# Viva Rocketry! Part 2

Fig. 1

A rocket with $n$ stages is a composition of $n$ single stage rocket (see Fig. 1) Each stage has its own casing, instruments and fuel. The $n$th stage houses the payload.

Fig. 2

The model is illustrated in Fig. 2, the $i^{th}$ stage having initial total mass $m_i$ and containing fuel $\epsilon_i m_i (0 < \epsilon_i <1, 1 \leq i \leq n)$. The exhaust speed of the $i^{th}$ stage is $c_i$. The flight of multi-stage rocket starts with the $1^{st}$ stage fires its engine and the rocket is lifted. When all the fuel in the $1^{st}$ stage has been burnt, the $1^{st}$ stage’s casing and instruments are detached. The remaining stages of the rocket continue the flight with $2^{nd}$ stage’s engine ignited. Generally, the rocket starts its $i^{th}$ stage of flight with final velocity achieved at the end of previous stage of flight. The entire rocket is propelled by the fuel in the $i^{th}$ casing of the rocket. When all the fuel for this stage has been burnt, the $i^{th}$ casing is separated  from the rest of the stages. The flight of the rocket is completed if $i=n$. Otherwise, it enters the next stage of flight. When all external forces are omitted, the governing equation of rocket’s $i^{th}$ stage flight (see “Viva Rocketry! Part 1” or “An alternate derivation of ideal rocket’s flight equation“) is

$0 = m_i(t) \frac{dv_i(t)}{dt} + c_i \frac{dm_i(t)}{dt}$

It can be written as

$\frac{dv_i(t)}{dt} = -\frac{c_i}{m_i(t)}\frac{dm_i(t)}{dt}\quad\quad\quad(1)$

Integrate (1) from $t_0$ to $t$,

$\int\limits_{t_0}^{t}\frac{dv_i(t)}{dt}\;dt = -c_i \int\limits_{t_0}^{t}\frac{1}{m_i(t)}\frac{dm_(t)}{dt}\;dt$

gives

$v_i(t) - v_i(t_0) = -c_i(\log(m_i(t))-\log(m_i(t_0)))=-c_i\log(\frac{m_i(t)}{m_i(t_0)})$.

At $t=t^*$ when $i^{th}$ stage’s fuel has been burnt, we have

$v_i(t^*) - v_i(t_0) = -c_i\log(\frac{m_i(t^*)}{m_i(t_0)})\quad\quad\quad(2)$

where

$m_i(t_0) = \sum\limits_{k=i}^{n} m_k +P$

and,

$m_i(t^*) = \sum\limits_{k=i}^n m_k - \epsilon_i m_i + P$.

Let $v^*_i = v_i(t^*), \; v^*_{i-1}$ the velocity of rocket at the end of ${i-1}^{th}$ stage of flight. Since $v_i(t_0) = v^*_{i-1}$, (2) becomes

$v^*_i - v^*_{i-1} = -c_i \log(\frac{ \sum\limits_{k=i}^n m_k - \epsilon_i m_i+P}{\sum\limits_{k = i}^{n} m_k + P}) = -c_i\log(1-\frac{\epsilon_i m_i}{\sum\limits_{k = i}^{n} m_k + P})$

i.e.,

$v^*_i = v^*_{i-1}-c_i\log(1-\frac{\epsilon_i m_i}{\sum\limits_{k = i}^{n} m_k + P}), \quad\quad 1\leq i \leq n, v_0=0\quad\quad\quad(3)$

For a single stage rocket ($n=1$), (3) is

$v_1^*=-c_1\log(1-\frac{\epsilon_1}{1+\beta})\quad\quad\quad(4)$

In my previous post “Viva Rocketry! Part 1“, it shows that given $c_1=3.0\;km\;s^{-1}, \epsilon_1=0.8$ and $\beta=\frac{1}{100}$, (4) yields $4.7 \;km\;s^{-1}$, a value far below $7.8\;km\;s^{-1}$, the required speed of an earth orbiting  satellite.

But is there a value of $\beta$ that will enable the single stage rocket to produce the speed a satellite needs? Let’s find out. Differentiate (4) with respect to $\beta$ gives

$\frac{dv_1^*}{d\beta} = - \frac{c_1 \epsilon_1}{(\beta+1)^2 (1-\frac{\epsilon_1}{\beta+1})} = - \frac{c_1\epsilon_1}{(\beta+1)^2(\frac{1-\epsilon_1+\beta}{\beta+1})} < 0$

since $c_1, \beta$ are positive quantities and $0< \epsilon_1 < 1$. It means $v_1^*$ is a monotonically decreasing function of $\beta$. Moreover,

$\lim\limits_{\beta \rightarrow 0}v_1^*=\lim\limits_{\beta \rightarrow 0}-c_1\log(1-\frac{\epsilon_1}{1+\beta}) = - c_1 \log(1-\epsilon_1)\quad\quad\quad(5)$

Given $c_1=3.0\;km\;s^{-1}, \epsilon_1=0.8$, (5) yields approximately

$4.8\;km\;s^{-1}$

Fig. 3

This upper limit implies that for the given $c_1$ and $\epsilon_1$, no value of $\beta$ will produce a speed beyond (see Fig. 4)
Let’s now turn to a two stage rocket ($n=2$) From (3), we have

$v_2^* = -c_1\log(1-\frac{\epsilon_1 m_1}{m_1+m_2+P}) - c_2\log(1-\frac{\epsilon_2 m_2}{m_2+P})\quad\quad\quad(6)$

If $c_1=c_2=c, \epsilon_1 = \epsilon_2 = \epsilon, m_1=m_2$ and $\frac{P}{m_1+m_2} = \beta$, then

$P = \beta (m_1+m_2) = 2m_1\beta = 2m_2\beta$.

Consequently,

$v_2^*=-c \log(1-\frac{\epsilon}{2(1+\beta)}) - c\log(1-\frac{\epsilon}{1+2\beta})\quad\quad\quad(7)$

When $c=3.0\;km\;s^{-1}, \epsilon=0.8$ and $\beta = \frac{1}{100}$,

$v_2^* \approx 6.1\;km\;s^{-1}$

Fig. 5

This is a considerable improvement over the single stage rocket ($v^*=4.7\; km\;s^{-1}$). Nevertheless, it is still short of producing the orbiting speed a satellite needs. In fact,

$\frac{dv_2^*}{d\beta} = -\frac{2c\epsilon}{(2\beta+2)^2(1-\frac{\epsilon}{2\epsilon+2})}-\frac{2c\epsilon}{(2\beta+1)^2(1-\frac{\epsilon}{2\beta+1})}= -\frac{2c\epsilon}{(2\beta+2)^2\frac{2\beta+2-\epsilon}{2\beta+2}}-\frac{2c\epsilon}{(2\beta+1)^2\frac{2\beta+1-\epsilon}{2\beta+1}} < 0$

indicates that $v_2^*$ is a monotonically decreasing function of $\beta$. In addition,

$\lim\limits_{\beta \rightarrow 0} v_2^*=\lim\limits_{\beta \rightarrow 0 }-c\log(1-\frac{\epsilon}{2+2\beta})-c\log(1-\frac{\epsilon}{1+2\beta})=-c\log(1-\frac{\epsilon}{2})-c\log(1-\epsilon)$.

Therefore, there is an upper limit to the speed a two stage rocket can produce. When $c=3.0\;km\;s^{-1}, \epsilon=0.8$, the limit is approximately

$6.4\;km\;s^{-1}$

Fig. 6

In the value used above, we have taken equal stage masses, $m_1 = m_2$. i.e., the ratio of $m_1 : m_2 = 1 : 1$. Is there a better choice for the ratio of $m_1:m_2$ such that a better $v_2^*$ can be obtained? To answer this question, let $\frac{m_1}{m_2} = \alpha$, we have

$m_1 = \alpha m_2\quad\quad\quad(8)$

Since $P = \beta (m_1+m_2)$, by (8),

$P = \beta (\alpha m_2 + m_2)\quad\quad\quad(9)$

Substituting (8), (9) into (6), $v_2^* = -c\log(1-\frac{\epsilon \alpha m_2}{\alpha m_2+m_2+\beta(\alpha m_2+m_2))}) - c\log(1-\frac{\epsilon m_2}{m_2 + \beta(\alpha m_2 + m_2)})$ $= -c\log(1-\frac{\epsilon \alpha}{\alpha+1+\beta(\alpha+1)})-c\log(1-\frac{\epsilon}{1+\beta(\alpha+1)})$ $= c\log\frac{(\alpha+1)(\beta+1)((\alpha+1)\beta+1)}{(1-\epsilon+(\alpha+1)\beta)((1-\epsilon)\alpha + (\alpha+1)\beta+1)}\quad\quad\quad(10)$ a function of $\alpha$. It can be written as

$v_2^*(\alpha) = c\log(w(\alpha))$

where

$w(\alpha) = \frac{(\alpha+1)(\beta+1)((\alpha+1)\beta+1)}{(1-\epsilon+(\alpha+1)\beta)((1-\epsilon)\alpha + (\alpha+1)\beta+1)}$

This is a composite function of $\log$ and $w$. Since $\log$ is a monotonic increasing function (see “Introducing Lady L“),

$(v_2^*)_{max} = c\log(w_{max})$

Here, $(v_2^*)_{max}, w_{max}$ denote the maximum of $v_2^*$ and $w$ respectively. To find $w_{max}$, we differentiate $w$,

$\frac{dw}{d\alpha} = \frac{(\beta+1)(\alpha^2\beta-\beta-1)(\epsilon-1)\epsilon}{(\epsilon-\alpha \beta-\beta-1)^2(\alpha\epsilon-\alpha \beta-\beta-\alpha-1)^2}=\frac{(\beta+1)(\alpha^2\beta-\beta-1)(\epsilon-1)\epsilon}{(1-\epsilon+\alpha \beta+\beta)^2(\alpha(1-\epsilon)+\alpha \beta+\beta)^2}\quad\quad\quad(11)$

Solving $\frac{dw}{d\alpha} = 0$ for $\alpha$ gives

$\alpha = - \sqrt{1+\frac{1}{\beta}}$ or $\sqrt{1+\frac{1}{\beta}}$.

Fig. 7

By (8), the valid solution is

$\alpha = \sqrt{1+\frac{1}{\beta}}$.

It shows that $w$ attains an extreme value at $\sqrt{1+\frac{1}{\beta}}$. Moreover, we observe from (11) that

$\alpha < \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} > 0$

and

$\alpha > \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} < 0$.

i.e., $w$ attains maximum at $\alpha=\sqrt{1+\frac{1}{\beta}}$. It follows that

$v_2^*$ attains maximum at $\alpha = \sqrt{1+\frac{1}{\beta}}$.

Therefore, to maximize the final speed given to the satellite, we must choose the ratio

$\frac{m_1}{m_2} = \sqrt{1+\frac{1}{\beta}}$.

With $\beta =\frac{1}{100}$, the optimum ratio $\frac{m_1}{m_2}=10.05$, showing that the first stage must be about ten times large than the second. Using this ratio and keep $\epsilon=0.8, c=3.0\;km\;s^{-1}$ as before, (10) now gives

$v_2 = 7.65\;km\;s^{-1}$,

a value very close to the required one.

Fig. 8

Setting $\beta = \frac{1}{128}$, we reach the goal:

$v_2^* = 7.8\;km\;s^{-1}$

Fig. 9

Fig. 10

At last, it is shown mathematically that provided the stage mass ratios ($\frac{P}{m_1+m_2}$ and $\frac{m_1}{m_2}$)are suitably chosen, a two stage rocket can indeed launch satellites into earth orbit.
Exercise 1. Show that $\alpha < \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} > 0$ and $\alpha > \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} < 0$. Exercise 2. Using the optimum $\frac{m_1}{m_2} = \sqrt{1+\frac{1}{\beta}}$ and $\epsilon=0.8, c=3.0\;km\;s^{-1}$, solving (10) numerically for $\beta$ such that $v_2^* = 7.8$.