^{18}Kg. So our figure is not that much out^{2}. However, looking at the recent liturature^{1} we noticed that they give a mean surface pressure (98305 Pa) which is much lower than the one we used (101325 Pa). If we use this in our calculations we get a mass of 5.11 x 10^{18}Kg which is much closer to the figure quoted in Wikipedia.

Since density is mass per unit volume (Ï?=m/V) then volume is mass divided by density (V =m/Ï?). The density of water is 1g/cm^{3} or 1 x 10^{3} Kg/m^{3}. This gives the volume of 5.15 x 10^{15}m^{3} or 5,150,000Km^{3}. This is a bit bigger than the Mediterranean^{3} (4,390,000Km^{3}) but much smaller than major seas and oceans – e.g. the Atlantic is 310,410,900Km^{3}

Could man have a significant affect on the Mediterranean by pollution – we already have.

^{1} The Mass of the Atmosphere: A Constraint on Global Analyses K. E. Trenberth and L.S. Smith, Journal of Climate, March 2005 (http://acd.ucar.edu/~lsmith/massERA40JC.pdf)

^{2} There are several inaccuracies with our ‘back of the envelope’ calculation. For instance it would be necessary to take into account the variation in the height above land since the air is much denser near sea level, the Earth is not totally spherical and the gravitational force decreases with height.

^{3} Volumes of the World’s Oceans from ETOPO1, NOAA (http://www.ngdc.noaa.gov/mgg/global/etopo1_ocean_volumes.html)

Hi Peter:

interesting post, thanks:

Personally I’d change…

“Could man have a significant affect….”

… with…

“Could man have a MEASURABLE effect…”

…and the answer is “yes”, of course… it is measurable since more than a century at least.

“Significant” doesn’t mean much, it is a relative and often subjective qualifier, in my opinion. What is significant to one may not be such for someone else.

“Measurable”, on the other hand, has a well defined and absolute meaning.

In the case of earth’s atmosphere, man can measure its effect on the atmosphere… man-made pollutants are easily identified even at very low concentrations… but how “significant” these man-made changes are is disputable.