<p>Ok, based on the comments above, this is where I am at... </p>

<ol>
<li>earth's radius is $6371 \text{ km}$</li>
<li>earth is a sphere</li>
<li>There is approximately $3 \cdot 10^{15}$ kg of CO2 in earth's atmosphere</li>
<li>The density of CO2 at 1atm is $\dfrac{1.977 \text{ kg}}{\text{ m}^3}$</li>
</ol>

<p>So, first, with assumption (3 and 4) we can compute the volume of CO2 in the atmosphere to be $\approx 1.52 \cdot 10^{15} \text{ m}^3$ or $1.52 \cdot 10^6 \text{ km}^3$.</p>

<p>So, now we can find the thickness of the layer by solving the equation</p>

<p>$\begin{align*}
1.52 \cdot 10^6 &amp;= \dfrac{4 \pi \cdot (6371 + x)^3}{3} - \dfrac{4 \pi (6371)^3}{3} \\
\end{align*}$</p>

<p>Where $x$ is the thickness of the layer of CO2 on earth's surface in km.</p>

<p>$\begin{align*}
1.52 \cdot 10^6 \text{ km}^3 &amp;= \dfrac{4 \pi \cdot (6371 + x)^3}{3} - \dfrac{4 \pi (6371)^3}{3} \\[.5pc]
\dfrac{3}{4\pi} \cdot 1.52 \cdot 10^6 \text{ km}^3 &amp;= (6371+x)^3 - 6371^3 \\
\dfrac{1.14}{\pi} \cdot 10^6 \text{ km}^3&amp;= (6371+x)^3 - 6371^3 \\
3.63 \cdot 10^5 +6371^3 \text{ km}^3 &amp;= (6371 + x)^3 \\
\sqrt[3]{3.63 \cdot 10^5 +6371^3 \text{ km}^3} &amp;= 6371 + x \\
6371.0029811 \text{ km} &amp;\approx 6371 + x \\
.0029811 \text{ km} &amp;\approx x
\end{align*}$</p>

<p>So, the layer of CO2 on earth's surface would be 2.98 meters thick...</p>

CC BY-SA 3.0
2016-06-15T13:00:37.810Z
That's about right. However, since you used a value of $3\cdot10^{15}$ kg (one significant digit), you should round that final result to 3 meters.
David Hammen
CC BY-SA 3.0
2016-06-15T13:02:36.007Z
An alternative approach is that the volume of a thin spherical shell of radius $r$ and height $h$ is approximately $4\pi r^2 h$, valid for $h\lll r$. You can see this by expanding $\frac43 \pi (r+h)^3 - \frac 43 \pi r^3 = \frac 43 \pi (3r^2h + 3rh^2 + h^3) = 4\pi r^2h(1+h/r+\frac13 (h/r)^2) \approx 4\pi r^2h$ for small $h$. Approximation is something 8th grade students should start learning. The calculation is much simpler with this approximation: $(1.52\times10^{15}\,\text{m}^3)/(4\pi\,(6371\,\text{km})^2) \approx 2.98\,\text{m}$
David Hammen
CC BY-SA 3.0
2016-06-15T13:11:51.697Z
Also note that the density value of 1.977 kg/m^3 is an empirical value that reflects the non-ideal nature of carbon dioxide. You'll get a slightly different value for the density of CO2 at standard temperature and pressure if you use the ideal gas law. I don't know if your students know about the ideal gas law. If they do, it might be a better learning experience to have them calculate that density value rather than taking it as a given. (They should arrive at 1.963 kg/m^3.)
David Hammen
CC BY-SA 3.0
2016-06-15T13:17:52.380Z
A couple of final comments: That value of $3\cdot 10^{15}$ kg is a bit outdated. Humanity has continued to pump CO2 into the atmosphere. The current value is about $3.16\cdot 10^{15}$ kg. Finally, as a teacher you should attach units to all of your calculations. Students regular make a mess of calculations because they get their units wrong. It is your job as a teacher to show them how to not make those mistakes.
David Hammen
CC BY-SA 3.0
2016-06-16T02:28:53.643Z
@DavidHammen yes, you are right about the units, Thanks for mentioning the gas law too.
Nick H
CC BY-SA 3.0
2016-06-15T02:42:38.300Z
2016-06-24T06:42:06.493Z
2016-06-24T06:42:06.493Z
Nick H
<p>A rough estimate is the portion of CO2 in the atmosphere (408 ppm) times the mean height (<a href="https://en.wikipedia.org/wiki/Atmosphere_of_Earth#Pressure_and_thickness" rel="nofollow noreferrer">5.6 km</a>) which gives 2.3 m.</p>

CC BY-SA 4.0
2021-01-11T19:14:10.123Z
Clever way to get a simple (albeit perhaps less accurate) answer!
Amphibio
CC BY-SA 4.0
2018-12-30T18:33:22.683Z
2018-12-30T18:33:22.683Z
Keith McClary
<p>I teach a Middle School (year 8) Geometry course and have a student attempting to answer this question. My assumptions:</p>

<ol>
<li>earth's radius is $3959 \text{ mi}$</li>
<li>earth is a sphere</li>
<li>There is approximately $3 \cdot 10^{15}$ kg of CO<sub>2</sub> in earth's atmosphere</li>
<li>The density of CO<sub>2</sub> at 1atm is $\dfrac{1562 \text{ kg}}{\text{cubic meter}}$</li>
<li>Conversion between cu meters and cu ft are $1:35.3147$. And cu ft to cu mi is $5280^3:1$</li>
</ol>

<p>So, first, with assumption (3 and 4) we can compute the volume of CO<sub>2</sub> in the atmosphere to be $\approx 1.92 \cdot 10^{12} \text{ m}^3$ or about $369 \text{ cu mi}$ of CO<sub>2</sub>.</p>

<p>So, now we can find the thickness of the layer by solving the equation</p>

<p>$\begin{align*}
369 &amp;= \dfrac{4 \pi \cdot (3959 + x)^3}{3} - \dfrac{4 \pi (3959)^3}{3} \\
\end{align*}$</p>

<p>Where $x$ is the thickness of the layer of CO<sub>2</sub> on earth's surface in miles.</p>

<p>$\begin{align*}
369 \cdot \dfrac{3}{4 \pi} &amp;= (3959+x)^3 - (3959)^3 \\[.5pc]
369 \cdot \dfrac{3}{4 \pi}+3959^3 &amp;= (3959+x)^3 \\[.5pc]
\sqrt[3]{369 \cdot \dfrac{3}{4 \pi}+3959^3} &amp;= 3959 + x \\[.5pc]
3959.0000018746 &amp;\approx 3959 + x \\[.5pc]
.00000018746 &amp;\approx x
\end{align*}$</p>

<p>So, the layer of CO<sub>2</sub> on earth's surface would be $\approx 1.87 \cdot 10^{-6} \text{ mi}$ or $.1187 \text{ in}$ in thickness. Which is less than half the thickness of an iPhone 6.</p>

<p><strong><em>My Question</em></strong>: This result seems incredibly small. Where did I go wrong? Or, is this reasonable. I have no intuition and would be interested in a second opinion!</p>

CC BY-SA 3.0
2016-06-14T10:02:29.040Z
Please note that [we use SI units here](http://meta.earthscience.stackexchange.com/q/318/100). If you must refer to quaint old units, please use the SI units as your primary units, and put the archaic stuff in parentheses afterwards. (oh, and you've missed the unit on the Earth's radius)
410 gone
CC BY-SA 3.0
2016-06-14T13:59:01.403Z
userLTK is right... your density of CO2 should be the same as the density of air at sea level... which is 3 orders of magnitude smaller than your number.
f.thorpe
CC BY-SA 3.0
2016-06-14T14:14:50.790Z
Another way to get a quick estimate, to check your answer, is to estimate the thickness of the Earth's atmosphere to some arbitrary level, say 100 mbars, The carbon dioxide fraction is currently about 404 ppm, so convert that to a fraction and multiply by the atmospheric thickness. Of course, CO2 is denser than air so you must make a density correction. You will find that the answer is massively more than you have calculated. Also, 'EnergyNumbers' is correct. The civilized world uses SI units!
Gordon Stanger
CC BY-SA 3.0
2016-06-15T02:23:29.047Z
@farrenthorpe I hadn't noticed that we took the density number from CO2 as a solid, but I guess that makes sense that a cubic meter wouldn't weigh 1562 kg as a gas! Thank you for pointing that out. Also, didn't realize that the SI units was a problem. Also, figured that this was the right place to ask the question and that someone might be interested in answer. But perhaps wrong spot?
Nick H
CC BY-SA 4.0
2018-12-30T19:47:30.667Z
I think it's appropriate to find the thickness of the CO as a solid. If it's a gas, how do you make it form a thin layer at the bottom of the atmosphere?
J Thomas
<atmosphere><atmosphere-modelling><homework>
Take all the CO2 in the atmosphere and put it into a layer on the surface of earth, how thick would the layer be?
Nick H
# Take all the CO2 in the atmosphere and put it into a layer on the surface of earth, how thick would the layer be?

(Asked by: username_0 on 2016-06-14)

I teach a Middle School (year 8) Geometry course and have a student attempting to answer this question. My assumptions:

1. earth's radius is $3959 \text{ mi}$
2. earth is a sphere
3. There is approximately $3 \cdot 10^{15}$ kg of CO2 in earth's atmosphere
4. The density of CO2 at 1atm is $\dfrac{1562 \text{ kg}}{\text{cubic meter}}$
5. Conversion between cu meters and cu ft are $1:35.3147$. And cu ft to cu mi is $5280^3:1$

So, first, with assumption (3 and 4) we can compute the volume of CO2 in the atmosphere to be $\approx 1.92 \cdot 10^{12} \text{ m}^3$ or about $369 \text{ cu mi}$ of CO2.

So, now we can find the thickness of the layer by solving the equation

$\begin{align\*}
369 &= \dfrac{4 \pi \cdot (3959 + x)^3}{3} - \dfrac{4 \pi (3959)^3}{3} \\
\end{align\*}$

Where $x$ is the thickness of the layer of CO2 on earth's surface in miles.

$\begin{align\*}
369 \cdot \dfrac{3}{4 \pi} &= (3959+x)^3 - (3959)^3 \\[.5pc]
369 \cdot \dfrac{3}{4 \pi}+3959^3 &= (3959+x)^3 \\[.5pc]
\sqrt[3]{369 \cdot \dfrac{3}{4 \pi}+3959^3} &= 3959 + x \\[.5pc]
3959.0000018746 &\approx 3959 + x \\[.5pc]
.00000018746 &\approx x
\end{align\*}$

So, the layer of CO2 on earth's surface would be $\approx 1.87 \cdot 10^{-6} \text{ mi}$ or $.1187 \text{ in}$ in thickness. Which is less than half the thickness of an iPhone 6.

***My Question***: This result seems incredibly small. Where did I go wrong? Or, is this reasonable. I have no intuition and would be interested in a second opinion!

**Tags:** atmosphere, atmosphere-modelling, homework

> **Comments:**
> **username_1**: Please note that [we use SI units here](http://meta.earthscience.stackexchange.com/q/318/100). If you must refer to quaint old units, please use the SI units as your primary units, and put the archaic stuff in parentheses afterwards. (oh, and you've missed the unit on the Earth's radius)  
> **username_2**: userLTK is right... your density of CO2 should be the same as the density of air at sea level... which is 3 orders of magnitude smaller than your number.  
> **username_3**: Another way to get a quick estimate, to check your answer, is to estimate the thickness of the Earth's atmosphere to some arbitrary level, say 100 mbars, The carbon dioxide fraction is currently about 404 ppm, so convert that to a fraction and multiply by the atmospheric thickness. Of course, CO2 is denser than air so you must make a density correction. You will find that the answer is massively more than you have calculated. Also, 'EnergyNumbers' is correct. The civilized world uses SI units!  
> **username_0**: @farrenthorpe I hadn't noticed that we took the density number from CO2 as a solid, but I guess that makes sense that a cubic meter wouldn't weigh 1562 kg as a gas! Thank you for pointing that out. Also, didn't realize that the SI units was a problem. Also, figured that this was the right place to ask the question and that someone might be interested in answer. But perhaps wrong spot?  
> **username_4**: I think it's appropriate to find the thickness of the CO as a solid. If it's a gas, how do you make it form a thin layer at the bottom of the atmosphere?  

---

## 2 answers

### Answer by username_0 on 2016-06-15. Score: 6

Ok, based on the comments above, this is where I am at...

1. earth's radius is $6371 \text{ km}$
2. earth is a sphere
3. There is approximately $3 \cdot 10^{15}$ kg of CO2 in earth's atmosphere
4. The density of CO2 at 1atm is $\dfrac{1.977 \text{ kg}}{\text{ m}^3}$

So, first, with assumption (3 and 4) we can compute the volume of CO2 in the atmosphere to be $\approx 1.52 \cdot 10^{15} \text{ m}^3$ or $1.52 \cdot 10^6 \text{ km}^3$.

So, now we can find the thickness of the layer by solving the equation

$\begin{align\*}
1.52 \cdot 10^6 &= \dfrac{4 \pi \cdot (6371 + x)^3}{3} - \dfrac{4 \pi (6371)^3}{3} \\
\end{align\*}$

Where $x$ is the thickness of the layer of CO2 on earth's surface in km.

$\begin{align\*}
1.52 \cdot 10^6 \text{ km}^3 &= \dfrac{4 \pi \cdot (6371 + x)^3}{3} - \dfrac{4 \pi (6371)^3}{3} \\[.5pc]
\dfrac{3}{4\pi} \cdot 1.52 \cdot 10^6 \text{ km}^3 &= (6371+x)^3 - 6371^3 \\
\dfrac{1.14}{\pi} \cdot 10^6 \text{ km}^3&= (6371+x)^3 - 6371^3 \\
3.63 \cdot 10^5 +6371^3 \text{ km}^3 &= (6371 + x)^3 \\
\sqrt[3]{3.63 \cdot 10^5 +6371^3 \text{ km}^3} &= 6371 + x \\
6371.0029811 \text{ km} &\approx 6371 + x \\
.0029811 \text{ km} &\approx x
\end{align\*}$

So, the layer of CO2 on earth's surface would be 2.98 meters thick...

> **Comments:**
> **username_5**: That's about right. However, since you used a value of $3\cdot10^{15}$ kg (one significant digit), you should round that final result to 3 meters.  
> **username_5**: An alternative approach is that the volume of a thin spherical shell of radius $r$ and height $h$ is approximately $4\pi r^2 h$, valid for $h\lll r$. You can see this by expanding $\frac43 \pi (r+h)^3 - \frac 43 \pi r^3 = \frac 43 \pi (3r^2h + 3rh^2 + h^3) = 4\pi r^2h(1+h/r+\frac13 (h/r)^2) \approx 4\pi r^2h$ for small $h$. Approximation is something 8th grade students should start learning. The calculation is much simpler with this approximation: $(1.52\times10^{15}\,\text{m}^3)/(4\pi\,(6371\,\text{km})^2) \approx 2.98\,\text{m}$  
> **username_5**: Also note that the density value of 1.977 kg/m^3 is an empirical value that reflects the non-ideal nature of carbon dioxide. You'll get a slightly different value for the density of CO2 at standard temperature and pressure if you use the ideal gas law. I don't know if your students know about the ideal gas law. If they do, it might be a better learning experience to have them calculate that density value rather than taking it as a given. (They should arrive at 1.963 kg/m^3.)  
> **username_5**: A couple of final comments: That value of $3\cdot 10^{15}$ kg is a bit outdated. Humanity has continued to pump CO2 into the atmosphere. The current value is about $3.16\cdot 10^{15}$ kg. Finally, as a teacher you should attach units to all of your calculations. Students regular make a mess of calculations because they get their units wrong. It is your job as a teacher to show them how to not make those mistakes.  
> **username_0**: @username_5 yes, you are right about the units, Thanks for mentioning the gas law too.  

---

### Answer by username_6 on 2018-12-30. Score: 1

A rough estimate is the portion of CO2 in the atmosphere (408 ppm) times the mean height ([5.6 km](https://en.wikipedia.org/wiki/Atmosphere_of_Earth#Pressure_and_thickness)) which gives 2.3 m.

> **Comments:**
> **username_7**: Clever way to get a simple (albeit perhaps less accurate) answer!  

---

