# Escape Velocity of Projectile fired Upwards/Proof 1

## Theorem

Let $P$ be a planet.

Let $P$ have an Acceleration Due to Gravity at its surface of $g$.

Let $P$ have a radius of $R$.

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

- $V = \sqrt {2 g R}$

## Proof

Let a projectile $B$ of mass $m$ be fired vertically upwards from the surface of $P$ at such a speed that it escapes the gravitational field of $P$ completely.

$F$ be the force exerted on the projectile by the gravitational field of $P$.

Let $x$ be the distance of $B$ from the surface of $P$ at time $t$.

We have:

- $F = -\dfrac k {x^2}$

where $k$ can be calculated at the surface of $P$ as:

\(\ds F\) | \(=\) | \(\ds m g\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac k{R^2}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds k\) | \(=\) | \(\ds R^2 m g\) |

Thus:

- $F = -\dfrac {m g R^2} {x^2}$

Let the $B$ be travelling away from $P$ with speed $v$ at time $t$.

Thus:

\(\ds F\) | \(=\) | \(\ds -m \frac {\d v} {\d t}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds -\frac {m g R^2} {x^2}\) | \(=\) | \(\ds m v \frac {\d v} {\d x}\) |

So:

- $\displaystyle -\int \frac {R^2 g} {x^2} \rd x = \int v \rd v$

This gives:

- $\dfrac {R^2 g} x = \dfrac {v^2} 2 + C$

Now when $x = R$, $v = v_0$, the launch velocity, which leads to:

- $C = R g - \dfrac {v_0^2} 2$

and so:

- $\dfrac {v^2} 2 = \dfrac {R^2 g} x - R g + \dfrac {v_0^2} 2$

For $\dfrac {v^2} 2 > 0$ we need:

- $\dfrac {v_0^2} 2 > R g \paren {1 - \dfrac R x}$

whatever $x$ may be.

So when $x \to \infty$ we have:

- $\dfrac {v_0^2} 2 > R g$

Hence:

- $v_0 > \sqrt {2 R g}$

$\blacksquare$