Tsiolkovsky rocket equation

Astrodynamics
Orbital mechanics
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Efficiency measures Payload fraction Propellant mass fraction Mass ratio

Rocket mass ratios versus final velocity calculated from the rocket equation

The Tsiolkovsky rocket equation, or ideal rocket equation is an equation that is useful for considering vehicles that follow the basic principle of a rocket: where a device that can apply acceleration to itself (a thrust) by expelling part of its mass with high speed and moving due to the conservation of momentum. Specifically, it is a mathematical equation that 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):

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

where:

$m_0$ is the initial total mass, including propellant,

$m_1$ is the final total mass,

$v_\text{e}$ is the effective exhaust velocity ($v_\text{e} = I_\text{sp} \cdot g_0$ where $I_\text{sp}$ is the specific impulse expressed as a time period and $g_0$ is the gravitational constant),

$\Delta v\$ is delta-v - the maximum change of speed of the vehicle (with no external forces acting).

Units used for mass or velocity do not matter as long as they are consistent.

The equation is named after Konstantin Tsiolkovsky who independently derived it and published it in his 1903 work.

History

This equation was independently derived by Konstantin Tsiolkovsky towards the end of the 19th century and is widely known under this name and ideal rocket equation. However a recently discovered pamphlet "A Treatise on the Motion of Rockets" by William Moore 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, 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 system as follows:

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

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

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

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

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

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

$V_e=V+v_e$

Solving yields:

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

and

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

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

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

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

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

or equivalently

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

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:

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

where $M_f$ is the 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.

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 does not apply to non-rocket systems such as gun launches, space elevators, launch loops, tether propulsion and air-breathing engines.

Delta-v is an important quantity in orbital mechanics. It quantifies how difficult it is to perform a given orbital maneuver. 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 $v_\text{e}$ must be very high, or some combination of all of these.

In practice, very-high delta-v has been achieved by a combination of 1) using very large rockets (increasing $m_0$), 2) using multiple stages (decreasing $m_1$), and 3) using rockets with very high exhaust velocities. The Saturn V rockets used in the Apollo space program is an example of a large and multistaged system. The ion thrusters used in long-distance unmanned probes is an example of a high exhaust velocity rocket.

The rocket equation shows a kind of "exponential decay" of mass $m_1$, not as a function of time, but as a function of delta-v produced. The delta-v that is the corresponding "half-life" is $v_\text{e} \ln 2 \approx 0.693 v_\text{e}$

Examples

Assume an exhaust velocity of 4.5 km/s and a $\Delta v$ of 9.7 km/s (Earth to LEO).

• Single stage to orbit rocket: $1-e^{-9.7/4.5}$ = 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 5.0 km/s; $1-e^{-5.0/4.5}$ = 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 4.7 km/s; $1-e^{-4.7/4.5}$ = 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 initial mass in the equation is the total mass of the rocket after discarding the previous stage, and the final 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

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

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% for fuel tanks and engines, and 88.9% for fuel. This would give

$\Delta v \ = v_\text{e} \ln(100/11.1) \ = 2.20 v_\text{e}.$

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

Because this is a variable-mass system, Newton's second law of motion cannot directly be applied because it is valid for constant mass systems only. It can cause confusion that the Tsiolkovsky rocket equation is similar to the relativistic force equation $F = dp/dt = m \; dv/dt + v \; dm/dt$. Using this formula with $m(t)$ as the varying mass of the rocket is mathematically equivalent to the derived Tsiolkovsky rocket equation, but this derivation is not correct.

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.

Any system with a non-constant mass must be treated as a variable-mass system.

See also

• Delta-v
• Delta-v budget
• Oberth effect applying delta-v in a gravity well increases the final velocity
• Specific impulse
• Spacecraft propulsion
• Mass ratio
• Working mass
• Relativistic rocket
• Reversibility of orbits
• Variable-mass systems

External links

• How to derive the rocket equation
• Relativity Calculator - Learn Tsiolkovsky's rocket equations
• Tsiolkovsky's rocket equations plot and calculator