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