# 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.

Peak Performance by Nicholas Gurewitch

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.

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}

Inserting rotational speed of the mountain peek and solving for $$r_p$$ yeilds

\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}

Inserting numbers we get

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}

Now, finding the mountain height.

r_m=r_p-r_e=42,167,401.9-6,378,100=35,789,301.9\;\text{m}

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

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

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