# Calculating the Mass of the Atmosphere (using a triple integral)

I have been given the task of calulating the mass of the atmosphere given the following (not quite correct) information. It is intended to be a thought experiment for an undergraduate course in atmospheric science. I recognize that I have chosen a difficult way of doing this but I believe it should still work, and I am confused as to where I may be going wrong in my reasoning.

Density of Atmosphere at sea level: $$3\times10^{19}$$ $$\frac{molecules}{m^3}$$.

Now, taking the composition of the atmosphere to be approximately 78.1% N2 (28.014 g/mol), 20.9% 02 (31.998 g/mol), and 0.9% Ar (39.948 g/mol), we get an average molecular weight of 28.926 g/mol.

Converting the density (Avogadro's and average molecular weight from above), we get a density of 0.00144 $$\frac{g}{m^3}$$.

Now I am assuming that the density of the atmosphere decreases exponentially proportionate to height. That is, $$d(h)=e^{\lambda h}$$ and calculated lambda using my density and the radius of the earth, $$d(6000000)=0.00144=e^{6000000\lambda}$$. Solving for the proportionality constant we get $$\lambda = -1.09\times10^{-6}$$. Note, prior to writing this question I had been using my original given density to calculate this proportionality constant and that was not valid. However, this is also wrong.

Taking the 'top' of the atmosphere to be 500 km above the Earth, and plugging this into a triple integral we get:

$$\int_0^{2\pi} \int_0^\pi \int_{6000000}^{6500000} r^2e^{(-1.09\times10^{-6}r)}sin\phi\, dr\, d\phi\, d\theta$$

I have been using Wolfram Alpha to compute, but have not been able to finagle an answer that is even close to correct.

Is there anything glaringly wrong with what I have done thus far? I am deeply suspicious of my treatment of the density function, and also of my units. Intuitively, this feels like a valid (albeit complicated) approach to the problem. I am not very interested in a simpler approach, but rather want to know where I have gone wrong. My calculus is also quite rusty.

This is my first StackExchange post, so thank you beforehand for any help.

• I wanted to come back and point out that the number density of air at sea level (which was given to me) is 6 orders of magnitude off. This is apparent in the 0.00144 g/m^3 calulated mass density (again, 6 orders of magnitude off). Commented Jan 19, 2019 at 19:49

Two issues that I see:

1. You calculated $$\lambda$$ incorrectly
2. You can simplify the integral

First, the Earth is not made up of gas. The exponential drop in density applies to distances above the Earth. If the density of the atmosphere is $$0.00144 \frac{g}{m^3}$$ at sea level, this implies:

$$Ce^{\lambda 0} = 0.00144$$

not

$$Ce^{\lambda 6,000,000} = 0.0014$$

Unfortunately, this means that you need one more data point (at some distance above sea level) to calculate $$\lambda$$ (since below sea level we have water/rock, not gas).

Second, there is a symmetry about $$r$$ if we treat the Earth as a sphere. You can take advantage of this to integrate the mass of the atmosphere using "shells" of radius $$6000000 + h$$ ($$h$$ is height above sea level). This reduces a triple integral to a single integral:

$$M_{atm} = \int_{6,000,000}^{6,500,000} Ce^{\lambda h}4\pi (6000000+h)^2dh$$

This is a realtively easy one for WolframAlpha...just need more data to get $$\lambda$$.

• Thanks! I knew it would be something silly. You are correct, the Earth is not made up of gas. Do you have any suggestions for how to go about calculating a correct value for lambda? I have been sitting here scratching my head over it. Does that invalidate my entire approach? I cannot think of a way to cause the exponential function to 'decay starting at 6,000,000. You mention a second data point but I do not see how that would help the situation. This is intended to be an extremely rough approximation, I can invent an approximate data point. Commented Jan 18, 2019 at 3:35
• @NLowe yes, it invalidates your approach. Why not set the density at 1000km above sea level to that of interplanetary space (5 atoms Hydrogen per cubic cm)? The equation to solve will be $0.00144e^{\lambda 1000000}=d_{space}$ Commented Jan 18, 2019 at 4:02