# Boosting rocket flight performance without calculus (Viva Rocketry! Part 2.1)

Fig. 1

Given

$v = c\log(\frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1})$

where $b>0, 00, a>0$, maximize $v$ with appropriate $a$.

The above optimization problem is solved using calculus (see “Viva Rocketry! Part 2“). However, there is an alternative that requires only high school mathematics with the help of a Computer Algebra System (CAS). This non-calculus approach places more emphasis on problem solving through mathematical thinking, as all symbolic calculations are carried out by the CAS (e.g., see Fig. 2). It also makes a range of interesting problems readily tackled with minimum mathematical prerequisites.

The fact that

$\log$ is a monotonic increasing function $\implies v_{max} = c\log(w_{max})$

where

$w = \frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)}$

or

$w(1-e+(a+1)b)((1-e)a+(a+1)b+1) - (a+1)(b+1)((a+1)b+1)=0\quad\quad\quad(1)$

(1) can be written as

$A_1 a^2 + B_1 a +C_1= 0$

where

$A_1 = -bew+b^2w+bw-b^2-b$

$B_1 = e^2w-2bew-2ew+2b^2w+3bw+w-2b^2-3b-1$,

$C_1 = -bew-ew+b^2w+2bw+w-b^2-2b-1$.

Since $A_1 = 0$ means

$-b e w + b^2 w + b w - b^2 - b =0$.

That is

$w = -\frac{b+1}{e-b-1}$.

Solve

$\frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = -\frac{b+1}{e-b-1}$

for $a$ gives $a = 0 \implies A_1 \neq 0$ if $a > 0$.

Hence, (1) is a quadratic equation. For it to have solution, its discriminant $B_1^2-4A_1C_1$ must be nonnegative, i.e.,

$(e^2-2be-2e+b+1)^2 w^2-2(b+1)(2be^2+e^2-2be-2e+b+1) w +(b+1)^2 \geq 0\quad(2)$

Consider

$(e^2-2be-2e+b+1)^2 w^2-2(b+1)(2be^2+e^2-2be-2e+b+1)w+(b+1)^2 = 0\quad(3)$

If $e^2-2be-2e+b+1 \neq 0$, (3) is a quadratic equation.

Solving (3) yields two solutions

$w_1 = -\frac{(b+1)(2\sqrt{b(b+1)}e^2-2be^2-e^2-2\sqrt{b(b+1)}e+2be+2e-b-1)}{(e^2-2be-2e+b+1)^2}$,

$w_2=\frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}$.

Since $0 < e < 1$,

$w_1 - w_2 = -\frac{4(b+1)\sqrt{b(b+1)}(e-1)e}{(e^2-2be-2e+b+1)^2} > 0\quad(4)$

(4) implies

$w_1 > w_2$

and, the solution to (2) is

$w \leq w_2$ or $w \ge w_1$

i.e.,

$w \leq \frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}\quad\quad\quad(4)$

or

$w \ge -\frac{(b+1)(2\sqrt{b(b+1)}e^2-2be^2-e^2-2\sqrt{b(b+1)}e+2be+2e-b-1)}{(e^2-2be-2e+b+1)^2}\quad\quad\quad(5)$

We prove that (4) is true by showing (5) is false:

Consider $w - w_1= 0$:

$\frac{(b+1)(e-1) e \cdot f(a)}{(1-e+ab+b)(a(1-e)+ab+b+1)(e^2-2be-2e+b+1)^2} = 0\quad\quad\quad(6)$

where

$f(a) = 2a\sqrt{b(b+1)}e^2+a^2be^2+be^2+e^2-2a^2b\sqrt{b(b+1)}$

$-4ab\sqrt{b(b+1)}e - 2b\sqrt{b(b+1)}e - 4a\sqrt{b(b+1)}e$

$-2\sqrt{b(b+1)}e-2a^2b^2e-4ab^2e-2a^2be-4abe-4be$

$-2e+2a^2b^2\sqrt{b(b+1)}+4ab^2\sqrt{b(b+1)} + 2b^2\sqrt{b(b+1)} +2a^2b\sqrt{b(b+1)}$

$+6ab\sqrt{b(b+1)} + 4b\sqrt{b(b+1)} + 2a\sqrt{b(b+1)} + 2\sqrt{b(b+1)}$

$+2a^2b^3 + 4ab^3 + 2b^3 +3a^2b^2 + 8ab^2 +5b^2+a^2b+4ab+4b+1$.

It can be written as

$A_2a^2 + B_2a + C_2\quad\quad\quad(7)$

where

$A_2 = be^2-2b\sqrt{b(b+1)}e-2b^2e-2be+2b^2\sqrt{b(b+1)}+2b\sqrt{b(b+1)}$

$+2b^3+3b^2+b$,

$B_2 = 2\sqrt{b(b+1)}e^2-4b\sqrt{b(b+1)}e -4\sqrt{b(b+1)}e$

$-4b^2e-4be+4b^2\sqrt{b(b+1)}+6b\sqrt{b(b+1)}+2\sqrt{b(b+1)}+4b^3+8b^2+4b$,

$C_2=be^2+e^2-2b\sqrt{b(b+1)}e-2\sqrt{b(b+1)}e-2b^2e-4be$

$-2e+2b^2\sqrt{b(b+1)}+4b\sqrt{b(b+1)}+2\sqrt{b(b+1)}+2b^3+5b^2+4b+1$.

Since $A_2 > 0$ (see Exercise 1) and,

solve (7) for $a$ yields

$a = -\sqrt{1+\frac{1}{b}}$.

It follows that for $a > 0, f(a) > 0$.

Consequently, $w-w_1$ is a negative quantity. i.e.,

$w-w1 < 0$

which tells that (5) is false.

Hence, when $e^2-2be-2e+b+1 \neq 0$, the global maximum $w_{max}$ is $w_2$.

Solving $w = w_2$ for $a$:

$\frac{(a+1)(b+1)((a+1)b+1}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = \frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}$,

we have

$a = \sqrt{1+ \frac{1}{b}}$.

Therefore,

$e^2-2be-2e+b+1 \neq 0 \implies w$ attains maximum at $a = \sqrt{1+ \frac{1}{b}}$.

In fact, $w$ attains maxima at $a = \sqrt{1+\frac{1}{b}}$ even when $e^2-2be-2e+b+1 = 0$, as shown below:

Solving $e^2-2be-2e+b+1 = 0$ for $e$, we have

$e_1= -\sqrt{b(b+1)}+b+1$ or $e_2 = \sqrt{b(b+1)} + b + 1$.

Only $e_1$ is valid (see Exercise-2),

When $e = e_1$,

$w(\sqrt{1+\frac{1}{b}} )- w(a) = - \frac{(b+1)g(a)}{4 \sqrt{b(b+1)} (\sqrt{b(b+1)}+ab) (a\sqrt{b(b+1)}+b+1)}\quad(8)$

where

$g(a) = (2a^2b+4ab+2b+2a+2)\sqrt{b(b+1)}-2a^2b^2-4ab^2-2b^2-a^2b-4ab-3b-1$

Solve quadratic equation $g(a) = 0$ for $a$ yields

$a = \sqrt{1+\frac{1}{b}}$.

The coefficient of $a^2$ in $g(a)$ is $2b\sqrt{b(b+1)}-2b^2-b$, a negative quantity (see Exercise-3).

The implication is that $g(a)$ is a negative quantity when $a \neq \sqrt{1 + \frac{1}{b}}$.

Hence, (8) is a positive quantity, i.e.,

$e^2-2be-2e+b+1 = 0, a \neq \sqrt{1+\frac{1}{b}} \implies w(\sqrt{1+\frac{1}{b}})-w(a) > 0$

We therefore conclude

$\forall 0 < e < 1, b > 0, w$ attains its maximum at $a = \sqrt{1+\frac{1}{b}}$.

Fig. 2

Exercise-1 Prove:$00 \implies$

$be^2-2b\sqrt{b(b+1)}e-2b^2e-2be+2b^2\sqrt{b(b+1)}+2b\sqrt{b(b+1)}+2b^3+3b^2+b > 0$

Exercise-2 Prove: $b > 0 \implies 0 <-\sqrt{b(b+1)} + b +1 <1$

Exercise-3 Prove: $b > 0 \implies 2b\sqrt{b(b+1)}-2b^2-b < 0$

# Prelude to Taylor’s theorem

As an application of derivative, we may prove the Binomial theorem that concerns the expansion of $(1+x)^n$ as a polynomial. Namely,

$(1+x)^n = 1+ a_1 x + a_2 x^2 + ... + a_n x^n\quad\quad\quad(1)$

where $a_k = \frac{n(n-1)(n-2)...(n-k+1)}{k!}, 1 \leq k \leq n, n \in N$.

There are two steps:

Step 1) Prove $(1+x)^n$ can be expressed as a polynomial $1+a_1 x + a_2 x^2 + ... + a_n x^n$, i.e.,

$(1+x)^n = 1 + \sum\limits_{i=1}^{n}a_i x^i$

where $a_k$s are constants.

Step 2) Show that

$a_k = \frac{n(n-1)(n-2)...(n-k+1)}{k!}$

We use mathematical induction first.

When $n = 1$,

$(1+x)^1 = 1+x = 1 +a_1 x$

where $a_1 = 1$.

Assume when $n=k, (1+x)^k$is a polynomial:

$(1+x)^k = 1+b_1 x + b_2 x + ... +b_k x^k\quad\quad\quad(2)$

When $n = k+1$,

$(1+x)^{k+1} = (1+x)^k (1+x)$.

By (2), it is

$(1+b_1 x+b_2 x^2+ ...+ b_k x^k) (1+x)$

$= (1+\sum\limits_{i=1}^{k} b_i x^i)(1+x)$

$= 1+ x + \sum\limits_{i=1}^{k}b_i x^i (1+x)$

$= 1+x + \sum\limits_{i=1}^{k}(b_i x^i +b_i x^{i+1})$

$= 1+x +\sum\limits_{i=1}^{k}b_i x^i + \sum\limits_{i=1}^{k} b_i x^{i+1}$

$= 1+x + (b_1 x + b_2 x^2 +... +b_k x^k) + (b_1 x^2 + ... + b_{k-1} x^k + b_k x^{k+1})$

$= 1+ (b_1+1)x + (b_2 + b_1) x^2 + ... + (b_k + b_{k-1}) x^k + b_k x^{k+1}$

$= 1 +a_1 x + a_2 x^2 + ... + a_k x^k + a_{k+1} x^{k+1}$

where $a_1 = b_1+1, a_2=b_2+b_1, ..., a_k = b_k + b_{k-1}, a_{k+1} = b_k$.

Once (1) is established, we proceed to step 2) to construct $a_k$:

From (1),

$\frac{d^k}{dx^k}(1+x)^n = \frac{d^k}{dx^k}(1+a_1 x + a_2 x^2 + ... + a_k x^k + ... +a_n x^n)$.

That is

$n(n-1)(n-2)...(n-k+1)(1+x)^{n-k} = k(k-1)(k-2)...1 \cdot {a_k} + \sum\limits_{i=1}^{n-k} c_i x^i\quad\quad\quad(3)$

where $c_i$s are constants.

Let $x = 0$, (3) becomes

$n(n-1)(n-2)...(n-k+1) \cdot 1= k(k-1)(k-2)...1 \cdot {a_k} + 0$

i.e.,

$n(n-1)(n-2)...(n-k+1) = k!\;a_k\quad\quad\quad(4)$

Solving (4) for $a_k$ gives

$a_k = \frac{n(n-1)(n-2)...(n-k+1)}{k!}$.

Since

$\frac{n(n-1)(n-2)...(n-k+1)}{k!} = \frac{n(n-1)(n-k+1)\boxed{(n-k)(n-k-1)...1}}{\boxed{(n-k)(n-k-1)...1}\;k!}=\frac{n!}{(n-k)!\;k!}$,

$a_k$ is often expressed as

$a_k = \frac{n!}{(n-k)!\;k!}$

# Piece of Pi

A while back, we deemed that it is utterly impractical to calculate the value of $\pi$ using the partial sum of Leibniz’s series due to its slow convergence (see “Pumpkin Pi“)

Fig. 1

As illustrated in Fig. 1, in order to determine each additional correct digit of $\pi$, the number of terms in the summation must increase by a factor of 10.

What we need is a fast converging series whose partial sum yields given number of correct digits with far fewer terms.

Looking back, we see that the origin of Leibniz’s series is the definite integral

$\frac{\pi}{4} = \int\limits_{0}^{1}\frac{1}{1+x^2} dx\quad\quad\quad(1)$

To find the needed new series, we consider a variation of (1), namely,

$\frac{\pi}{6} = \int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+x^2} dx\quad\quad\quad(2)$

Given the fact (see “Pumpkin Pi“) that

$\frac{1}{1+x^2} = \sum\limits_{k=1}^{n}(-1)^{k+1}x^{2k-2}+\frac{(-1)^n x^{2n}}{1+x^2}\quad\quad\quad(3)$

We proceed to integrate (3) with respect to $x$ from $0$ to $\frac{1}{\sqrt{3}}$,

$\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+x^2} dx=\int\limits_{0}^{\frac{1}{\sqrt{3}}}\sum\limits_{k=1}^{n}(-1)^{k+1}x^{2k-2} dx + \int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{(-1)^n x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\int\limits_{0}^{\frac{1}{\sqrt{3}}}x^{2k-2} dx + (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\frac{x^{2k-1}}{2k-1}\bigg|_{0}^{\frac{1}{\sqrt{3}}}+ (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

As result, (2) becomes

$\frac{\pi}{6}=\sum\limits_{k=1}^{n}(-1)^{k+1}\frac{(\frac{1}{\sqrt{3}})^{2k-1}}{2k-1}+(-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\frac{{(\frac{1}{3}})^k}{(2k-1)\frac{1}{\sqrt{3}}} + (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

$= \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)} + (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

i.e.,

$\frac{\pi}{6} - \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)} = (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx\quad\quad\quad(4)$

By (4),

$|\frac{\pi}{6} - \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}| = |(-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx|=\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx<\int\limits_{0}^{\frac{1}{\sqrt{3}}}x^{2n}dx$

$=\frac{x^{2n+1}}{2n+1}\bigg|_0^{\frac{1}{\sqrt{3}}}$

$=\frac{1}{3^n \sqrt{3} (2n+1)}$

which gives

$|\sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)} - \frac{\pi}{6}| < \frac{1}{3^n \sqrt{3} (2n+1)}$.

And so

$-\frac{1}{3^n \sqrt{3} (2n+1)}<\sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}-\frac{\pi}{6}<\frac{1}{3^n \sqrt{3} (2n+1)}\quad\quad\quad(5)$

Since $\lim\limits_{n\rightarrow \infty}\frac{1}{3^n \sqrt{3} (2n+1)}=0$, (5) implies

$\lim\limits_{n\rightarrow \infty} \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}-\frac{\pi}{6}= 0$.

Hence,

$\lim\limits_{n\rightarrow \infty} \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}=\frac{\pi}{6}$.

It follows that the value of $\pi$ can be approximated by the partial sum of a new series

$6\sqrt{3}\sum\limits_{k=1}^{\infty}\frac{(-1)^{k+1}}{3^k(2k-1)}$

Let’s compute it with Omega CAS Explorer (see Fig. 2, 3)

Fig. 2

Fig. 2 shows the series converges quickly. The sum of the first 10 terms yields the first 6 digits!

Fig. 3

Totaling the first 100 terms of the series gives the first 49 digits of $\pi$ (see Fig. 3)

Exercise 1. Show that $\lim\limits_{n\rightarrow \infty}\frac{1}{3^n \sqrt{3} (2n+1)}=0$.

Exercise 2. Can we use $\frac{\pi}{3} = \int\limits_{0}^{\sqrt{3}}\frac{1}{1+x^2}dx$ to compute the value of $\pi$ in a similar fashion? Explain.

# 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 (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

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

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 $\beta$.

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

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. 7

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. 8

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. 9

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

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

Fig. 10

Fig. 11

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 $v_2^* = 7.8$.

# When rocket ejects its propellant at a variable rate (Viva Rocketry! Part 1.4)

A rocket is programmed to burn and ejects its propellant at the variable rate $\alpha \cdot k \cdot e^{-kt}$, where $k$ and $\alpha$ are positive constants. The rocket is launched vertically from rest. Neglecting all external forces except gravity, show that the final speed given to the payload, of mass $P$, when all the fuel has been burnt is

$v= -c \log(1-{\frac{\epsilon m_0}{m_0+P}}) + {\frac{g}{k}\log(1-{\frac{\epsilon m_0}{\alpha}}})$.

Here $c$ is the speed of the propellant relative to the rocket, $m_0$ the initial rocket mass, excluding the payload. The initial fuel mass is $\epsilon m_0$.

From my previous post “An alternative derivation of rocket’s flight equation (Viva Rocketry! Part 1.3)“, we know in our present context,

$\frac{dm}{dt} = - \; \alpha k e^{-kt}\quad\quad\quad(1)$

Integrate (1) from $0$ to $t$,

$m(t)-m(0) = \int\limits_{0}^{t}{ -\alpha k e^{-kt}} dt = \alpha e^{-kt}\bigg|_{0}^{t}=\alpha e^{-kt}-\alpha$

Since $m(0) = m_0 + P$,

$m(t) = m_0 + P -\alpha +\alpha e^{-kt}$.

The rocket’s flight equation now is

$-mg = m \frac{dv}{dt} + c\cdot (-\alpha k e^{-kt})$

i.e.,

$\frac{dv}{dt} = \frac{c\alpha k e^{-kt}}{m_0+P-\alpha+\alpha e^{-kt}} -g\quad\quad\quad(2)$

When all the fuel has been burnt at time $t^*$,

$m(t^*) = (1-\epsilon) m_0 + P$.

That is:

$m_0 + P - \alpha + \alpha e^{-kt^*} = (1-\epsilon) m_0 + P\quad\quad\quad(3)$.

Solve (3) for $t^*$,

$t^* = -\frac{1}{k}\log(1-\frac{\epsilon m_0}{\alpha})$

Integrate (2) from $0$ to $t^*$, we have

$v(t^*) - v(0) = -c \log(m_0+P-\alpha + \alpha e^{-kt}) \bigg|_{0}^{t^*}- gt\bigg|_{0}^{t^*}$

Since $v(0) = 0$,

$v(t^*) = -c \log(\frac{m_0 + P -\alpha + \alpha e^{(-k) \cdot ({-\frac{1}{k}\log(1-\frac{\epsilon m_0}{\alpha}))}}} {m_0 + P -\alpha + \alpha e^{-k\cdot 0}}) - g\cdot( -\frac{1}{k}\log(1-\frac{\epsilon m_0}{\alpha}) - 0)$

$= -c \log(\frac{m_0+P-\alpha +\alpha (1-\frac{\epsilon m_0}{\alpha})}{m_0+P}) + \frac{g}{k}\log(1-\frac{\epsilon m_0}{\alpha})$

$= -c \log(\frac{m_0+P-{\epsilon m_0})}{m_0+P}) + \frac{g}{k}\log(1-\frac{\epsilon m_0}{\alpha})$

gives the final speed

$v(t^*) = -c \log(1-\frac{\epsilon m_0}{m_0+P}) + \frac{g}{k}\log(1-\frac{\epsilon m_0}{\alpha})$

Exercise 1. Using Omega CAS Explorer, solve $(1), (2), (3)$ for $m(t), v(t), t^*$ respectively.

Exercise 2. Before firing, a single stage rocket has total mass $m_0$, which comprises the casing, instruments etc, with mass $m_c$, and the fuel. The fuel is programmed to burn and to be ejected at a variable rate such that the total mass of the rocket $m(t)$ at any time $t$, during which the fuel is being burnt, is given by

$m(t) = m_0 e^{\frac{-kt}{m_0}}$

where $k$ is a constant.

The rocket is launched vertically from rest. Neglect all external forces except gravity, show that the height $h$ attained at the instant the fuel is fully consumed is

$h = \frac{1}{2}(\frac{m_0}{k} \cdot \log{\frac{m_0}{m_c}})^2(\frac{ck}{m_0}-g)$

$c$ being the exhaust speed relative to the rocket.

# An alternative derivation of ideal rocket’s flight equation (Viva Rocketry! Part 1.3)

I will derive the ideal rocket’s flight equation differently than what is shown in “Viva Rocketry! Part 1

Let

$\Delta m$ –  the mass of the propellant

$m$ – the mass of the rocket at time $t$

$v$ – the speed of the rocket and $\Delta m$ at time $t$

$u$ – the speed of the ejected propellant, relative to the rocket

$p_1$ – the magnitude of $\Delta m$‘s momentum

$p_2$ – the magnitude of rocket’s momentum

For the propellant:

$\Delta p_1 = \Delta m \cdot (\boxed {v +\Delta v -u} ) - \Delta m \cdot v$

$=\Delta m\cdot v + \Delta m\cdot \Delta v -\Delta m\cdot u - \Delta m \cdot v$

$= -\Delta m \cdot u + \Delta m \cdot \Delta v$

where $\boxed {v +\Delta v -u}$ is the speed of $\Delta m$ at $t+\Delta t$ (see “A Thought Experiment on Velocities”)

By Newton’s second law,

$F_1 = \lim\limits_{\Delta t \rightarrow 0}\frac{\Delta p_1}{\Delta t}$

$= \lim\limits_{\Delta t \rightarrow 0} \frac{-\Delta m \cdot u + \Delta m \cdot \Delta v}{\Delta t}$.

For the rocket:

$\Delta p_2 = (m-\Delta m) \cdot (v +\Delta v) - (m-\Delta m)\cdot v$

$= (m-\Delta m)(v +\Delta v-v)$

$= (m-\Delta m)\cdot {\Delta v}$

$= m \cdot \Delta v - \Delta m \cdot \Delta v$

$F_2 = \lim\limits_{\Delta t \rightarrow 0} \frac{\Delta p_2}{\Delta t}$

$= \lim\limits_{\Delta t \rightarrow 0} \frac{m \cdot \Delta v - \Delta m \cdot \Delta v}{\Delta t}$.

By Newton’s third law,

$F_2 = -F_1$.

Therefore,

$\lim\limits_{\Delta t \rightarrow 0} \frac{m \cdot \Delta v - \Delta m \cdot \Delta v}{\Delta t} = - \lim\limits_{\Delta t \rightarrow 0} \frac{-\Delta m \cdot u + \Delta m \cdot \Delta v}{\Delta t}$

That is,

$\lim\limits_{\Delta t \rightarrow 0} \frac{m \cdot \Delta v - \Delta m \cdot \Delta v}{\Delta t} + \lim\limits_{\Delta t \rightarrow 0} \frac{-\Delta m \cdot u + \Delta m \cdot \Delta v}{\Delta t} = 0$.

It implies

$\lim\limits_{\Delta t \rightarrow 0} \frac{m \cdot \Delta v - \Delta m \cdot \Delta v -\Delta m \cdot u + \Delta m \cdot \Delta v}{\Delta t} = 0$

or,

$\lim\limits_{\Delta t \rightarrow 0} \frac{m \cdot \Delta v -\Delta m \cdot u}{\Delta t} = 0$.

Since

$\lim\limits_{\Delta t \rightarrow 0} \frac{m \cdot \Delta v -\Delta m \cdot u }{\Delta t}= \lim\limits_{\Delta t \rightarrow 0} {\frac{m\cdot \Delta v -u \cdot \Delta m}{\Delta t}}=m \cdot \lim\limits_{\Delta t \rightarrow 0}\frac{\Delta v}{\Delta t}-u \cdot \lim\limits_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t}$,

$\frac{dv}{dt} = \lim\limits_{\Delta t \rightarrow 0}\frac{\Delta v}{\Delta t}$,

and

$\lim\limits_{\Delta t \rightarrow 0}\frac{\Delta m}{\Delta t}= \lim\limits_{\Delta t \rightarrow 0} \frac{m(t) - m(t+\Delta t)}{\Delta t}= -\lim\limits_{\Delta t \rightarrow 0} \frac{m(t+\Delta t) - m(t)}{\Delta t} = -\frac{dm}{dt}$

we have

$m \cdot \frac{dv}{dt} + u \cdot \frac{dm}{dt} = 0$,

the ideal rocket’s flight equation obtained before in “Viva Rocketry! Part 1“.

# Does gravity matter ? (Viva Rocketry! Part 1.2)

A single rocket expels its propellant at a constant rate $k$.

Assuming constant gravity is the only external force, show that the equation of motion is

$(p+m_0-kt)\frac{dv}{dt}=ck-(p+m_0-kt)g$

where $v$ is the rocket’s speed, $c$ the speed of the propellant relative to the rocket, $p$ the payload mass, and $m_0$ the initial rocket mass.

If the rocket burn is continuous, show that the burn time is $\frac{\epsilon m_0}{k}$ and deduce that the final speed given to the payload is

$v=-c \log(1-\frac{\epsilon m_0}{m_0+p}) - \frac{g\epsilon m_0}{k}$

where $1-\epsilon$ is the structural factor of the rocket.

Estimate the percentage reduction in the predicted final speed due to the inclusion of the gravity term if

$\epsilon=0.8, \frac{p}{m_0}=\frac{1}{100}, c=3.0\;km\;s^{-1}, m_0=10^5\;kg$, and $k=5 \times 10^3\;kg\;s^{-1}$.

Find an expression for the height reached by the rocket during the burn and estimate its value using the data above.

Let’s recall the governing equation of rocket’s flight derived in “Viva Rocketry! Part 1“, namely,

$F = m\frac{dv}{dt} + u\frac{dm}{dt}$.

In the present context, $m = m_0 + p- k t$. It implies that

$\frac{dm}{dt}=-k$

and,

$F=-mg=-(m_0+p-kt)g$.

With $u = c$, we have

$-(p+m_0-kt)g = (p+m_0-kt)\frac{dv}{dt}+c\cdot(-k)$,

i.e.,

$(p+m_0-kt)\frac{dv}{dt}=ck-(p+m_0-kt)g$

or

$\frac{dv}{dt}=\frac{ck}{m_0+p-kt}-g$.

The structural factor $1-\epsilon$ indicates the amount of fuel is $\epsilon m_0$. Since the fuel is burnt at a constant rate $k$, it must be true that at burnt out time $T$,

$\epsilon m_0 = kT$.

Therefore,

$T=\frac{\epsilon m_0}{k}$.

The solution to initial-value problem

$\begin{cases} \frac{dv}{dt}=\frac{ck}{m_0+p-kt}-g\\ v(0) = 0 \end{cases}$

tells the speed of the rocket during its flight while fuel is burnt (see Fig. 1):

$v = -c \log(1-\frac{kt}{m_0+p})-gt\quad\quad\quad(1)$

Fig. 1

Evaluate (1) at burnt out time gives the final speed of the payload:

$v_1=-c \log(1-\frac{\epsilon m_0}{m_0+p}) - \frac{g\epsilon m_0}{k}\quad\quad\quad(2)$

Notice the first term of (2) is the burnt out velocity without gravity (see “Viva Rocketry! Part 1“)

It follows that the percentage reduction in the predicted final speed due to the inclusion of gravity is

$\frac{\frac{g \epsilon m_0} {k}}{-c \log(1-{\epsilon \over {1+\frac{p}{m_0}}})}\quad\quad\quad(3)$

Using the given values which are typical, the estimated value of (3) (see Fig. 2) is

$0.003\%$.

Fig. 2

This shows the results obtained without taking gravity into consideration can be regarded as a reasonable approximation and the characteristics of rocket flight indicated in “Viva Rocketry! Part 1” are valid.

Since $v = \frac{dy}{dt}$, (1) can be written as

$\frac{dy}{dt} = -c \log(1-\frac{kt}{m_0+p})-gt$

To find the distance travelled while the fuel is burnt, we solve yet another initial-value problem:

$\begin{cases}\frac{dy}{dt} = -c \log(1-\frac{kt}{m_0+p})-gt \\ y(0) = 0 \end{cases}$

Fig. 3

The solution (see Fig. 3) is

$y= -\frac{1}{2}g t^2 + ct - c\cdot (t-\frac{m_0+p}{k}) \cdot \log(1-\frac{kt}{m_0+p})$.

Hence, the height reached at the burnt out time $t=\frac{\epsilon m_0}{k}$ is

$h = -\frac{g\epsilon^2 m_0^2}{2k^2}+\frac{c\epsilon m_0}{k}+\frac{c}{k}\cdot (p+(1-\epsilon)m_0) \cdot \log(1-\frac{\epsilon m_0}{m_0+p})$.

Using the given values, we estimate that $h \approx 27 \; km$ (see Fig. 4)

Fig. 4

Exercise 1: Find the distance the rocket travelled while the fuel is burnt by solving the following initial-value problem:

$\begin{cases}\frac{d^2y}{dt^2} =\frac{ck}{p+m_0-kt}-g \\ y(0) = 0, y'(0)=0 \end{cases}$