# Motion of Rocket in Outer Space/Proof 1

## Theorem

Let $B$ be a rocket travelling in outer space.

Let the velocity of $B$ at time $t$ be $\mathbf v$.

Let the mass of $B$ at time $t$ be $m$.

Let the exhaust velocity of $B$ be constant at $\mathbf b$.

Then the equation of motion of $B$ is given by:

$m \dfrac {\mathrm d \mathbf v} {\mathrm d t} = - \mathbf b \dfrac {\mathrm d m} {\mathrm d t}$

## Proof

$(1): \quad \mathbf w \dfrac {\mathrm d m} {\mathrm d t} + \mathbf F = m \dfrac {\mathrm d \mathbf v} {\mathrm d t}$

where:

$\mathbf F$ is the external force being applied
$\mathbf w$ is the velocity of the added mass relative to $B$.

In this scenario:

there is no external force and so $\mathbf F = \mathbf 0$
the velocity of the added mass relative to $B$ is $-\mathbf b$.

Thus $(1)$ becomes:

$-\mathbf b \dfrac {\mathrm d m} {\mathrm d t} = m \dfrac {\mathrm d \mathbf v} {\mathrm d t}$

$\blacksquare$