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 C02.

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!

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!

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
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 C02.

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!

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 C02.

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!