# Déjà vu!

The stages of a two stage rocket have initial masses $m_1$ and $m_2$ respectively and carry a payload of mass $P$. Both stages have equal structure factors and equal relative exhaust speeds. If the rocket mass, $m_1+m_2$, is fixed, show that the condition for maximal final speed is

$m_2^2 + P m_2 = P m_1$.

Find the optimal ratio $\frac{m_1}{m_2}$ when $\frac{P}{m_1+m_2} = b$.

According to multi-stage rocket’s flight equation (see “Viva Rocketry! Part 2“), the final speed of a two stage rocket is

$v = -c \log(1-\frac{e \cdot m_1}{m_1 + m_2 +P})- c\log(1-\frac{e \cdot m_2}{m_2+P})$

Let $m_0 = m_1 + m_2$, we have

$m_1 = m_0-m_2$

and,

$v = -c \log(1-\frac{e \cdot (m_0-m_2)}{m_0 +P})- c\log(1-\frac{e \cdot m_2}{m_2+P})$

Differentiate $v$ with respect to $m_2$ gives

$v' = \frac{c(e-1)e(2m_2 P-m_0 P +m_2^2)}{(P+m_2)(P-e m_2+m_2)(P+e m_2-e m_0+m_0)}= \frac{c(e-1)e(2m_2-m_0 P+m_2^2)}{(P+m_2)(P+(1-e)m_2)(P+e m_2 + (1-e)m_0)}\quad\quad\quad(1)$

It follows that $v'=0$ implies

$2m_2 P-m_0 P + m_2^2 =0$.

That is, $2 m_2 P - (m_1+m_2) P + m_2^2 = (m_2-m_1) P + m_2^2=0$. i.e.,

$m_2^2 + P m_2 = P m_1\quad\quad\quad(2)$

It is the condition for an extreme value of $v$. Specifically, the condition to attain a maximum (see Exercise-2)

When $\frac{P}{m_1+m_2} = b$, solving

$\begin{cases} (m_2-m_1)P+m_2^2=0\\ \frac{P}{m_1+m_2} = b\end{cases}$

yields two pairs:

$\begin{cases} m_1=\frac{(\sqrt{b}\sqrt{b+1}+b+1)P}{b}\\m_2= -\frac{\sqrt{b^2+b}P+bP}{b}\end{cases}$

and

$\begin{cases} m_1= - \frac{(\sqrt{b}\sqrt{b+1}-b-1)P}{b}\\ m_2=\frac{\sqrt{b^2+b}P-bP}{b}\end{cases}\quad\quad\quad(3)$

Only (3) is valid (see Exercise-1)

Hence

$\frac{m_1}{m_2} = -\frac{(\sqrt{b}\sqrt{b+1}-b-1)P}{\sqrt{b^2+b}P-b P} = \frac{\sqrt{b+1}}{\sqrt{b}} = \sqrt{1+\frac{1}{b}}$

The entire process is captured in Fig. 2.

Fig. 2

Exercise-1 Given $b>0, 00$, prove:

1. $- \frac{(\sqrt{b}\sqrt{b+1}-b-1)P}{b}>0$
2. $\frac{\sqrt{b^2+b}P-bP}{b}>0$

Exercise-2 From (1), prove the extreme value attained under (2) is a maximum.