# Motion of Rocket in Outer Space/Proof 2

## 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:

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

## Proof

$\mathbf F = \dfrac \d {\d t} \paren {m \mathbf v}$

At time $t + \Delta t$, let:

the mass of $B$ be $m + \Delta m$
the velocity of $B$ be $\mathbf v + \Delta \mathbf v$.

The fuel is being consumed, so the increase in mass of the fuel during time $\Delta t$ is $-\Delta m$.

Thus the exhaust products, therefore of mass $-\Delta m$, are expelled at a velocity $-\mathbf b$ relative to $B$.

Thus this material is actually moving at a velocity $\mathbf v - \mathbf b$.

By Conservation of Momentum, the total linear momentum is constant.

Thus:

 $\displaystyle m \mathbf v$ $=$ $\displaystyle \paren {m + \Delta m} \paren {\mathbf v + \Delta \mathbf v} + \paren {-\Delta m} \paren {\mathbf v - \mathbf b}$ $\displaystyle$ $=$ $\displaystyle m \mathbf v + m \Delta \mathbf v + \paren {\Delta m} \mathbf v + \Delta m \Delta \mathbf v - \paren {-\Delta m} \mathbf v + \Delta m \mathbf b$ $\displaystyle \leadsto \ \$ $\displaystyle m \Delta \mathbf v$ $=$ $\displaystyle - \Delta m \paren {\mathbf b + \Delta \mathbf v}$ $\displaystyle \leadsto \ \$ $\displaystyle m \frac {\Delta \mathbf v} {\Delta t}$ $=$ $\displaystyle - \frac {\Delta m} {\Delta t} \paren {\mathbf b + \Delta \mathbf v}$ $\displaystyle \leadsto \ \$ $\displaystyle m \frac {\d \mathbf v} {\d t}$ $=$ $\displaystyle - \mathbf b \frac {\d m} {\d t}$ letting $\Delta t \to 0$

$\blacksquare$