Motion of Rocket in Outer Space/Proof 1
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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 equation of motion of $B$ is given by:
- $m \dfrac {\d \mathbf v} {\d t} = - \mathbf b \dfrac {\d m} {\d t}$
Proof
From Motion of Body with Variable Mass:
- $(1): \quad \mathbf w \dfrac {\d m} {\d t} + \mathbf F = m \dfrac {\d \mathbf v} {\d t}$
where:
- $\mathbf F$ is the external force being applied
- $\mathbf w$ is the velocity of the added mass relative to $B$.
In this scenario:
- there is no external force and so $\mathbf F = \mathbf 0$
- the velocity of the added mass relative to $B$ is $-\mathbf b$.
Thus $(1)$ becomes:
- $-\mathbf b \dfrac {\d m} {\d t} = m \dfrac {\d \mathbf v} {\d t}$
$\blacksquare$