# 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.