- #1
IsakVern
- 8
- 0
If you derive the equation for orbital velocity you get
\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?
\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?