# Acceleration of Rocket in Outer Space

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## 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:

- $\mathbf a = \dfrac 1 m \paren {-\mathbf b \dfrac {\d m} {\d t} }$

## Proof

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

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

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

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

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

$\blacksquare$

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.8$: Rocket Propulsion in Outer Space