Escape Velocity in terms of Gravitational Constant

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Theorem

Let $P$ be a planet.

Let $P$ have:

a mass of $M$
a radius of $R$.


Then the escape velocity of $P$ is given by:

$V = \sqrt {\dfrac {2 M G} R}$

where $G$ is the gravitational constant.


Proof

Let $b$ be a particle with mass $m$.

Let $K_i$ be the kinetic energy of $b$ at the surface of $P$.

Let $U_i$ be the potential energy of $b$ at the surface of $P$.

Let $K_f$ be the limit of the kinetic energy of $b$ as it leaves $P$.

Let $U_f$ be the limit of the potential energy of $b$ as it leaves $P$.

Let $v_f$ be the limit of the speed of $b$ as it leaves $P$.

Let $r_f$ be the limit of the distance of $b$ from the center of $P$ as $b$ leaves $P$.

Let $v_e$ be the speed of $b$ at the surface of $P$.


The speed of $b$ becomes arbitarily small as it leaves $P$.

Thus:

$v_f = 0$

The distance of $b$ to the center of $P$ becomes arbitarily large as it leaves $P$.

Thus:

$r_f = \infty$


By Kinetic Energy of Motion:

\(\ds K_i\) \(=\) \(\ds \frac 1 2 m {v_e}^2\)
\(\ds K_f\) \(=\) \(\ds \frac 1 2 m {v_f}^2\)
\(\ds \) \(=\) \(\ds \frac 1 2 m \paren 0^2\)
\(\ds \) \(=\) \(\ds 0\)


By Gravitational Potential Energy of Position for Bodies Far From Center:

\(\ds U_i\) \(=\) \(\ds -\frac {G M m} R\)
\(\ds U_f\) \(=\) \(\ds 0\)


Now:

\(\ds K_i + U_i\) \(=\) \(\ds K_f + U_f\) Principle of Conservation of Energy
\(\ds \frac 1 2 m {v_e}^2 - \frac {G M m} R\) \(=\) \(\ds 0\) substitution
\(\ds v_e\) \(=\) \(\ds \sqrt {\frac {2 G M} R}\) after algebra

$\blacksquare$


Sources

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