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{\bf Question}

An extremely well constructed rocket has mass ratio (initial to
final mass) of 10.  A new fuel is developed that has an exhaust
velocity of $4500$ms$^{-1}$.  The fuel burns at a constant rate
for 300s.  Calculate the maximum velocity of this single stage
rocket assuming constant acceleration due to gravity.  If the
escape velocity of a particle from the earth is $11.2$kms$^{-1}$,
can this rocket reach the moon?

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{\bf Answer}

PICTURE

By Newton's 2nd law:
\begin{eqnarray*}
-m(t)g & = & {\rm rate\ of\ change\ of\ momentum} \\ & = & m(t)
\frac{dv}{dt} + u \frac{dm}{dt} \\  \frac{dv}{dt} & = & - g -
\frac{u}{m} \frac{dm}{dt} \\ {\rm Integrating\ w.r.t.}\ t:\ \ \int
dv & = & \int -g \, dt - \int \frac{u}{m} \, dm \\ v & = & -gt - u
\ln |m| + c \\ & & {\rm using\ the\ fact\ that\ }t = 0, v = 0, m =
m_0 \\ \Rightarrow c & = & u \ln |m_0| \\ {\rm therefore\ \ } v(t)
& = & -gt + u\ln|m_0| - u \ln|m| \\ v(t) & = & -gt + u \ln
\frac{m_0}{m(t)} \\  \end{eqnarray*}

In this case $m(t) = m_0 - \alpha t$

Where the burn rate $\alpha = \frac{0.9 m_0}{r}$, and where $r =
300$s

We must first check that it lifts off.

i.e. Thrust $ =  u \alpha = 4.5 \times 10^3 \times
\frac{9.0m_0}{300} = 13.3m_0{\rm ms}^{-1} \geq m_0g$

(where $g$ = 9.81 ms$^{-2}$). Therefore the rocket does lift off.

Now $v(0) = 0.$  Are there any maxima of $v(t)$ on the range $0<t<
r$?

$\ds \frac{dv}{dt} = -g + \frac{\alpha n}{m(t)} = g \left[
\frac{\alpha n}{m(t)g} - 1 \right] > 0$

as $\alpha u > m_0g > m(t) g$ from the lift-off condition above.

Hence there are no internal maxima, so the maximum velocity is at
$t = r;$

$v_{\rm max} = -g 30 + u \ln 10 = 7400ms^{-1}$.

Thus the final velocity is less than escape velocity of
11.2kms$^{-1}$ and so the rocket cannot reach the moon.



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