# Mass of the earth's atmosphere given the density as a function of h

1. Feb 19, 2014

### swashbuckler77

1. The problem statement, all variables and given/known data
An exponential model for the density of the earth’s atmosphere says that if the temperature of the atmosphere were constant, then the density of the atmosphere as a function of height, h (in meters), above the surface of the earth would be given by δ(h) = 1.28e(−0.000124h) kg/m3.

Write and evaluate a sum that approximates the mass of the portion of the atmosphere from h = 0 to h = 100 m (i.e., the ﬁrst 100 meters above sea level). Assume the radius of the earth is 6400 km.

2. Relevant equations
Since this is like a density distribution problem but in three dimensions, I surmised that now, instead of using mass=2∏∫r*δ(r) dr from a to b (the circumference of a circle and the formula provided in the textbook's example problem), I will use mass=4∏∫h2*δ(h) dh, which is the surface area of a sphere with density as a function δ(h) distributed over 6400000 to 6400100, because going from circumference of a circle to surface area of a sphere seems like a logical transition from 2D to 3D.

3. The attempt at a solution
mass=4∏∫h2*δ(h) dh from 6400000 to 6400100

=4∏∫h2*1.28e(−0.000124h) dh from 6400000 to 6400100

=5.12∏∫h2*e(−0.000124h) dh from 6400000 to 6400100

I'm unsure if I am going in the right direction because this looks like a very ugly integral with a lot of zeroes that make me rather uncomfortable.

Last edited: Feb 20, 2014
2. Feb 20, 2014

### maajdl

It is not a very hugly integral.
It is easy to do by integration by parts.
However, check that the variation of density over 100 meters is very small.
For a physics course, you could as well assume it is constant or it varies linearly (Taylor or interpolation).
Anyway, if you want to see the effect of the exponential dependence over only 100m, you need to care for all the digits.

I checked that:

- the error is 1% if you assume constant density
- the error is 0.005% by Taylor expansion
- the error is 0.001% by linear interpolation