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, 0<e<1, P>0, 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.

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