### Rocket Equation

The Tsiolkovsky rocket equation, or ideal rocket equation, describes the motion of vehicles that follow the basic principle of a rocket: a device that can apply acceleration to itself (a thrust) by expelling part of its mass with high speed and move due to the conservation of momentum. The equation relates the delta-v (the maximum change of speed of the rocket if no other external forces act) with the effective exhaust velocity and the initial and final mass of a rocket (or other reaction engine).

For any such maneuver (or journey involving a number of such maneuvers):

$\Delta v = v_\text\left\{e\right\} \ln \frac \left\{m_0\right\} \left\{m_1\right\}$

where:

$m_0$ is the initial total mass, including propellant,
$m_1$ is the final total mass,
$v_\text\left\{e\right\}$ is the effective exhaust velocity,
$\Delta v\$ is delta-v - the maximum change of velocity of the vehicle (with no external forces acting),
$\ln$ refers to the natural logarithm function.

(The equation can also be written using the specific impulse instead of the effective exhaust velocity by applying the formula $v_\text\left\{e\right\} = I_\text\left\{sp\right\} \cdot g_0$ where $I_\text\left\{sp\right\}$ is the specific impulse expressed as a time period and $g_0$ is Standard Gravity.)

The equation is named after Konstantin Tsiolkovsky who independently derived it and published it in his 1903 work.[1] The equation had been derived earlier by the British mathematician William Moore.

## History

This equation was independently derived by Konstantin Tsiolkovsky towards the end of the 19th century and is widely known under his name or as the 'ideal rocket equation'. However, a recently discovered pamphlet "A Treatise on the Motion of Rockets" by William Moore[2] shows that the earliest known derivation of this kind of equation was in fact at the Royal Military Academy at Woolwich in England in 1813,[3] and was used for weapons research.

## 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 ($F_i\,$) to the change in linear momentum of the whole system (including rocket and exhaust) as follows:

$\sum F_i = \lim_\left\{\Delta t \to 0\right\} \frac\left\{P_2-P_1\right\}\left\{\Delta t\right\}$

where $P_1\,$ is the momentum of the rocket at time t=0:

$P_1 = \left\left( \left\{m + \Delta m\right\} \right\right)V$

and $P_2\,$ is the momentum of the rocket and exhausted mass at time $t=\Delta t\,$:

$P_2 = m\left\left(V + \Delta V \right\right) + \Delta m V_e$

and where, with respect to the observer:

 $V\,$ is the velocity of the rocket at time t=0 $V+\Delta V\,$ is the velocity of the rocket at time $t=\Delta t\,$ $V_e\,$ is the velocity of the mass added to the exhaust (and lost by the rocket) during time $\Delta t\,$ $m+\Delta m\,$ is the mass of the rocket at time t=0 $m\,$ is the mass of the rocket at time $t=\Delta t\,$

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

$V_e=V-v_e$

Solving yields:

$P_2-P_1=m\Delta V-v_e\Delta m\,$

and, using $dm=-\Delta m$, since ejecting a positive $\Delta m$ results in a decrease in mass,

$\sum F_i=m\frac\left\{dV\right\}\left\{dt\right\}+v_e\frac\left\{dm\right\}\left\{dt\right\}$

If there are no external forces then $\sum F_i=0$ and

$m\frac\left\{dV\right\}\left\{dt\right\}=-v_e\frac\left\{dm\right\}\left\{dt\right\}$

Assuming $v_e\,$ is constant, this may be integrated to yield:

$\Delta V\ = v_e \ln \frac \left\{m_0\right\} \left\{m_1\right\}$

or equivalently

$m_1=m_0 e^\left\{-\Delta V\ / v_e\right\}$      or      $m_0=m_1 e^\left\{\Delta V\ / v_e\right\}$      or      $m_0 - m_1=m_1 \left(e^\left\{\Delta V\ / v_e\right\} - 1\right)$

where $m_0$ is the initial total mass including propellant, $m_1$ the final total mass, and $v_e$ 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 $m_0 - m_1$ is the total mass of propellant expended, and hence:

$M_f = 1-\frac \left\{m_1\right\} \left\{m_0\right\}=1-e^\left\{-\Delta V\ / v_\text\left\{e\right\}\right\}$

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

$\Delta V\$ (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,[4] with $\Delta v$ again standing for the rocket's final velocity (after burning off all its fuel and being reduced to a rest mass of $m_1$) in the inertial frame of reference where the rocket started at rest (with the rest mass including fuel being $m_0$ initially), and $c$ standing for the speed of light in a vacuum:

$\frac\left\{m_0\right\}\left\{m_1\right\} = \left\left[\frac\left\{1 + \left\{\frac\left\{\Delta v\right\}\left\{c\right\}\right\}\right\}\left\{1 - \left\{\frac\left\{\Delta v\right\}\left\{c\right\}\right\}\right\}\right\right]^\left\{\frac\left\{c\right\}\left\{2v_e\right\}\right\}$

Writing $\frac\left\{m_0\right\}\left\{m_1\right\}$ as $R$, a little algebra allows this equation to be rearranged as

$\frac\left\{\Delta v\right\}\left\{c\right\} = \frac\left\{R^\left\{\frac\left\{2v_e\right\}\left\{c\right\}\right\} - 1\right\}\left\{R^\left\{\frac\left\{2v_e\right\}\left\{c\right\}\right\} + 1\right\}$

Then, using the identity $R^\left\{\frac\left\{2v_e\right\}\left\{c\right\}\right\} = \exp \left\left[ \frac\left\{2v_e\right\}\left\{c\right\} \ln R \right\right]$ (here "exp" denotes the exponential function; see also Natural logarithm as well as the "power" identity at Logarithm#Logarithmic identities) and the identity $\tanh x = \frac\left\{e^\left\{2x\right\} - 1\right\} \left\{e^\left\{2x\right\} + 1\right\}$ (see Hyperbolic function), this is equivalent to

$\Delta v = c \cdot \tanh \left\left(\frac \left\{v_e\right\}\left\{c\right\} \ln \frac\left\{m_0\right\}\left\{m_1\right\} \right\right)$

## Applicability

The rocket equation captures the essentials of rocket flight physics in a single short equation. It also holds true for rocket-like reaction vehicles whenever the effective exhaust velocity is constant; and can be summed or integrated when the effective exhaust velocity varies. It takes only the propulsive force of the engine into account, neglecting aerodynamic or gravitational forces on the vehicle. As such, it cannot be used by itself to accurately calculate the propellant requirement for launch from (or powered descent to) a planet with an atmosphere, and does not apply to non-rocket systems such as aerobraking, gun launches, space elevators, launch loops, or tether propulsion.

Also, the equation strictly applies only to a theoretical impulsive maneuver, in which the propellant is discharged and delta-v applied instantaneously. Orbital maneuvers involving significantly large delta-v (such as translunar injection) still are under the influence of gravity for the duration of the propellant discharge, which influences the vehicle's velocity. The equation is most accurately applied to relatively small delta-v maneuvers such as those involved in fine-tuning space rendezvous, or mid-course corrections in translunar or interplanetary flights where the gravity field is relatively weak.

Nevertheless, the equation is useful for estimating the propellant requirement to perform a given orbital maneuver, assuming a required delta-v. To achieve a large delta-v, either $m_0$ must be huge (growing exponentially as delta-v rises), or $m_1$ must be tiny, or $v_e$ must be very high, or some combination of all of these. In practice, very high delta-v has been achieved by a combination of

• very large rockets (increasing $m_0$ with more fuel)
• staging (decreasing $m_1$ by throwing out the previous stage)
• very high exhaust velocities (increasing $v_e$)

## Examples

Assume an exhaust velocity of 4,500 meters per second (15,000 ft/s) and a $\Delta v$ of 9,700 meters per second (32,000 ft/s) (Earth to LEO, including $\Delta v$ to overcome gravity and aerodynamic drag).

• Single stage to orbit rocket: $1-e^\left\{-9.7/4.5\right\}$ = 0.884, therefore 88.4% of the initial total mass has to be propellant. The remaining 11.6% is for the engines, the tank, and the payload. In the case of a space shuttle, it would also include the orbiter.
• Two stage to orbit: suppose that the first stage should provide a $\Delta v$ of 5,000 meters per second (16,000 ft/s); $1-e^\left\{-5.0/4.5\right\}$ = 0.671, therefore 67.1% of the initial total mass has to be propellant to the first stage. The remaining mass is 32.9%. After disposing of the first stage, a mass remains equal to this 32.9%, minus the mass of the tank and engines of the first stage. Assume that this is 8% of the initial total mass, then 24.9% remains. The second stage should provide a $\Delta v$ of 4,700 meters per second (15,000 ft/s); $1-e^\left\{-4.7/4.5\right\}$ = 0.648, therefore 64.8% of the remaining mass has to be propellant, which is 16.2%, and 8.7% remains for the tank and engines of the second stage, the payload, and in the case of a space shuttle, also the orbiter. Thus together 16.7% is available for all engines, the tanks, the payload, and the possible orbiter.

## Stages

In the case of sequentially thrusting rocket stages, the equation applies for each stage, where for each stage the final mass in the equation is the total mass of the rocket after discarding the previous stage, and the initial mass in the equation is the total mass of the rocket just before discarding the stage concerned. For each stage the specific impulse may be different.

For example, if 80% of the mass of a rocket is the fuel of the first stage, and 10% is the dry mass of the first stage, and 10% is the remaining rocket, then



\begin{align} \Delta v \ & = v_\text{e} \ln { 100 \over 100 - 80 }\\

          & = v_\text{e} \ln 5 \\
& = 1.61 v_\text{e}. \\


\end{align}

With three similar, subsequently smaller stages with the same $v_e$ for each stage, we have

$\Delta v \ = 3 v_\text\left\{e\right\} \ln 5 \ = 4.83 v_\text\left\{e\right\}$

and the payload is 10%*10%*10% = 0.1% of the initial mass.

A comparable SSTO rocket, also with a 0.1% payload, could have a mass of 11.1% for fuel tanks and engines, and 88.8% for fuel. This would give

$\Delta v \ = v_\text\left\{e\right\} \ln\left(100/11.2\right) \ = 2.19 v_\text\left\{e\right\}.$

If the motor of a new stage is ignited before the previous stage has been discarded and the simultaneously working motors have a different specific impulse (as is often the case with solid rocket boosters and a liquid-fuel stage), the situation is more complicated.

## Common misconceptions

When viewed as a variable-mass system, a rocket cannot be directly analyzed with Newton's second law of motion because the law is valid for constant-mass systems only.[5][6][7] It can cause confusion that the Tsiolkovsky rocket equation looks similar to the relativistic force equation $F = dp/dt = m \; dv/dt + v \; dm/dt$. Using this formula with $m\left(t\right)$ as the varying mass of the rocket seems to derive Tsiolkovsky rocket equation, but this derivation is not correct. Notice that the effective exhaust velocity $v_e$ doesn't even appear in this formula.

A simple counter example is to consider a rocket travelling with a constant velocity $v$ with two maneuvering thrusters pointing out on either side, with both firing such that their forces cancel each other out. In such a case the rocket would be losing mass and an incorrect application of $F = dp/dt$ would result in a non-zero but non-accelerating force, leading to nonsensical answers. However when including the mass of the exhaust products, the system becomes a closed system where the derivation is correct, as $F = dp/dt = 0$ in the x direction, where x is the direction one thruster is pointing and -x is the direction of the other thruster.