System of Variable Mass - The Rocket

Rocket

Propulsion of rocket is based on the conservation of linear momentum. When the fuel is burnt inside the combustion chamber, the large quantity of heat so produced greatly raises the pressure inside the chamber. As a result, the burnt up gases issue out in the form of high velocity stream called the jet. In consequence, the rocket is propelled forward:

Let, \(m\) be the instanteneous mass of the rocket and \(V\) be the instanteneous velocity of the rocket with respect to lab frame of reference. Then, rate of change of mass of the rocket is,\begin{align*} \frac{\mathrm{dM} }{\mathrm{d} t}=-\alpha-----1\end{align*} Here, the negative sign indicates that the mass decreases with time. Now,\begin{align*} \mathrm{dM}=-\alpha\mathrm{d} t \end{align*} Integrating, we get,\begin{align*} \int_{M_{o}}^{M}\mathrm{dM}=-\alpha\int_{0}^{t}\mathrm{d} t\end{align*}\begin{align*} or, M-M_{o}=-\alpha t\end{align*}\begin{align*} or, M=M_{o}(1-\frac{\alpha t}{M_{o}})\end{align*}\begin{align*} or, M=M_{o}(1-\beta t)-----2\end{align*}\begin{align*} where, \beta=\frac{\alpha}{M_{o}}-----3 \end{align*} Let, \(\overrightarrow{v}\) be the exhaust velocity relative to the rocket, then, velocity of jet in lab frame is \(\overrightarrow{V}-\overrightarrow{v}-----4\). Rate of change of momentum of the jet in lab frame is \(\frac{\mathrm{dM} }{\mathrm{d} t}(\overrightarrow{V}-\overrightarrow{v})\). Then, accordding to the third law of motion of Newton, force acting on the rocket to propel it forward is\(F=\frac{\mathrm{dM} }{\mathrm{d} t}(\overrightarrow{V}-\overrightarrow{v})\). Also, instatnteneous force acting on the rocket is,\begin{align*} F=\frac{\mathrm{d(M\overrightarrow{V})} }{\mathrm{d} t}=\frac{\mathrm{dM} }{\mathrm{d} t}(\overrightarrow{V}-\overrightarrow{v})\end{align*}\begin{align*} or, M\frac{\mathrm{dV} }{\mathrm{d} t}=-v\frac{\mathrm{dM} }{\mathrm{d} t}\end{align*}\begin{align*} or,\mathrm{dV}=-v\frac{\mathrm{dM} }{M}\end{align*}\begin{align*} or, \int_{V_{o}}^{V}\mathrm{dV}=-v\int_{M_{o}}^{M}\frac{\mathrm{dM} }{M}\end{align*}\begin{align*} or, V-v_{o}=-v\ln \frac{M}{M_{o}}\end{align*}\begin{align*} or, V=v_{o}+v\ln \frac{M_{o}}{M}-----5 \end{align*} It is the instanteneous velocity of the rocket in lab frame and \(v_{o}\) is the initial velocity of rocket at any state. Since, \(M_{o}>M\) for \(t>0\), equation 5 indicates that the velocity of rocket increases with time. Again,\begin{align*} or, V=v_{o}+v\ln \frac{M_{o}}{M_{o}(1-\beta t)}\end{align*}\begin{align*} or, V=v_{o}-v\ln (1-\beta t)-----6 \end{align*} This is the instantenous velocity of the rocket when its velocity is not considered. If the weight of the rocket is taken into account then, we get the equation \(V=v_{o}-v\ln (1-\beta t)-gt \).

Multistage Rocket

A single stage rocket that can achieve escape velocity has not been developed yet. To send the rocket successfully outside the gravitational field of the earth, the combination of two or more than two single rockets is necessary. This combination is called the multistage rocket. In a multistage rocket, the velocity of rocket goes on increasing at every stage. When this velocity reaches or exceeds the escape velocity, then the rocket escapes out of the gravitational field of the earth. Initial velocity at any stage is equal to the final velocity at the previous stage.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.