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The rocket equation

We have now seen why rockets have a force acting on them when an onboard motor fires.

raket_1

The gaseous combustion products are pushed backwards at high speed. This acceleration of gaseous mass creates a force (K), in such a way that the particles gain speed (v_\text{e}) out through the narrow channel. The counter force (F) pushes the rocket forwards.

Let us take a closer look at the momentum of the system and how it is transferred. Imagine the same rocket with the motor firing. The rocket has a mass of M+\Delta M. It releases a small part of that mass, called \Delta M, over a short time span of \Delta t. The momentum of the system at time t=0 is P=(M+\Delta M)V, where V is the velocity of the rocket. When the small mass \Delta M is accelerated from the rocket to a velocity of v_\text{e}, the momentum of the system, that is the rocket with momentum P_{r,\Delta t}=M(V + \Delta V) and the momentum of the small mass P_{\Delta M}=-\Delta M\, v_\text{e}, must be the same as at t=0. This means that the rocket has lost some mass but gained velocity.
The change in momentum of the system, that is, the rocket and the small expelled mass, is

\Delta P=M\,\Delta V-v_\text{e} \hspace{0.1em} \Delta M.

Since there are no external forces on the system, the change of momentum on the system must be zero, and hence M\,\Delta V=v_\text{e} \,\Delta M. For small time periods \text{d}t, we kan write d’s instead of deltas:

M \, \text{d}V=-v_\text{e} \, \text{d}M

Note that \text{d}M=-\Delta M, since the overall mass of the rocket decreases when \Delta M is ejected.

Remembering Newton’s second law, and that the sum of the forces acting on the rocket equals the change in time of momentum, we can write

\displaystyle \Sigma F = \frac{\text{d}P}{\text{d}t} = \frac{M \text{d}V + v_\text{e} \hspace{0.1em} \text{d}M}{\text{d}t} = 0,

rocketprinciple

The small mass dm is pushed from the rocket to a velocity of v_\text{e}. The rocket has lost some mass and gained velocity.

which gives us

\displaystyle M \hspace{0.1em} \frac{\text{d}V}{\text{d}t} = -v_\text{e} \frac{\text{d}M}{\text{d}t}.

Integrating both sides of this equation,

\displaystyle \int^{V_\text{f}}_{V_\text{i}} \text{d}V = -v_\text{e} \int^{M_\text{f}}_{M_\text{i}} \frac{1}{M} \ \text{d}M,

we have the following relation between change in velocity and mass:

\displaystyle \Delta V = v_\text{e} \hspace{0.1em} \ln \left( \frac{M_\text{i}}{M_\text{f}} \right),

where M_\text{i} is the initial mass before the rocket motor starts to burn and M_\text{f} is the final mass of the rocket after the burn stopped. The difference between the two, \Delta M, is the fuel burned. This equation is often called the (ideal) rocket equation, or also sometimes the Tsiolkovsky rocket equation, after one of the scientists who first derived it.

In the integration of the equations we assumed v_\text{e} to be a constant in time. It turns out that this is usually a very good approximation. The rocket motor takes in a certain amount of fuel and oxidizer and accelerates the molecules from the chemical reaction to a given velocity. Note that the actual exhaust velocity v_\text{e} has a second term that comes from the pressure difference from the rocket pressure and the ambient pressure. We have neglected v_\text{e} in a moment.

We must take a look at other simplifications we have done. We assumed that there were no other forces acting on the system. This means that the rocket equation is not valid on a launch vehicle since the gravitational and aerodynamically forces is non-negligible. Also we assumed that the burn happens in a short timespan. This is a good approximation for orbital manoeuvres of satellites, which we will discuss in the next chapter.

See a quite interesting TEDx talk on the rocket equation and the difficulties of reaching space here:

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This article is a part of a pre-course program used by NAROM in different courses, for example Fly a Rocket!