Category Archives: Physics

What is the shape of a hanging rope?

Question: What is the shape of a flexible rope hanging from nails at each end and sagging under the gravity?

Answer:

First, observe that no matter how the rope hangs, it will have a lowest point $A$ (see Fig. 1)

Fig. 1

It follows that the hanging rope can be placed in a coordinate system whose origin coincides with the lowest point $A$ and the tangent to the rope at $A$ is horizontal:

Fig. 2

At $A$ , the rope to its left exerts a horizontal force. This force (or tension), denoted by $T_0$ , is a constant:

Fig. 3

Shown in Fig. 3 also is an arbitrary point $B$ with coordinates $(x, y)$ on the rope. The tension at $B$ , denoted by $T_1$ , is along the tangent to the rope curve. $\theta$ is the angle $T_1$ makes with the horizontal.

Since the section of the rope from $A$ to $B$ is stationary, the net force acting on it must be zero. Namely, the sum of the horizontal force, and the sum of the vertical force, must each be zero:

$\begin{cases}T_1cos(\theta)=T_0\quad\quad\quad(1)\\ T_1\sin(\theta) = \rho gs\;\;\quad\quad(2)\end{cases}$

where $\rho$ is the hanging rope’s mass density and $s$ its length from $A$ to $B$ .

Dividing (2) by (1), we have

$\frac{T\sin(\theta)}{T\cos(\theta)} = \tan(\theta) = \frac{\rho g}{T_0}s\overset{k=\frac{\rho g}{T_0}}{\implies} \tan(\theta) = ks.\quad\quad\quad(3)$

Since

$\tan(\theta) = \frac{dy}{dx}$ , the slope of the curve at $B$ ,

and

$s = \int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}$ ,

we rewrite (3) as

$\frac{dy}{dx} = k \int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx$

and so,

$\frac{d^2y}{dx^2}=k\cdot \frac{d}{dx}(\int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx)=k\sqrt{1+(\frac{dy}{dx})^2}$

i.e.,

$\frac{d^2y}{dx^2}=k\sqrt{1+(\frac{dy}{dx})^2}.\quad\quad\quad(4)$

To solve (4), let

$p = \frac{dy}{dx}$ .

We have

$\frac{dp}{dx} = k\sqrt{1+p^2} \implies \frac{1}{\sqrt{1+p^2}}\frac{dp}{dx}=k.\quad\quad\quad(5)$

Integrate (5) with respect to $x$ gives

$\log(p+\sqrt{1+p^2}) = kx + C_1\overset{p(0)=y'(0)=0}{\implies} C_1 = 0.$

i.e.,

$\log(p+\sqrt{1+p^2}) = kx.\quad\quad\quad(6)$

Solving (6) for $p$ yields

$p = \frac{dy}{dx} =\sinh(kx).\quad\quad\quad(7)$

Integrate (7) with respect to $x$ ,

$y = \frac{1}{k} \cosh(kx) + C_2\overset{y(0)=0,\cosh(0)=1}{\implies}C_2=-\frac{1}{k}$ .

Hence,

$y = \frac{1}{k}\cosh(kx)-\frac{1}{k}$ .

Essentially, it is the hyperbolic cosine function that describes the shape of a hanging rope.

Exercise-1 Show that $\int \frac{1}{\sqrt{1+p^2}} dp = \log(p + \sqrt{1+p^2})$ .

Exercise-2 Solve $\log(p+\sqrt{1+p^2}) = kx$ for $p$ .

From Dancing Planet to Kepler’s Laws

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“This most beautiful system of the sun, planets, and comets, could only proceed from the counsel and dominion of an intelligent powerful Being” – Sir. Issac Newton

When I was seven years old, I had the notion that all planets dance around the sun along a wavy orbit (see Fig. 1).

Fig. 1

Many years later, I took on a challenge to show mathematically the orbit of my ‘dancing planet’ . This post is a long overdue report of my journey.

Shown in Fig. 2 is the sun and a planet in a x-y-z coordinate system. The sun is at the origin. The moving planet’s position is being described by $x=x(t), y=y(t), z=z(t)$ .

Fig. 2 $r=\sqrt{x^2+y^2+z^2}, F=G\frac{Mm}{r^2}, F_z=-F\cos(c)=-F\cdot\frac{z}{r}$

According to Newton’s theory, the gravitational force sun exerts on the planet is

$F=-G\cdot M \cdot m \cdot \frac{1}{r^2}(\frac{x}{r},\frac{y}{r}, \frac{z}{r})=-\mu\cdot m \cdot \frac{1}{r^3}\cdot(x, y, z)$

where $G$ is the gravitational constant, $M, m$ the mass of the sun and planet respectively. $\mu = G\cdot M$ .

By Newton’s second law of motion,

$\frac{d^2x}{dt^2} = -\mu\frac{x}{r^3},\quad\quad\quad(0-1)$

$\frac{d^2y}{dt^2} = -\mu\frac{y}{r^3},\quad\quad\quad(0-2)$

$\frac{d^2z}{dt^2} = -\mu\frac{z}{r^3}.\quad\quad\quad(0-3)$

$y \cdot$ (0-3) $- z \cdot$ (0-2) yields

$y\frac{d^2z}{dt^2}-z\frac{d^2y}{dt^2} = -\mu\frac{yz}{r^3}+ \mu\frac{yz}{r^3}=0$ .

Since

$y\frac{d^2z}{dt^2}-z\frac{d^2y}{dt^2} = \frac{dy}{dt}\frac{dz}{dt}+y\frac{d^2z}{dt^2}-\frac{dz}{dt}\frac{dy}{dt}-z\frac{d^2y}{dt^2}=\frac{d}{dt}(y\frac{dz}{dt}-z\frac{dy}{dt})$ ,

it must be true that

$\frac{d}{dt}(y\frac{dz}{dt}-z\frac{dy}{dt}) = 0$ .

i.e.

$y\frac{dz}{dt}-z\frac{dy}{dt}=A\quad\quad\quad(0-4)$

where $A$ is a constant.

Similarly,

$z\frac{dx}{dt}-x\frac{dz}{dt}= B,\quad\quad\quad(0-5)$

$x\frac{dy}{dt}-y\frac{dx}{dt}= C\quad\quad\quad(0-6)$

where $B,C$ are constants.

Consequently,

$Ax=xy\frac{dz}{dt} - xz\frac{dy}{dt}$ ,

$By=yz\frac{dx}{dt} - xy\frac{dz}{dt}$ ,

$Cz=xz\frac{dy}{dt}-yz\frac{dx}{dt}$ .

Hence

$Ax + By +Cz=0.\quad\quad\quad(0-7)$

If $C \ne 0$ then by the following well known theorem in Analytic Geometry:

“If A, B, C and D are constants and A, B, and C are not all zero, then the graph of the equation Ax+By+Cz+D=0 is a plane“,

(0-7) represents a plane in the x-y-z coordinate system.

For $C=0$ , we have

$\frac{d}{dt}(\frac{y}{x})=\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2}\overset{(0-6)}{=}\frac{C}{x^2}=\frac{0}{x^2}=0$ .

It means

$\frac{y}{x}=k$

where $k$ is a constant. Simply put,

$y=k x$ .

Hence, (0-7) still represents a plane in the x-y-z coordinate system (see Fig. 3(a)).

Fig. 3

The implication is that the planet moves around the sun on a plane (see Fig. 4).

Fig. 4

By rotating the axes so that the orbit of the planet is on the x-y plane where $z \equiv 0$ (see Fig. 3), we simplify the equations (0-1)-(0-3) to

$\begin{cases} \frac{d^2x}{dt^2}=-\mu\frac{x}{r^3} \\ \frac{d^2y}{dt^2}=-\mu\frac{y}{r^3}\end{cases}.\quad\quad\quad(1-1)$

It follows that

$\frac{d}{dt}((\frac{dx}{dt})^2 + (\frac{dy}{dt})^2)$

$= 2\frac{dx}{dt}\cdot\frac{d^2x}{dt^2} + 2 \frac{dy}{dt}\cdot\frac{d^2y}{dt^2}$

$\overset{(1-1)}{=}2\frac{dx}{dt}\cdot(-\mu\frac{x}{r^3})+ 2\frac{dy}{dt}\cdot(-\mu\frac{y}{r^3})$

$= -\frac{\mu}{r^3}\cdot(2x\frac{dx}{dt}+2y\frac{dy}{dt})$

$= -\frac{\mu}{r^3}\cdot\frac{d(x^2+y^2)}{dt}$

$= -\frac{\mu}{r^3}\cdot\frac{dr^2}{dt}$

$= -\frac{\mu}{r^3} \cdot 2r \cdot \frac{dr}{dt}$

$= -\frac{2\mu}{r^2} \cdot \frac{dr}{dt}$ .

i.e.,

$\frac{d}{dt}((\frac{dx}{dt})^2 + (\frac{dy}{dt})^2) = -\frac{2\mu}{r^2} \cdot \frac{dr}{dt}$ .

Integrate with respect to $t$ ,

$(\frac{dx}{dt})^2+(\frac{dy}{dt})^2 = \frac{2\mu}{r} + c_1\quad\quad\quad(1-2)$

where $c_1$ is a constant.

We can also re-write (0-6) as

$x\frac{dy}{dt}-y\frac{dx}{dt}=c_2\quad\quad\quad(1-3)$

where $c_2$ is a constant.

Using polar coordinates

$\begin{cases} x= r\cos(\theta) \\ y=r\sin(\theta) \end{cases},$

Fig. 5

we obtain from (1-2) and (1-3) (see Fig. 5):

$(\frac{dr}{dt})^2 + (r\frac{d\theta}{dt})^2-\frac{2\mu}{r} = c_1,\quad\quad\quad(1-4)$

$r^2\frac{d\theta}{dt} = c_2.\quad\quad\quad(1-5)$

If the speed of planet at time $t$ is $v$ then from Fig. 6,

Fig. 6

$v = \lim\limits_{\Delta t \rightarrow 0}\frac{\Delta l}{\Delta t} = \lim\limits_{\Delta t\rightarrow 0}\frac{l_r}{\Delta t}\overset{l_r=r\Delta \theta}{=}\lim\limits_{\Delta t \rightarrow 0}\frac{r\cdot \Delta \theta}{\Delta t}=r\cdot\lim\limits_{\Delta t\rightarrow 0}\frac{\Delta \theta}{\Delta t}=r\cdot\frac{d\theta}{dt}$

gives

$v = r\frac{d\theta}{dt}.\quad\quad\quad(1-6)$

Suppose at $t=0$ , the planet is at the greatest distance from the sun with $r=r_0, \theta=0$ and speed $v_0$ . Then the fact that $r$ attains maximum at $t=0$ implies $(\frac{dr}{dt})_{t=0}=0$ . Therefore, by (1-4) and (1-5),

$(\frac{dr}{dt})^2_{t=0} + (r\frac{d\theta}{dt})^2_{t=0}-\frac{s\mu}{r} = 0+ v_0^2-\frac{2\mu}{r}=c_1,$

$r (r\frac{d\theta}{dt})_{t=0}=r_0v_0=c_2.$

i.e.,

$c_1=v_0^2-\frac{2\mu}{r_0},\quad\quad\quad(1-7)$

$c_2=v_0 r_0.\quad\quad\quad(1-8)$

We can now express (1-4) and (1-5) as:

$\frac{dr}{dt} = \pm \sqrt{c_1+\frac{2\mu}{r}-\frac{c_2^2}{r^2}},\quad\quad\quad(1-9)$

$\frac{d\theta}{dt} = \frac{c_2}{r^2}.\quad\quad\quad(1-10)$

Let

$\rho = \frac{c_2}{r}\quad\quad\quad(1-11)$

then

$\frac{d\rho}{dr} = -\frac{c_2}{r^2},\quad\quad\quad(1-12)$

$r=\frac{c_2}{\rho}.\quad\quad\quad(1-13)$

By chain rule,

$\frac{d\theta}{dt} = \frac{d\theta}{d\rho}\cdot\frac{d\rho}{dr}\cdot\frac{dr}{dt}$ .

Thus,

$\frac{d\theta}{d\rho} = \frac{\frac{d\theta}{dt}}{ \frac{d\rho}{dr} \cdot \frac{dr}{dt}}$

$\overset{(1-10), (1-12), (1-9)}{=} \frac{\frac{c_2}{r^2}}{ (-\frac{c_2}{r^2})\cdot(\pm\sqrt{c_1+\frac{2\mu}{r}-\frac{c_2^2}{r^2}}) }$

$\overset{(1-11)}{=} \mp\frac{1}{\sqrt{c_1-\rho^2+2\mu(\frac{\rho}{c_2})}}$

$= \mp\frac{1}{\sqrt{c_1+(\frac{\mu}{c_1})^2-\rho^2+2\mu(\frac{\rho}{c_2}) -(\frac{\mu}{c_2})^2}}$

$= \mp\frac{1}{\sqrt{c_1+(\frac{\mu}{c_1})^2-(\rho^2-2\mu(\frac{\rho}{c_2}) +(\frac{\mu}{c_2})^2)}}$

$= \mp\frac{1}{\sqrt{c_1+(\frac{\mu}{c_1})^2-(\rho-\frac{\mu}{c_2})^2}}$ .

That is,

$\frac{d\theta}{d\rho} = \mp\frac{1}{\sqrt{c_1+(\frac{\mu}{c_1})^2-(\rho-\frac{\mu}{c_2})^2}}.\quad\quad\quad(1-14)$

Since

$c_1+(\frac{\mu}{c_2})^2\overset{(1-7)}{=}v_0^2-\frac{2\mu}{r_0}+(\frac{\mu}{v_0r_0})^2=(v_0-\frac{\mu}{v_0r_0})^2$ ,

we let

$\lambda = \sqrt{c_1 + (\frac{\mu}{c_2})^2}=\sqrt{(v_0-\frac{\mu}{v_0r_0})^2}=|v_0-\frac{\mu}{v_0r_0}|$ .

Notice that $\lambda \ge 0$ .

By doing so, (1-14) can be expressed as

$\frac{d\theta}{d\rho} =\mp \frac{1}{\sqrt{\lambda^2-(\rho-\frac{\mu}{c_2})^2}}$ .

Take the first case,

$\frac{d\theta}{d\rho} = -\frac{1}{\sqrt{\lambda^2-(\rho-\frac{\mu}{c_2})^2}}$ .

Integrate it with respect to $\rho$ gives

$\theta + c = \arccos(\frac{\rho-\frac{\mu}{c_2}}{\lambda})$

where $c$ is a constant.

When $r=r_0, \theta=0$ ,

$c = \arccos(1)=0$ or $c = \arccos(-1) = \pi$ .

And so,

$\theta = \arccos(\frac{\rho-\frac{\mu}{c_2}}{\lambda})$ or $\theta+\pi = \arccos(\frac{\rho-\frac{\mu}{c_2}}{\lambda})$ .

For $c = 0$ ,

$\lambda\cos(\theta) = \rho-\frac{\mu}{c_2}$ .

By (1-11), it is

$\frac{c_2}{r}-\frac{\mu}{c_2} = \lambda \cos(\theta).\quad\quad\quad(1-15)$

Fig. 7

Solving (1-15) for $r$ yields

$r=\frac{c_2^2}{c_2 \lambda \cos(\theta)+\mu}=\frac{\frac{c_2^2}{\mu}}{\frac{c_2}{\mu}\lambda \cos(\theta)+1}\overset{p=\frac{c_2^2}{\mu}, e=\frac{c_2 \lambda}{\mu}}{=}\frac{p}{e \cos(\theta) + 1}$ .

i.e.,

$r = \frac{p}{e \cos(\theta) + 1}.\quad\quad\quad(1-16)$

Studies in Analytic Geometry show that for an orbit expressed by (1-16), there are four cases to consider depend on the value of $e$ :

We can rule out parabolic and hyperbolic orbit immediately for they are not periodic. Given the fact that a circle is a special case of an ellipse, it is fair to say:

The orbit of a planet is an ellipse with the Sun at one of the two foci.

In fact, this is what Kepler stated as his first law of planetary motion.

Fig. 8

For $c=\pi$ ,

$\theta + \pi = \arccos(\frac{\rho-\frac{\mu}{c_2}}{\lambda})$

from which we obtain

$r=\frac{c_2^2}{c_2 \lambda \cos(\theta+\pi)+\mu}=\frac{\frac{c_2^2}{\mu}}{\frac{c_2}{\mu}\lambda\cos(\theta+\pi)+1}\overset{p=\frac{c_2^2}{\mu}, e=\frac{c_2 \lambda}{\mu}}{=}\frac{p}{e \cos(\theta+\pi) + 1}.\quad\quad(1-17)$

This is an ellipse. Namely, the result of rotating (1-16) by hundred eighty degrees or assuming $r$ attains its minimum at $t=0$ .

The second case

$\frac{d\theta}{d\rho} = +\frac{1}{\sqrt{\lambda^2-(\rho-\frac{\mu}{c_2})^2}}$

can be written as

$-\frac{d\theta}{d\rho} = -\frac{1}{\sqrt{\lambda^2-(\rho-\frac{\mu}{c_2})^2}}$ .

Integrate it with respect to $\rho$ yields

$-\theta + c = \arccos(\frac{\rho-\frac{\mu}{c_2}}{\lambda})$

from which we can obtain (1-16) and (1-17) again.

Fig. 9

Over the time duration $\Delta t$ , the area a line joining the sun and a planet sweeps an area $A$ (see Fig. 9).

$A = \int\limits_{t}^{t+\Delta t}\frac{1}{2}r\cdot v\;dt \overset{(1-6)}{=} \int\limits_{t}^{t+\Delta t}\frac{1}{2}r\cdot r\frac{d\theta}{dt}\;dt=\int\limits_{t}^{t+\Delta t}\frac{1}{2}r^2\frac{d\theta}{dt}\;dt\overset{(1-5)}{=}\int\limits_{t}^{t+\Delta t}\frac{1}{2}c_2\;dt = \frac{1}{2}c_2\Delta t$ .

It means

$A = \frac{1}{2}c_2\Delta t\quad\quad\quad(2-1)$

or that

$\frac{A}{\Delta t} = \frac{1}{2}c_2$

is a constant. Therefore,

A line joining the Sun and a planet sweeps out equal areas during equal intervals of time.

This is Kepler’s second law. It suggests that the speed of the planet increases as it nears the sun and decreases as it recedes from the sun (see Fig. 10).

Fig. 10

Furthermore, over the interval $T$ , the period of the planet’s revolution around the sun, the line joining the sun and the planet sweeps the enire interior of the planet’s elliptical orbit with semi-major axis $a$ and semi-minor axis $b$ . Since the area enlosed by such orbit is $\pi ab$ (see “Evaluate a Definite Integral without FTC“), setting $\Delta t$ in (2-1) to $T$ gives

${\frac{1}{2}c_2 T} = {\pi a b} \implies {\frac{1}{4}c_2^2 T^2}={\pi^2 a^2 b^2} \implies T^2 = \frac{4\pi^2 a^2 b^2}{c_2^2} \implies \frac{T^2}{a^3} = \frac{4\pi^2b^2}{c_2^2a}. (3-1)$

While we have $p = \frac{c_2^2}{\mu}$ in (1-16), it is also true that for such ellipse, $p=\frac{b^2}{a}$ (see “An Ellipse in Its Polar Form“). Hence,

$\frac{b^2}{a}=\frac{c_2^2}{\mu}\implies c_2^2=\frac{\mu b^2}{a}.\quad\quad\quad(3-2)$

Substituting (3-2) for $c_2^2$ in (3-1),

$\frac{T^2}{a^3} = \frac{4\pi^2 b^2}{(\frac{\mu b^2}{a})a}=\frac{4\pi^2}{\mu} \overset{\mu=GM}{=}\frac{4\pi^2}{GM}.\quad\quad\quad(3-3)$

Thus emerges Kepler’s third law of planetary motion:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Established by (3-3) is the fact that the proportionality constant is the same for all planets orbiting around the sun.

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Déjà vu!

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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:

$- \frac{(\sqrt{b}\sqrt{b+1}-b-1)P}{b}>0$
$\frac{\sqrt{b^2+b}P-bP}{b}>0$

Exercise-2 From (1), prove the extreme value attained under (2) is a maximum.

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

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Fig. 1

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 $e$ and equal relative exhaust speed $c$ . The rocket mass, $m_1+m_2$ is fixed and $\frac{P}{m_1+m_2}=b$ .

According to multi-stage rocket’s flight equation (see “Viva Rocketry! Part 2“), the final speed of a two-stage rocket is

$v = -c \log(1-\frac{em_1}{m_1+m_2+P}) - c\log(1-\frac{em_2}{m_2+P})$

Let $a = \frac{m1}{m_2}$ , it becomes

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

where $a>0, b>0, c>0, 0< e < 1$ . We will maximize $v$ with an appropriate choice of $a$ .

That is, 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 $a>0, b>0, c>0, 0<e<1$ . Maximize $v$ with an appropriate value of $a$ .

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

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

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

(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.,

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

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

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: $0<e<1, b>0 \implies$

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

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

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

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 $n$ th 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)

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 $v_2^* = 7.8$ .

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

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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)

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USS_Cape_St._George_(CG_71)_fires_a_tomahawk_missile_in_support_of_OIF.jpg

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)

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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.,

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

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

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):

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

Screen Shot 2018-12-03 at 4.48.48 PM.png

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

$0.003\%$ .

Screen Shot 2018-12-03 at 3.43.03 PM.png

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:

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

Screen Shot 2018-12-03 at 4.54.24 PM.png

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)

Screen Shot 2018-12-03 at 3.34.13 PM.png

Fig. 4

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

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

Thunderbolt (Viva Rocketry! Part 1.1)

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Fig. 1

Shown in Fig. 1 is an experimental car propelled by a rocket motor. The drag force (air resistance) is given by $R = \beta v^2$ . The initial mass of the car, which includes fuel of mass $m_f$ , is $m_0$ . The rocket motor is burning fuel at the rate of $q$ with an exhaust velocity of $u$ relative to the car. The car is at rest at $t=0$ . Show that the velocity of the car is given by, for $0 \leq t \leq T$ ,

$v(t) = \mu \cdot \frac{1-({m \over m_0})^{\frac{2\beta \mu}{q}}}{1+({m \over m_0})^{\frac{2\beta \mu}{q}}}$ ,

where $m=m_0-qt, \mu^2=\frac{qu}{\beta}$ , and $T=\frac{m_f}{q}$ is the time when the fuel is burnt out.

We have derived the governing equation of rocket flight in “Viva Rocketry! Part 1“, namely,

$F = m \frac{dv}{dt} + u \frac{dm}{dt}\quad\quad\quad(1)$

From $m=m_0-qt$ , we have

$\frac{dm}{dt} = -q$ .

Apply air resistance $R=\beta v^2$ as the external force, (1) becomes

$-\beta v^2 = (m_0 - q t) \frac{dv}{dt} - u q$ .

And the car is at rest initially implies

$v(0)=0$ .

It follows that the motion of the car can be modeled by an initial-value problem

$\begin{cases} -\beta v^2 = (m_0 - q t) \frac{dv}{dt} - u q \\ v(0) = 0 \end{cases}\quad\quad\quad(2)$

It suffices to show that the given $v(t)$ is the solution to this initial-value problem:

Screen Shot 2018-11-28 at 9.28.46 AM.png

Fig. 2

An alternative is obtaining the stated $v(t)$ through solving (2).

Screen Shot 2018-11-27 at 4.40.18 PM.png

Fig. 3

The fact that $m = m_0 -qt, (-1)^{2 \frac{\sqrt{b}\sqrt{u}}{\sqrt{q}}} = 1$ simplifies the result considerably,

$\frac{\sqrt q \sqrt u}{\sqrt \beta}\cdot \frac{m_0^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}-m^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}{m_0^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}+m^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}\quad\quad\quad(3)$

Divide both the numerator and denominator of (3) by $m_0^{\frac {2 \sqrt{\beta} \sqrt{u}}{\sqrt{q}}}$ then yields

$\frac{\sqrt q \sqrt u}{\sqrt \beta}\cdot \frac{1-(\frac{m}{m_0})^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}{1+(\frac{m}{m_0})^{\frac{2 \sqrt \beta \sqrt u}{\sqrt q}}}$

which is equivalent to the given $v(t)$ since $\mu^2=\frac{qu}{\beta}$ .

At time $t=T$ , the fuel is burnt out. It means

$m_0-m_f = m_0 - qT$ .

Therefore,

$T = \frac{m_f}{q}$

Exercise 1: Solve (2) manually.

Hint: The differential equation of (2) can be written as $\frac{1}{uq - \beta v^2} \frac{dv}{dt} = \frac{1}{m_0 - q t}$ .

Exercise 2: For $m_0=900\;kg, m_f=450\;kg, q=15\;kg/sec, u=500\;m/sec, \beta=0.3$ , what is the burnout velocity of the car?

Viva Rocketry! Part 1

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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.

Screen Shot 2018-11-09 at 9.37.30 AM.png

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.,

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

The 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.,

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

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 $t_1$

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

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

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

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

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

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

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.

Screen Shot 2018-11-08 at 3.47.09 PM.png

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 $R_{\oplus} = 6371\;km$ .