Monthly Archives: December 2018

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.

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

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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“.