December 2018

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“, 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.

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“.

Month: December 2018

When rocket ejects its propellant at a variable rate

An alternative derivation of ideal rocket’s flight equation