A Labscale Hybrid Rocket Motor for Instrumentation Studies

A rocket is a cylindrical projectile that can be propelled to a great height or distance by the combustion of its contents, used typically as a firework or signal, and used for scientific purposes as an engine to carry payloads including satellites.

The propulsion of a rocket is achieved by ejecting fuel at very high velocities opposite to the desired direction of motion. This propulsion is governed by Newton’s third law, which states as follows:

Newton’s Third Law

Every action has an equal and opposite reaction. That is, if a particular amount of force is applied on an object in a given direction, the object in return will exert the same amount of force in the opposite direction.

This clip from the cop drama “Brooklyn 99” shows the operation of a fire extinguisher-propelled roller chair cart, which operates by the same principles that govern rocket motion. Extinguisher fluid is ejected out the back at high velocity, which gives the roller chair cart significant momentum in the opposite direction.

Up to this point, we have come to understand how rockets fly in concept. Let us now derive the central equation for the motion of a thrust-propelled rocket. The calculations that govern rocket motion are somewhat complicated, so let us proceed from the basics.

The motion of a rocket is essentially an effect of the conservation of momentum. That is, for a given isolated system, the total momentum will remain constant. Thus, if the main part of the rocket gains any speed in a given direction, it can only come from ejecting fuel with some velocity in the opposite direction. This movement occurs such that the gain in momentum by the rocket is balanced perfectly by the momentum imparted to the fuel that is ejected.

The simplest way to analyze the motion is to consider a rocket in the moment before, and after, the release of a packet of fuel of mass $\Delta m$. For simplicity, we consider the frame of an observer who travels with the initial velocity of the rocket.

Prior to combusting the fuel and ejecting it, the whole rocket is cruising along with constant momentum M\vec{V}_\text{ship}MVship​.

To boost the rocket velocity, a packet of fuel of mass $\Delta m$ (boxed in black in the diagram below) is combusted and ejected backward with velocity \vec{u}u relative to the rocket, which reduces the rocket mass by $\Delta m$, and increases the rocket velocity by an amount \Delta \vec{v}_\text{ship}Δvship​.

This is the situation illustrated in the diagram below:

Because there is no outside force acting on the system (with the system taken to be the rocket and its fuel), the change in total momentum must be zero. If we subtract the total momentum before the combustion from the total momentum after combustion, we have

\begin{aligned} 0 &= \Delta p \\ &= p_t – p_0 \\ &= \Delta m \left(\vec{V}_\text{ship} – \vec{u}\right) + \left(\vec{V}_\text{ship} + \Delta \vec{v}_\text{ship} \right) \left(M – \Delta m\right) – \vec{V}_\text{ship} M \\ &= \vec{V}_\text{ship} \left(\Delta m – \Delta m\right) – \vec{u}\Delta m + M \Delta \vec{v}_\text{ship} + \vec{V}_\text{ship}\left(M – M\right) \\ &= -\vec{u} \Delta m + M \Delta \vec{v}_\text{ship}, \end{aligned}0​=Δp=pt​−p0​=Δm(Vship​−u)+(Vship​+Δvship​)(M−Δm)−Vship​M=Vship​(Δm−Δm)−uΔm+MΔvship​+Vship​(M−M)=−uΔm+MΔvship​,​

which yields the following relationship between the changes in mass and velocity:

\vec{u}\Delta m = M \Delta \vec{v}_\text{ship}.uΔm=MΔvship​.