# Acceleration of Rocket in Outer Space

## 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 acceleration of $B$ at time $t$ is given by:

$\mathbf a = \dfrac 1 m \left({- \mathbf b \dfrac {\mathrm d m} {\mathrm d t} }\right)$

## Proof

From Motion of Rocket in Outer Space, the equation of motion of $B$ is given by:

$(1): \quad m \dfrac {\mathrm d \mathbf v} {\mathrm d t} = - \mathbf b \dfrac {\mathrm d m} {\mathrm d t}$

By definition, the acceleration of $B$ is its rate of change of velocity:

$\mathbf a = \dfrac {\mathrm d \mathbf v} {\mathrm d t}$

The result follows by substituting $\mathbf a$ for $\dfrac {\mathrm d \mathbf v} {\mathrm d t}$ and dividing by $m$.

$\blacksquare$