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:

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

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