Thunderbolt (Viva Rocketry! Part 1.1)

Viva Rocketry! Part 1

In this post, we will first look at the main characteristics of rocket flight, and then examine the feasibility of launching a satellite as the payload of a rocket into an orbit above the earth.

A rocket accelerates itself by ejecting part of its mass with high velocity.

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Fig. 1

Fig. 1 shows a moving rocket. At time $t+\Delta t$ , the mass $\Delta m$ leaves the rocket in opposite direction. As a result, the rocket is being propelled away with an increased speed.

Let

$m(\square), m_{\square}$ – the mass of rocket at time ${\square}$

$\vec{v}_{\square}$ – the velocity of rocket at time $\square$

$v(\square), v_{\square}$ – the magnitude of $\vec{v}_{\square}$

$\vec{v}^*_{t+\Delta t}$ – the velocity of ejected mass $\Delta m$ at $t + \Delta t$

$v^*_{t+\Delta t}$ – the magnitude of $\vec{v}^*_{t+\Delta t}$

$u$ – the magnitude of $\Delta m$ ‘s velocity relative to the rocket when it is ejected. It is time invariant.

From Fig. 1, we have

$\Delta m = m_t - m_{t + \Delta t}$ ,

$\vec{v}_t = v_{t}$ ,

$\vec{v}_{t + \Delta t} = v_{t + \Delta t}$

and most notably, the relationship between $v^*_{t+\Delta t}, v_{t+\Delta t}$ and $u$ (see “A Thought Experiment on Velocities”):

$v^*_{t+\Delta t} = u - v_{t + \Delta t}$ .

It follows that

$\vec{v}^*_{t+\Delta t} = -v^*_{t+\Delta t} = v_{t + \Delta t} - u$ ,

momentum at time $t$ : $\vec{p}(t) = m_t \vec{v}_t = m_t v_t$

and,

momentum at time $t+\Delta t$ : $\vec{p}(t+\Delta t) = m_{t+\Delta t}\vec{v}_{t+\Delta t} + {\Delta m} \vec{v}^*_{t+\Delta t}=m_{t+\Delta t}\vec{v}_{t+\Delta t} + (m_t - m_{t+\Delta t}) \vec{v}^*_{t+\Delta t}$ $= m_{t+\Delta t}v_{t+\Delta t} + (m_t - m_{t+\Delta t})(v_{t+\Delta t}-u)$ .

Consequently, change of momentum in $\Delta t$ is $\vec{p}(t+\Delta t)- \vec{p}(t) = m_t (v_{t + \Delta t} - v_t) + u (m_{t + \Delta t} - m_t)$ .

Apply Newton’s second law of motion to the whole system,

$\vec{F}= {d \over dt} \vec{p}(t)$

$= \lim\limits_{\Delta t \rightarrow 0} {{\vec{p}(t+\Delta t) - \vec{p}(t)} \over \Delta t}$

$= \lim\limits_{\Delta t \rightarrow 0} { {m_t (v_{t + \Delta t} - v_t) + u (m_{t + \Delta t} - m_t)} \over {\Delta t} }$

$= \lim\limits_{\Delta t \rightarrow 0} {m_t {{v_{t + \Delta t} - v_t} \over {\Delta t}} + {u {{m_{t + \Delta t} - m_t} \over {\Delta t}}}}$

$= m_t \lim\limits_{\Delta t \rightarrow 0}{(v_{t+\Delta t} - v_t) \over {\Delta t}} + u \lim\limits_{\Delta t \rightarrow 0}{(m_{t +\Delta t} - m_t) \over \Delta t}$

That is,

$\vec{F} = m(t) {d \over dt} v(t) + u {d \over dt} m(t)$

where $\vec{F}$ is the sum of external forces acting on the system.

To get an overall picture of the rocket flight, we will neglect all external forces.

Without any external force, $\vec{F} = 0$ . Therefore

$0 = m(t) {d \over dt} v(t) + u {d \over dt} m(t)$

i.e.,

${d \over dt} v(t) = -{u \over m(t)} {d \over dt} m(t)\quad\quad\quad(1)$

That fact that $u, m(t)$ in (1) are positive quantities shows as the rocket loses mass ( ${d \over dt} m(t) < 0$ ), its velocity increases ( ${d \over dt} v(t) > 0$ )

Integrate (1) with respect to $t$ ,

$\int {d \over dt} v(t)\;dt = -u \int {1 \over m(t)} {d \over dt} m(t)\;dt$

gives

$v(t) = -u \log(m(t)) + c$

where $c$ is the constant of integration.

At $t = 0, v(0)=0, m(0) = m_1 + P$ where $m_1$ is the initial rocket mass (liquid or solid fuel + casing and instruments, exclude payload) and $P$ the payload.

It means $c = u \log(m_1+P)$ .

As a result,

$v(t) = -u \log(m(t)) + u \log(m_1+P)$

$= -u (\log(m(t) - \log(m_1+P))$

$= -u \log({m(t) \over m_1+P})$

i.e.,

$v(t) = -u \log({m(t) \over m_1+P})\quad\quad\quad(2)$

Since $m_1$ is divided into two parts, the initial fuel mass $\epsilon m_1 (0 < \epsilon < 1)$ , and the casing and instruments of mass $(1-\epsilon)m_1$ , $m(0)$ can be written as

$m(0) = \epsilon m_1 + ( 1 - \epsilon) m_1 + P$

When all the fuel has burnt out at $t_1$ ,

$m(t_1) = (1 - \epsilon)m_1 + P$

By (2), the rocket’s final speed at $t_1$

$v(t_1) = -u \log({m(t_1) \over {m_1+P}})$

$= -u \log({(1-\epsilon)m_1+P \over {m_1 + P}})$

$= -u \log({{m_1 + P -\epsilon m_1} \over {m_1+P}})$

$= -u \log(1-{{\epsilon m_1} \over {m_1+P}})$

$= -u \log(1-{\epsilon \over {1 + {P \over m_1}}})$

$= -u \log(1-{\epsilon \over {1 + \beta}})$

where $\beta = {P \over m_1}$ .

In other words,

$v(t_1) =-u \log(1-{\epsilon \over {1 + \beta}})\quad\quad\quad(3)$

Hence, the final speed depends on three parameters

$u, \epsilon$ and $\beta$

Typically,

$u = 3.0\;km\;s^{-1}, \epsilon = 0.8$ and $\beta = 1/100$ .

Using these values, (3) gives

Screen Shot 2018-11-05 at 9.21.14 PM.png

$v_1 = 4.7\;km\;s^{-1}\quad\quad\quad(4)$

This is an upper estimate to the typical final speed a single stage rocket can give to its payload. Neglected external forces such as gravity and air resistance would have reduced this speed.

With (4) in mind, let’s find out whether a satellite can be put into earth’s orbit as the payload of a single stage rocket.

We need to determine the speed that a satellite needs to have in order to stay in a circular orbit of height $h$ above the earth, as illustrated in Fig. 2.

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Fig. 2

By Newton’s inverse square law of attraction, The gravitational pull on satellite with mass $m_{s}$ is

${\gamma \; m_{s} M_{\oplus} \over (R_{\oplus} + h)^2}$

where universal gravitational constant $\gamma = 6.67 \times 10^{-11}$ , the earth’s mass $M_{\oplus} = 5.9722 \times 10^{24}\; kg$ , and the earth’s radius $R_{\oplus} = 6371\;km$ .

For a satellite to circle around the earth with a velocity of magnitude $v$ , it must be true that

${\gamma \; m_{s} M_{\oplus} \over (R_{\oplus} + h)^2} = {m_{s} v^2 \over (R_{\oplus}+h) }$

i.e,

$v = \sqrt{\gamma \; M_{\oplus} \over (R_{\oplus}+h)}$

On a typical orbit, $h = 100\;km$ above earth’s surface,

Screen Shot 2018-11-02 at 10.04.41 AM.png

$v = 7.8\;km\cdot s^{-1}$

This is far in excess of (4), the value obtained from a single stage rocket.

The implication is that a typical single stage rocket cannot serve as the launching vehicle of satellite orbiting around earth.

We will turn to multi-stage rocket in “Viva Rocketry! Part 2”.

What moves fast, will slow down, Part One

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This post aims to explain mathematically how populations change.

Our first attempt is based on ideas put forward by Thomas Malthus’ article “An Essay on the Principle of Population” published in 1798.

Let $p(t)$ denotes total population at time $t$ .

Assume in a small interval $\Delta t$ , births and deaths are proportional to $p(t)$ and $\Delta t$ . i.e.

births = $a \cdot p(t) \Delta t$

deaths = $b \cdot p(t) \Delta t$

where $a, b$ are constants.

It follows that the change of total population during time interval $\Delta t$ is

$p(t+\Delta t) - p(t) = a\cdot p(t)\Delta t - b \cdot p(t)\Delta t = r\cdot p(t)\Delta t$

where $r = a - b$ .

Dividing by $\Delta t$ and taking the limit as $\Delta t \rightarrow 0$ , we have

$\lim\limits_{\Delta \rightarrow 0} {p(t+\Delta t) - p(t) \over \Delta t} = r \cdot p(t)$

which is

${d \over dt} p(t) = r \cdot p(t)\quad\quad\quad(1)$

a first order differential equation.

Since (1) can be written as

${1 \over p(t)} {d \over dt} p(t) = r$ ,

integrate with respect to $t$ ; i.e.

$\int {1 \over p(t)}{d \over dt} p(t)dt = \int {r} dt$

leads to

$\log p(t) = r\cdot t + c$

where $c$ is the constant of integration.

If at $t=0, p(0) = p_0$ , we have

$c = \log p_0$

and so

$p(t) = p_0 e^{r\cdot t}\quad\quad\quad(2)$

The result of our first attempt shows that the behavior of the population depends on the sign of constant $r$ . We have exponential growth if $r > 0$ , exponential decay if $r < 0$ and no change if $r = 0$ .

The world population has been on a upward trend ever since such data is collected (see “World Population by Year“)

Qualitatively, our model (2) with $r>0$ indicates this trend. However, it also predicts the world population would grow exponentially without limit. And that, is most unlikely to occur, since there are so many limitation factors to growth: lack of food, insufficient energy, overcrowding, disease and war.

Therefore, it is doubtful that model (1) is The One.

Our second attempt makes a modification to (1). It takes the limitation factors into consideration by replacing constant $r$ in (1) with a function $r(t)$ . Namely,

$r(t) = \gamma - \alpha \cdot p(t)\quad\quad\quad(3)$

where $\gamma$ and $\alpha$ are both positive constants.

Replace $r$ in (1) with (3),

${d \over dt} p(t) = (\gamma - \alpha \cdot p(t)) p(t) = \gamma (1 - {p(t) \over {\gamma \over \alpha}}) p(t)\quad\quad\quad(4)$

Since $r(t)$ is a monotonic decreasing function, it shows as population grows, the growth slows down due to the limitation factors.

Let $p_{\infty} = {\gamma \over \alpha}$ ,

${d \over dt} p(t) = \gamma (1- {p(t) \over p_{\infty}}) p(t)\quad\quad\quad(5)$

This is the Logistic Differential Equation.

Written differently as

${d \over dt} p(t) - \gamma \cdot p(t) = -{\gamma \over p_{\infty}} p(t)^2$ ,

the Logistic Differential Equation is also a Bernoulli’s equation (see “Meeting Mr. Bernoulli“)

Let’s understand (5) geometrically without solving it.

Two constant functions, $p(t) = 0$ or $p_{\infty}$ are solutions of (5), since

${d \over dt} 0 = \gamma (1-{0\over p_{\infty}}) 0 = 0$

and

${d \over dt} p_{\infty} = \gamma (1-{p_{\infty} \over {p_{\infty}}}) p_{\infty} = 0$ .

Plot $p(t)$ vs. ${d \over dt} p(t)$ in Fig. 1, the points, $0$ and $p_{\infty}$ , are where the curve of ${d \over dt} p(t)$ intersects the axis of $p(t)$ .

Screen Shot 2018-10-23 at 9.42.20 AM.png

Fig. 1

At point $A$ where $p(t) > p_{\infty}$ , since ${d \over dt} p(t) < 0$ , $p(t)$ will decrease; i.e., $A$ moves left toward $p_{\infty}$ .

Similarly, at point $B$ where $p(t) < p_{\infty}, {d \over dt} p(t) > 0$ implies that $p(t)$ will increase and $B$ moves right toward $p_{\infty}$ .

The model equation can also tell the manner in which $p(t)$ approaches $p_{\infty}$ .

Let $p = p(t)$ ,

${d^2 \over dt^2} p(t) = {d \over dt}({d \over dt} p)$

$= {d \over dp} ({d \over dt}p) \cdot {d \over dt} p$

$= {d \over d p}(\gamma (1-{p \over p_{\infty}})p)\cdot {d \over dt }p$

$= \gamma(1 - {2 p\over p_{\infty}})\cdot \gamma (1-{p \over p_{\infty}})p$

$= \gamma^2 p ({{2 p} \over p_{\infty}} -1)({p \over p_{\infty}}-1)$

As an equation with unknown $p$ , $\gamma^2 p ({{2 p} \over p_{\infty}} -1)({p \over p_{\infty}}-1)=0$ has three zeros:

$0, {p_{\infty} \over 2}$ and $p_{\infty}$ .

Therefore,

${d^2 \over dt^2}p > 0$ if $p > p_{\infty}$ ,

${d^2 \over dt^2} p < 0$ if ${p_{\infty} \over 2} < p < p_{\infty}$

and

${d^2 \over dt^2} p > 0$ if $p < {p_{\infty} \over 2}$ .

Consequently $p(t)$ , the solution of initial-value problem

$\begin{cases} {d \over dt} p(t) = \gamma (1-{p(t) \over p_{\infty}}) p(t) \\ p(0)=p_0 \end{cases}\quad\quad(6)$

where $p_0 \neq 0, p_{\infty}$ behaves in the manner illustrated in Fig. 2.

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Fig. 2

If $p_0 > p_{\infty}, p(t)$ approaches $p_{\infty}$ on a concave curve. Otherwise, when ${p_{\infty} \over 2} \leq p_0 < p_{\infty}, p(t)$ moves along a convex curve. For $p_0 < {p_{\infty} \over 2}$ , the curve is concave first. It turns convex after $p(t)$ reaches ${p_{\infty} \over 2}$ .

Next, let’s solve the initial-value problem analytically for $p_0 \neq 0, p_{\infty}$ .

Instead of using the result from “Meeting Mr. Bernoulli“, we will start from scratch.

At $t$ where $p(t) \neq 0, p_{\infty}$ , we re-write (5) as

${1 \over p(t)(1-{p(t) \over p_{\infty}}) }{d \over dt} p(t) = \gamma$ .

Expressed in partial fraction,

$({1 \over p(t)} + {{1 \over p_{\infty}} \over {1-{p(t) \over p_{\infty}}}}) {d \over dt} p(t) = \gamma$ .

Integrate it with respect to $t$ ,

$\int ({1 \over p(t)} + {{1 \over p_{\infty}} \over {1-{p(t) \over p_{\infty}}}}) {d \over dt} p(t) dt = \int \gamma dt$

gives

$\log p(t) - \log (1-{p(t) \over p_{\infty}}) = \gamma t + c$

where $c$ is the constant of integration.

i.e.,

$\log {p(t) \over {1-{p(t) \over p_{\infty}}}} = \gamma t + c$ .

Since $p(0) = p_0$ , we have

$c = {\log {p_{0} \over {1-{p_0 \over p_{\infty}}}}}$

and so

$\log ({{p(t) \over {1-{p(t) \over p_{\infty}}}} \cdot {{1-{p_0 \over p_{\infty}}}\over p_0}} )=\gamma t$ .

Hence,

${{p(t) \over {1 - {p(t) \over p_\infty}}}= {{p_0 \cdot e^{\gamma t}} \over {1-{p_0 \over p_\infty}}}}$ .

Solving for $p(t)$ gives

$p(t) = { p_{\infty} \over {1+({p_{\infty} \over p_0}-1)e^{-\gamma \cdot t}}}\quad\quad\quad(7)$

We proceed to show that (7) expresses the value of $p(t)$ , the solution to (6) where $p_0 \neq 0, p_{\infty}$ , for all $t$ ‘s (see Fig.3)

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Fig. 3

From (7), we have

$\lim\limits_{t \rightarrow \infty} p(t) = p_{\infty}$ .

It validates Fig. 1.

(7) also indicates that none of the curves in Fig. 2 touch horizontal line $p(t) = p_{\infty}$ .

If this is not the case, then there exists at least one instance of $t$ where $p(t) = p_{\infty}$ ; i.e.,

${p_{\infty} \over {1+({p_{\infty} \over p_0}-1)e^{-\gamma \cdot t}}} = p_{\infty}$ .

It follows that

${({p_{\infty} \over {p_0}} - 1) e^{-\gamma t}} = 0$

Since ${e^{-\gamma t}} > 0$ (see “Two Peas in a Pod, Part 2“), it must be true that

$p_0 = p_{\infty}$ .

But this contradicts the fact that (7) is the solution of the initial-value problem (6) where $p_0 \neq 0,p_\infty$ .

Reflected in Fig.1 is that $A$ and $B$ will not become $p_{\infty}$ . They only move ever closer to it.

Last, but not the least,

${\lim \limits_{t \rightarrow \infty}} {d \over dt} p(t) = \gamma (1-{{ \lim\limits_{t \rightarrow \infty} p(t)} \over p_{\infty}}) {\lim\limits_{t \rightarrow \infty} p(t)} = \gamma (1 - {p_{\infty} \over p_{\infty}}) p_{\infty} = 0$ .

Hence the title of this post.

You say, “y” I say, “y[x]”

Meeting Mr. Bernoulli

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The differential equation

${d \over dx} y + f(x) y = g(x) y^{\alpha}\quad\quad\quad(1)$

where $\alpha \neq 0, 1$ and $g(x) \not \equiv 0$ , is known as the Bernoulli’s equation.

When $\alpha$ is an integer, (1) has trivial solution $y(x) \equiv 0$ .

To obtain nontrivial solution, we divide each term of (1) by $y^{\alpha}$ to get,

$\boxed{y^{-\alpha}{d \over dx}y} + f(x) y^{1-\alpha} = g(x)\quad\quad\quad(2)$

Since ${d \over {dx}}({{1 \over {1-\alpha}}y^{1-\alpha}}) ={1 \over {1-\alpha}}\cdot (1-\alpha) y^{1-\alpha-1}{d \over dx}y=\boxed{y^{-\alpha}{d \over dx}y}$ ,

(2) can be expressed as

${d \over dx} ({{1 \over {1-\alpha}} y^{1-\alpha}}) + f(x) y^{1-\alpha} = g(x)$

which is

${{1 \over {1-\alpha}} {d \over dx} y^{1-\alpha}} + f(x) y^{1-\alpha} = g(x)$ .

Multiply $1-\alpha$ throughout,

${d \over dx} y^{1-\alpha} + (1-\alpha) f(x) y^{1-\alpha} = (1-\alpha) g(x)\quad\quad\quad(3)$

Let $z = y^{1-\alpha}$ , (3) is transformed to a first order linear equation

${d \over dx} z + (1-\alpha) f(x) z = (1-\alpha) g(x)$ ,

giving the general solution of a Bernoulli’s equation (see Fig. 1)

Screen Shot 2018-09-28 at 4.36.05 PM.png

Fig. 1

For a concrete example of Bernoulli’s equation, see “What moves fast, will slow down“

Pandora’s Box

My shot at Harmonic Series

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To prove Beer Theorem 2 (see “Beer theorems and their proofs“) is to show that the Harmonic Series $1 + {1 \over 2} + {1 \over 3} + ...$ diverges.

Below is my shot at it.

Yaser S. Abu-Mostafa proved a theorem in an article titled “A differentiation test for absolute convergence” (see Mathematics Magazine 57(4), 228-231)

His theorem states that

Let $f$ be a real function such that ${d^2 f} \over {dx^2}$ exists at $x = 0$ . Then $\sum\limits_{n=1}^{\infty} f({1 \over n})$ converges absolutely if and only if $f(0) = f'(0)=0$ .