Why is the escape velocity equal to the square root of 2 times the orbital velocity?

[IsakVern]

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?

 

[vanhees71]

To escape to infinity the total energy of the body must at least be $0$, i.e.,
$$\frac{m}{2} v^2-\frac{G M m}{R} \geq 0,$$
where $R$ is the radius of the Earth, $M$ its mass, and $G$ Newton's gravitational constant; $v$ is the initial velocity which gets
$$v \geq v_{\text{escape}}=\sqrt{\frac{2 G M}{R}}=\sqrt{2gR},$$
where $g=M G/R^2$ is the gravitational acceleration at the surface of the earth.

What you call $v_{\text{orbit}}$ is the velocity of a body in a circular orbit at radius $\tilde{R}$ (I call it $\tilde{R}$ to distinguish it from the radius of the Earth). Then setting the centripetal force equal to the gravitational force indeed gives
$$\frac{m v_{\text{orb}}^2}{\tilde{R}}=\frac{G M}{\tilde{R}^2} \; \Rightarrow\; v_{\text{orb}}=\sqrt{\frac{G M}{\tilde{R}}}.$$

 

[kuruman]

It's an algebraic necessity, not coincidence, if you want to know the speed required so that an object leaves the Earth's surface and reaches infinity with zero kinetic energy. The total energy for an orbiting satellite is $E =KE+PE=-\frac{1}{2}PE$. To have escape velocity from the surface you need the total energy to be zero. That means doubling the kinetic energy of the bound orbit. To do that you must double $v^2$ which means multiplying $v$ by $\sqrt{2}$.

 

[Ibix]

Just to expand slightly on @vanhees71's maths, remember that both cases involve the same basic values - $G$, $M$, $R$ and $m$, and you are combining them to make a velocity. Dimensional analysis tells you immediately that the results have to be the same to within a numerical factor.

 

[PeroK]

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?

The algebraic "coincidence" is that when you set the centripetal force equal to the central force, you get something of the form:
$$\frac{v^2}{r} = \frac{k}{r^2}$$
The $r$ cancels on the denominator, which is similar to integrating the force to get the potential. You'll get a factor of $n -1$ for any force of the form $1/r^n$. In this case $n = 2$, so the factor is $1$.

Then, for the escape velocity you are setting $\frac 1 2 v^2$ equal to the potential which is where the factor of $\sqrt 2$ comes in.