# 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?

Gold Member
2022 Award
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
$$\frac{m v_{\text{orb}}^2}{\tilde{R}}=\frac{G M}{\tilde{R}^2} \; \Rightarrow\; v_{\text{orb}}=\sqrt{\frac{G M}{\tilde{R}}}.$$

• Ibix
Homework Helper
Gold Member
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}##.

• vanhees71 and Ibix
2022 Award
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.

• vanhees71 and PeroK
$$\frac{v^2}{r} = \frac{k}{r^2}$$
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