# Escape Velocity Derivation

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### Escape Velocity Derivation

Escape velocity derivation in the context of Newtonian gravity is straightforward. In a gravitational field the escape velocity of an object is an indicator of the kinetic energy it needs to become gravitationally unbound from the system. If the initial energy could take the object to “infinity,” then it would indeed become unbound, and the kinetic energy required to do this would need to be at least equal to the absolute value of the potential energy. In other words

\[

\frac{1}{2}m v^{2} > \frac{GMm}{r}

\]

Note that all of this *assumes the object receives no additional energy other than it’s initial kinetic energy, and the gravitational potential energy*. In other words, the concept of escape velocity does not apply to, for example, powered flight.

Suppose that the mass trying to escape is $m$ and that there is only one other mass, $M$. In the case that $m \ll M$, the escape velocity is

\[

v = \sqrt{\left(\frac{2GM}{r}\right)}

\]

or

\[

v \sim 42 \left(\frac{M}{M_{\odot}}\right)^{\frac{1}{2}}

\left(\frac{1 {\rm AU}}{r}\right)^{\frac{1}{2}} \ \ {\rm km \ s^{-1}}

\]

or

\[

v \sim 11.2 \left(\frac{M}{M_{\rm E}}\right)^{\frac{1}{2}}

\left(\frac{r_{\rm E}}{r}\right)^{\frac{1}{2}} \ \ {\rm km \ s^{-1}}

\]

or

\[

v \sim 94,200 \left(\frac{M}{M_{\odot}}\right)^{\frac{1}{2}}

\left(\frac{1 {\rm AU}}{r}\right)^{\frac{1}{2}} \ \ {\rm miles \ per \ hour}

\]

or

\[

v \sim 25,000 \left(\frac{M}{M_{\rm E}}\right)^{\frac{1}{2}}

\left(\frac{r_{\rm E}}{r}\right)^{\frac{1}{2}} \ \ {\rm miles \ per \ hour}

\]

where 1 AU is 1 astronomical unit, and $M_{\odot}$ is a solar mass, $r_{\rm E}$ is the radius of the Earth, and $M_{\rm E}$ is the mass of the Earth.

Another convenient form (again for $m \ll M$) that expresses the escape velocity as a fraction of the speed of light, $c$, is

\[

\frac{v}{c} = \sqrt{ \frac{2r_{g}}{r} }

\]

where $r_{g}$ is the gravitational radius, for which convenient forms are

\[

\begin{align*}

r_{g} & = 1.4822 \times 10^{13}M_{8} \ {\rm cm} \\

\\

r_{g} & = 1.4822 (M/M_{\odot}) \ {\rm km} \\

\\

r_{g} & \sim M_{8} \ {\rm AU}, \\

\end{align*}

\]

where $M_{8}$ is the central mass in units of $10^{8}$ solar masses.

In the case that one of the masses is not negligible compared to the other, simply use the total mass ($M+m$) in place of $M$ in the above formulas for escape velocity.