Tsiolkovsky Rocket Equation - Derivation

Derivation

Consider the following system:

In the following derivation, "the rocket" is taken to mean "the rocket and all of its unburned propellant".

Newton's second law of motion relates external forces to the change in linear momentum of the system as follows:

where is the momentum of the rocket at time t=0:

and is the momentum of the rocket and exhausted mass at time :

and where, with respect to the observer:

is the velocity of the rocket at time t=0

is the velocity of the rocket at time

is the velocity of the mass added to the exhaust (and lost by the rocket) during time

is the mass of the rocket at time t=0

is the mass of the rocket at time

The velocity of the exhaust in the observer frame is related to the velocity of the exhaust in the rocket frame by (since exhaust velocity is in the negative direction)

Solving yields:

and, using, since ejecting a positive results in a decrease in mass,

If there are no external forces then and

Assuming is constant, this may be integrated to yield:

or equivalently

or or

where is the initial total mass including propellant, the final total mass, and the velocity of the rocket exhaust with respect to the rocket (the specific impulse, or, if measured in time, that multiplied by gravity-on-Earth acceleration).

The value is the total mass of propellant expended, and hence:

where is the mass fraction (the part of the initial total mass that is spent as reaction mass).

(delta v) is the integration over time of the magnitude of the acceleration produced by using the rocket engine (what would be the actual acceleration if external forces were absent). In free space, for the case of acceleration in the direction of the velocity, this is the increase of the speed. In the case of an acceleration in opposite direction (deceleration) it is the decrease of the speed. Of course gravity and drag also accelerate the vehicle, and they can add or subtract to the change in velocity experienced by the vehicle. Hence delta-v is not usually the actual change in speed or velocity of the vehicle.

If special relativity is taken into account, the following equation can be derived for a relativistic rocket, with again standing for the rocket's final velocity (after burning off all its fuel and being reduced to a rest mass of ) in the inertial frame of reference where the rocket started at rest (with the rest mass including fuel being initially), and standing for the speed of light in a vacuum:

Writing as, a little algebra allows this equation to be rearranged as

Then, using the identity (here "exp" denotes the exponential function; see also Natural logarithm as well as the "power" identity at Logarithm#Logarithmic identities) and the identity (see Hyperbolic function), this is equivalent to