# Velocity 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 velocity of $B$ at time $t$ is given by:

- $\map {\mathbf v} t = \map {\mathbf v} 0 + \mathbf b \ln \dfrac {\map m 0} {\map m t}$

where $\map {\mathbf v} 0$ and $\map m 0$ are the velocity and mass of $B$ at time $t = 0$.

## Proof

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

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

Hence:

\(\displaystyle \int_0^t \dfrac {\d \mathbf v} {\d t} \rd t\) | \(=\) | \(\displaystyle -\int_0^t \mathbf b \frac 1 m \dfrac {\d m} {\d t} \rd t\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \int_{t \mathop = 0}^t \rd \mathbf v\) | \(=\) | \(\displaystyle -\int_{t \mathop = 0}^t \mathbf b \dfrac {\d m} m\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \bigintlimits {\mathbf v} 0 t\) | \(=\) | \(\displaystyle -\mathbf b \bigintlimits {\ln m} 0 t\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {\mathbf v} t - \map {\mathbf v} 0\) | \(=\) | \(\displaystyle -\mathbf b \paren {\map {\ln m} t - \map {\ln m} 0}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {\mathbf v} t\) | \(=\) | \(\displaystyle \map {\mathbf v} 0 + \mathbf b \ln \dfrac {\map m 0} {\map m t}\) |

$\blacksquare$

## Sources

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