Peak Performance: Falling of the earth

I was reading Nicholas Gurewitch’s comic “The Perry Bible Fellowship” when I stumbled upon this gem of a strip. Being the geek I am, I of course just had to figure out how tall such a mountain would need to be.

So, here we go. From Wikipedia we can define some stuff.

Ohh, almost forgot, I’m assuming the climber is at the equator and probably a bunch of other stuff.

The climber is “weightless” when the centripetal force and gravitational force are equal.
\begin{equation}
F=G\to\frac{m_p v_p^2}{r_p} =\gamma \frac{m_e m_p}{r_p^2}\to\frac{v_p^2}{r_p} =\gamma\frac{m_e}{r_p^2}\to v_p^2=\gamma \frac{m_e}{r_p}
\end{equation}

Inserting rotational speed of the mountain peek and solving for \(r_p\) yeilds
\begin{equation}
\left(r_p\frac{v_e}{r_e} \right)^2=\gamma \frac{m_e}{r_p}\to r_p^3=\gamma m_e \left(\frac{r_e}{v_e}\right)^2\to r_p=\sqrt[3]{\gamma m_e\left(\frac{r_e}{v_e}\right)^2}
\end{equation}

Inserting numbers we get
\begin{equation}
r_p=\sqrt[3]{(6.67428\cdot10^{-11}) (5.9736\cdot10^{24})\left(\frac{6,378,100}{1,674,400}\right)^2}=42,167,401.9\;\text{m}
\end{equation}

Now, finding the mountain height.
\begin{equation}
r_m=r_p-r_e=42,167,401.9-6,378,100=35,789,301.9\;\text{m}
\end{equation}

which is ~10% of the way up to the moon.

\begin{equation}
\frac{r_m}{r_{moon}} =\frac{35,789,301.9}{384,399,000}=0.093\approx10\%
\end{equation}

Kinda sweet, eh? I guess I can now say Q … E … D. :p