# Homework Help: Deriving/Calculating the Mass of Earth's atmosphere

1. Sep 21, 2011

### guitarstorm

1. The problem statement, all variables and given/known data

Consider an isothermal atmosphere with temperature $T_{0}$. Show that the mass of this atmosphere on a planet with the same radius at earth, a, and the same gravity, g, is greater than $\frac{4\pi a^{2}p_{s}}{g}$.

If the scale height of this isothermal atmosphere is 7 km, by how much does the atmospheric mass exceed the value given in the text on page 27?

2. Relevant equations

$m_{a}=\frac{4\pi a^{2}p_{s}}{g}$

$H=\frac{RT_{0}}{g}$

$\rho (z)=\frac{p_{s}}{RT_{0}}e^{-z/H}$

3. The attempt at a solution

Basically, our professor wants us to disprove the atmospheric mass formula given in our textbook. He said in lecture that the actual mass is greater because the method in the textbook doesn't account for the mass between the columns due to the nature of spherical geometry.

I'm trying two different methods.

First, I thought maybe multiplying by the solid angle would account for this missed mass, and when multiplying $\frac{4\pi a^{2}p_{s}}{g}$ by $\omega =\frac{4\pi a^{2}}{a^{2}}$, I get $\frac{16\pi^{2} a^{2}p_{s}}{g}$.

Plugging in a = 6.37x10^6 m, $p_{s}$=100000 Pa, and g=9.81 m/s^2, I get a mass of 6.53x10^19 kg, greater than the 5.20x10^18 kg obtained by the original equation... However I'm not sure if my logic is quite right here.

The other method would be to use a shell method instead... But my calculus skills are rusty and I'm not sure how to go about this.

I know that to get the volume you just integrate $4\pi a^{2}$ to get $\frac{4}{3}\pi a^{3}$.

Volume multiplied by density would give the mass... However, I believe this has to be integrated over a certain height to account for the depth of the atmosphere... This is where I'm unsure of how to proceed.

Last edited: Sep 21, 2011
2. Sep 21, 2011

### guitarstorm

I've attempted to set up the integral to calculate the volume of the entire atmosphere...

$\int_{\theta =0}^{2\pi }\int_{\phi =0}^{2\pi}\int_{a}^{\infty }4\pi a^{2}\sin \phi\, da\, d\phi\, d\theta$

Are my limits of integration for a correct? What's throwing me off is that we're integrating from Earth's surface to the top of the atmosphere, so do we take it from a=the radius of the Earth to a=infinity, or a=0 to a=infinity?

After this, I think I have to integrate the density equation to account for the decrease of rho with height... Does it matter if I do this separately or combined with the volume equation?

This is quite frustrating because I can visualize what I have to do, but I don't know how to represent it mathematically/physically...

3. Sep 21, 2011

### guitarstorm

Running through the integration in my previous post didn't work out, as it became zero when I integrated phi and evaluated it from 0 to 2*pi... Guess I'll take another stab at this question tomorrow.

4. Sep 22, 2011

### D H

Staff Emeritus
They are not correct. Longitude (your theta) does vary from 0 to 2*pi, but latitude runs from -pi/2 to pi/2. That sin phi term looks like you are using colatitude instead of latitude, so the range is 0 to pi.

5. Sep 22, 2011

### D H

Staff Emeritus
Your professor is correct in that that formula for mass isn't exactly correct.

It is too big.

$\frac{4\pi \, (6378\,\text{km})^2 \, (1\,\text{bar})} {9.80665\,\text{m/s}^2} \approx 5.213\,\times 10^{18}\,\text{kg}$

The mass of the atmosphere is about $5.1480\,\times 10^{18}\,\text{kg}$. Source: Trenberth, Kevin E., Lesley Smith, 2005: The Mass of the Atmosphere: A Constraint on Global Analyses. J. Climate, 18, 864–875.

Last edited: Sep 22, 2011
6. Sep 22, 2011

### guitarstorm

No, actually he said the mass is too small given that equation... He showed an image where you integrate in cylindrical columns from the surface to the top of the atmosphere, and due to the curvature of the Earth, they diverge as you get higher up so that there are spaces in between the columns... Thus extra mass that is unaccounted for, so it has to be greater than that.

7. Sep 22, 2011

### guitarstorm

When I run through the integration, I simply get what I had... So apparently I didn't even need to integrate to get the volume, and just assume $V=\frac{4\pi a^{3}}{3}$.

But then I have the density distribution, which I believe I have to integrate from zero to infinity?

8. Sep 22, 2011

### dynamicsolo

I believe that your source is calculating the mass by multiplying the observed surface atmospheric pressure by the surface area of Earth. That is why this number offers the opposite conclusion to the one the instructor wishes to draw. The stated problem is for an isothermal atmosphere (which, of course, is not the real atmosphere).

9. Sep 22, 2011

### dynamicsolo

You have omitted the given density function, which also goes into your integrand in order to obtain the isothermal atmospheric mass. I have corrected the "latitude" integration limits, but what you have written is just a volume integral; you would get an infinite result if you actually calculate with that.

The infinitesimal mass at every radius is the density at that radius times the surface area at that radius times the infinitesimal radius, so you have the right idea, but you left out the factor $\rho (z)=\frac{p_{s}}{RT_{0}}e^{-z/H}$ *...

* z here is why you are calling "a" in your integral

10. Sep 22, 2011

### dynamicsolo

See above for the rest. You want to integrate from a = RE, the Earth's radius, to infinity. What it looks like you've gotten with your volume integral is the result of integrating from zero to a , the sphere's radius, so naturally you got the volume of a sphere.

(It would avoid potential confusion if you chose a to be the Earth's radius -- or even RE -- and used r for the radial coordinate in your integral.)

11. Sep 22, 2011

### guitarstorm

I set up the integral as $\int_{a}^{\infty }4\pi a^{2}\rho\, da$, but when I evaluated it, it came out to zero.

I spoke to my professor and he said my error is in the density function I plugged in...

I was plugging it in as $\rho =\frac{p_{s}}{Hg}e^{-a/H}$, while the textbook has it as $\rho =\frac{p_{s}}{Hg}e^{-z/H}$.

I figured z is the same thing as a, since we're using a in this case to represent the radius of the Earth, but then he said let z=a-r, at which point I got even more confused, because now there's 4 different terms to represent length, distance, etc?!

12. Sep 22, 2011

### dynamicsolo

This is why I suggested using "r" as the radial coordinate in your integral. (And I missed it earlier today, but you didn't need to put $4 \pi$ in your triple integral, because the "latitude" and "longitude" integrals are what give you the $2 · 2 \pi$ . If you are using a as Earth's radius, the integral would be

$$4\pi \cdot \int_{a}^{\infty} r^{2} \rho(z) dr .$$

Your professor is correct that the variables are mismatched, so we should use z = r - a (z being altitude above Earth's mean radius), which makes the integral

$$4\pi \cdot \int_{a}^{\infty} r^{2} \cdot \frac{p_{s}}{gH} \cdot e^{-(r-a)/H} dr$$

$$= \frac{4\pi p_{s}}{gH} \cdot \int_{a}^{\infty} r^{2} \cdot e^{-r/H} \cdot e^{a/H} dr = \frac{4\pi p_{s}}{gH} \cdot e^{a/H} \cdot \int_{a}^{\infty} r^{2} \cdot e^{-r/H} dr .$$

You have a (Type I) improper integral of a product of a power-law function times an "exponential decay" function, which will give a positive number. I see that the temperature T0 was substituted out, but you can always put it back in at the end if your result is supposed to be a function of the (uniform) atmospheric temperature. (I see, though, that your instructor wants you to plug in H = 7000 m. And you know that R in the scale height expression is the "universal gas constant", right? We've had radius mentioned so much, it would be easy to get confused...)

13. Sep 23, 2011

### guitarstorm

Yeah, I used the scale height formula $H=\frac{RT_{0}}{g}$ and rearranged to plug in for $T_{0}$ because he didn't give a specific temperature value.

The only problem I see arising is that even with the previous integral I evaluated, I couldn't get an answer because when I tried to evaluate $e^{-a/H}$, plugging in 6.37x10^6 for a and 7000 for H, my calculator gave me zero. And if I try to evaluate the positive one, it says overflow... But I think that's because the numbers are too large/small for my calculator to handle. I just plugged it into Wolfram Alpha and got actual answers, a number times 10^-396 and 10^395, so I guess I should be all set now.

Thanks for the assistance!

14. Sep 23, 2011

### dynamicsolo

Reading through this again, I think your instructor has misled himself as to the cause of the discrepancy. What the expression for the mass of the atmosphere in the book is based on is that the weight force of the atmosphere applied to the surface of the Earth is estimated as the surface atmospheric pressure times the surface area of Earth (this is reasonable, since F = P · A ). So the weight of the atmosphere is estimated by $p_{s} \cdot 4\pi \cdot a^{2}$. If the acceleration due to gravity is taken as a constant g over the entire height of the atmosphere, then since W = mg , the mass of the atmosphere is estimated as $\frac{4\pi a^{2} p_{s}}{g}$. (And that is what D_H was remarking about earlier, but the difference between the two numbers he gives is small and could be due to choices of values for a , ps, and g , or the (unstated) model used by the quoted source.)

The local value of g decreases with altitude, though, so the "effective value" of g in this formula actually becomes some few percent smaller over the "height" of the atmosphere. If one simply "plugs in" 9.81 m/sec2 , that formula will underestimate the atmospheric mass by a few percent.

Of course, a more serious discrepancy will arise from the fact that the temperature and chemical composition of the atmosphere are not constant, with temperature dropping over the height of the troposphere (above that, the temperature rise in the stratosphere is less important because the density of gas is already rather small). So a model based on an isothermal atmosphere gives too large a scale height and suggests that the density is falling more gradually than it actually does, since H is smaller for lower temperatures at the same mean molecular weight of gas. (I believe H = 7 km. is an "averaged value", but is not really a constant with altitude.)

Last edited: Sep 23, 2011
15. Sep 23, 2011

### dynamicsolo

That's fine -- although you can back-figure what is being used for T0...

Yes, e(a/H) is a huge number, but the improper integral is going to make the nightmare go away:

we need to integrate this "by parts", which gives

$$\int_{a}^{\infty} r^{2} \cdot e^{-r/H} dr = -H \cdot [ r^{2}e^{-r/H} + 2Hre^{-r/H} + 2H^{2}e^{-r/H} ] \vert_{a}^{\infty} = -H \cdot [ 0 - ( a^{2}e^{-a/H} + 2Hae^{-a/H} + 2H^{2}e^{-a/H} ) ]$$

$$= He^{-a/H} \cdot ( a^{2} + 2Ha + 2H^{2} ) ,$$

where the limits at infinity of all the terms in the anti-derivative are zero (and I hope the mathematicians on PF don't fuss at me too much -- I'm more a physicist and write the upper evaluation limit on the anti-derivative as "infinity" as a shorthand for taking the limit...)

The expression for the isothermal atmospheric mass is then

$$\frac{4\pi p_{s}}{gH} \cdot e^{a/H} \cdot \int_{a}^{\infty} r^{2} \cdot e^{-r/H} dr = \frac{4\pi p_{s}}{gH} \cdot e^{a/H} \cdot He^{-a/H} \cdot ( a^{2} + 2Ha + 2H^{2} )$$

$$= \frac{4\pi p_{s}}{g} \cdot ( a^{2} + 2Ha + 2H^{2} ) .$$

All of the terms in parentheses are positive, so we indeed see that the isothermal atmospheric mass is larger than $\frac{4\pi p_{s}}{g} a^{2} ,$ though only slightly...

Last edited: Sep 23, 2011