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 Derivation of the ideal rocket equation which describes the change in
 velocity as a function of the exit velocity of the rocket and the change
in mass of the rocket during the burn.

The forces on a rocket change dramatically during a typical flight. This figure shows a derivation of the change in velocity during powered flight while accounting for the changing mass of the rocket. During powered flight the propellants of the propulsion system are constantly being exhausted from the nozzle. As a result, the weight of the rocket is constantly changing. In this derivation, we are going to neglect the effects of aerodynamic lift and drag. We can add these effects to the final answer.

Let us begin with Newton's second law of motion, shown in blue on the figure:

d (M u) / dt = F net

where M is the mass of the rocket, u is the velocity of the rocket, F net is the net external force on the rocket and the symbol d / dt denotes that this is a differential equation in time t. The only external force which we will consider is the thrust from the propulsion system.

On the web page describing the specific impulse, the thrust equation is given by:

F = mdot * Veq

where mdot is the mass flow rate, and Veq is the equivalent exit velocity of the nozzle which is defined to be:

Veq = V exit + (p exit - p0) * Aexit / mdot

where V exit is the exit velocity, p exit is the exit pressure, p0 is the free stream pressure, and A exit is the exit area of the nozzle. Veq is also related to the specific impulse Isp:

Veq = Isp * g0

where g0 is the gravitational constant. m dot is mass flow rate and is equal to the change in the mass of the propellants mp on board the rocket:

mdot = d mp / dt

Substituting the expression for the thrust into the motion equation gives:

d (M u) / dt = V eq * d mp / dt

d (M u) = Veq d mp

Expanding the left side of the equation:

M du + u dM = Veq d mp

Assume we are moving with the rocket, then the value of u is zero:

M du = Veq d mp

Now, if we consider the instantaneous mass of the rocket M, the mass is composed of two main parts, the empty mass me and the propellant mass mp. The empty mass does not change with time, but the mass of propellants on board the rocket does change with time:

M(t) = me + mp (t)

Initially, the full mass of the rocket mf contains the empty mass and all of the propellant at lift off. At the end of the burn, the mass of the rocket contains only the empty mass:

M initial = mf = me + mp

M final = me

The change on the mass of the rocket is equal to the change in mass of the propellant, which is negative, since propellant mass is constantly being ejected out of the nozzle:

dM = - d mp

If we substitute this relation into the motion equation:

M du = - Veq dM

du = - Veq dM / M

We can now integrate this equation:

delta u = - Veq ln (M)

where delta represents the change in velocity, and ln is the symbol for the natural logarithmic function. The limits of integration are from the initial mass of the rocket to the final mass of the rocket. Substituting for these values we obtain:

delta u = Veq ln (mf / me)

This equation is called the ideal rocket equation. There are several additional forms of this equation which we list here: Using the definition of the propellant mass ratio MR

MR = mf / me

delta u = Veq * ln (MR)

or in terms of the specific impulse of the engine:

delta u = Isp * g0 * ln (MR)

If we have a desired delta u for a maneuver, we can invert this equation to determine the amount of propellant required:

MR = exp (delta u / (Isp * g0) )

where exp is the exponential function.

If you include the effects of gravity, the rocket equation becomes:

delta u = Veq ln (MR) - g0 * tb

where tb is the time for the burn.

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Editor: Tom Benson
NASA Official: Tom Benson
Last Updated: June 12, 2014 Oct 22 2015

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