Return to Missiles## A derivation of "the rocket equation" from Newton's laws.

byPeter Baum

peter underscore baum at verizon dot netRegarding the relationship between rocket velocity and exhaust velocity: The force driving a rocket forward is an example of Newton's Third Law of Motion

(to every action (force) there is always an equal and contrary reaction (force)). The gas goes in one direction and the rocket in the opposite direction.It may be helpful in getting an intuitive sense of what is going on by considering the situation to be one in which someone in a rocket is throwing out little masses of gas, golf balls, rocks, or whatever. The rocket is going at a certain velocity and the object that is thrown out has the same velocity as the rocket before it is thrown. When the object is thrown out the rear of the rocket, it will increase the velocity of the rocket. It doesn't matter how fast the rocket is going when the object is expelled; it is the force needed to achieve the relative exit speed for a given mass of gas that increases the speed of the rocket.

Roughly speaking, it will be the rocket mass times the rocket speed that is equal to the gas mass times gas speed by the general principle of

The Conservation of Momentum. If, for example, the mass of the gases is large relative to the mass of the rocket, then the rocket can eventually travel at a speed greater than the exhaust gases. This is only roughly speaking because the fuel that becomes exhaust gas late in the flight will have previously been accelerated by exhaust gases earlier in the flight.To make these intuitive ideas more precise, the following shows how to derive the rocket equation. (There are two places in the derivation that I found to be a little tricky. The two reference books I looked at glossed over these points so I'll do my best to explain the derivation in a way that makes sense to me. I welcome general comments in this regard.)

First I will develop expressions for the change in momentum that occurs.

LetM(t)be the mass of the rocket at timet. Now we want an expression that gives the momentum change for the rocket. We have to be careful at this point. Supposedmis a small quantity of exhaust gas that is ejected after a small time intervaldthas elapsed. Then the "rocket" had massM(t)at timetand has massM(t) - dmat timet + dt. The change in momentum is caused by the exhaust of the massdm. That change is a force that is applied toM(t) - dmto accelerate the "rocket", not toM(t)since at the end of timet + dt, the "rocket" has massM(t) - dm.Let

V(t)= the velocity of the rocket at time t

V(t) + dV= the velocity of the rocket at time t +dt.

U= the exhaust speed (considered to be a constant)

Momentum of "rocket" less dmTime Momentum t(M(t) - dm) V(t)t + dt(M(t) - dm) (V(t) + dV)

Momentum of dmTime Momentum tdm V(t)t + dtdm (V(t) - U)In the expression for the momentum of dm, the velocity is relative to the earth, although for the purpose of finding the change in momentum, one could perhaps more easily work with velocity relative to the rocket.

By subtracting the momentum at time

tfrom the momentum at timet+dtthe change in momentum can be determined:

rocket momentum change =dV(M(t) - dm)

gas momentum change =- dm UNotice that the rocket gains momentum (expressed as a positive value) and the gas loses momentum and is expressed as a negative value. Since momentum is conserved,

dV(M(t) - dm) - dm U = 0, which can be written as

dV = (dm U)/(M(t) - dm)Since

dM(t)is the change in the mass of the "rocket" between timetandt + dtand is the loss of massdm, we have

dM(t) = - dmso that from the previous equation

dV = -U dM(t)/(M(t) +dM(t))The next step is to integrate both sides of this equation. The left side is integrated with respect to velocity and the right side with respect to mass. Integrating

dV(with respect to velocity) fromVto_{0}Vgives the value_{final}V._{final}- V_{0}It will be

-U dM(t)/(M(t) +dM(t))that will be integrated with respect to mass fromMto_{rocket+fuel}M. Note that_{rocket}Vis the velocity when the rocket is fully loaded with fuel and the mass is_{0}Mand that_{rocket+fuel}Vis the velocity when the fuel is exhausted so that the rocket has mass_{final}M._{rocket}Assume for the moment that the integral of

-U dM(t)/(M(t) + dM(t))is the same as the integral of-U dM(t)/M(t). (Intuitively, we are looking at the area under the curve of1/(x+dx)where in the process we makedxarbitrarily small so that we are looking at the area under the curve of1/xin the limit. In some presentations, this simplification is done before the integration is performed.)If we assume a constant exhaust speed,

Uis a constant, the indefinite integral is-U ln(M(t))and the value in the range specified is

-U ln(M_{rocket}) + U ln(M_{rocket+fuel}) = -U ln(M_{rocket}/ M_{rocket+fuel})

and combining with the result for velocity give

Vwhich can be rewritten as_{final}- V_{0}= -U ln(M_{rocket}/ M_{rocket+fuel})

V_{final}= V_{0}+ U ln( M_{rocket+fuel}/ M_{rocket})Since the expression is in terms of velocity, we can also add the expression

-gtto the right hand side of the equation, wheregis the constant of acceleration by gravity andtthe time. This accounts for the force of gravity if the rocket is going straight up. It assumesgis constant; a more complex expression is needed if we account for the fact that the force of gravity actually varies with altitude.I hope this explanation is helpful. I realize there is still room for more detail and precision and welcome further elaboration by others.

-P

Originated January 7, 1999

Updated January 6, 2004

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