Monthly Archives: January 2019

Viva Rocketry! Part 2

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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 nth stage houses the payload.

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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 (Viva Rocketry! Part 1.3)“) 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}

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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)

 

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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}

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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}

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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}}.

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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.

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Fig. 8

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

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

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Fig. 9

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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

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