- #1

IsakVern

- 8

- 0

\begin{equation}

v_{orbit} = \sqrt{\frac{GM}{R}}

\end{equation}

and for escape velocity you get

\begin{equation}

v_{escape} = \sqrt{\frac{2GM}{R}}=\sqrt{2}\,v_{orbit}

\end{equation}

I'm wondering if there is a logical/geometrical explanation for why the escape velocity is exactly square root 2 times the orbital velocity, or if this is just an algebraic coincidence?