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!
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...