*p*, varied with altitude,

*y*according to,

*p*= 10

^{5}e

^{−y/104}

Furthermore, we concluded that the density of the atmosphere, ρ, was related to pressure according to,

ρ =

*p*/

*f*, with

*f*= 10

^{5}Nm/kg

When we combine these two equations, we arrive at,

ρ = e

^{−y/104}

Consider our infinitesimal atmospheric element, which you can see here. Its infinitesimal mass, d

*m*, is given by,

d

*m*=

*A*ρd

*y*=

*A*e

^{−y/104}d

*y*

Let's imagine going from sea-level to infinity, adding together the masses of all the infinitesimal elements as we go. That is to say: let us integrate d

*m*with respect to

*y*for

*y*= 0 to ∞, and let's also assume

*A*= 1 m

^{2}, so we obtain the mass of air above one square meter.

_{0}∫

^{∞}

*A*e

^{−y/104}d

*y*= 10

^{4}kg

So, there we have it: the mass of air above one square meter at sea-level is 10,000 kg, or ten tonnes. The pressure of the atmosphere at sea-level is equal to the weight of the atmosphere above each square meter. Indeed, the same is true at any altitude: the pressure at any altitude is equal to the weight per square meter of the atmosphere above that altitude. This conclusion is important for our future efforts to determine the absorbing power of the atmosphere. If we have radiation at altitude

*y*, we know what mass of air this radiation must pass through to get to outer space.

Suppose the atmosphere were incompressible, like water, but we had just as much of it as we do now. This atmosphere's total mass is the same, but its density would be 1 kg/m

^{3}all the way through. It's height would be 10 km. As it is, we have 68% of the atmosphere contained in the first 10 km and 99% contained in the first 46 km.

In a future post, we'll see if we can determine the altitude at which the upper atmosphere becomes so thin that it is transparent to its own long-wave radiation.

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