Tsiolkovsky rocket equation

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Tsiolkovsky's rocket equation, named after Konstantin Tsiolkovsky who independently derived it and published in his 1903 work,^[1] considers the principle of a rocket: a device that can apply an acceleration to itself (a thrust) by expelling part of its mass with high speed in the opposite direction, due to the conservation of momentum.

1 Derivation
2 History
3 Stages
4 Energy
5 Examples
6 See also
7 References

[edit] Derivation

It says that for any maneuver or any journey involving a number of maneuvers:

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

or equivalently

$m_1=m_0 e^{-\Delta v\ / v_e}$ or $m_0=m_1 e^{\Delta v\ / v_e}$

where $m 0$ is the initial total mass, and $m 1$ 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).

$1-\frac {m_1} {m_0}=1-e^{-\Delta v\ / v_e}$ 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.

The equation is obtained by integrating the conservation of momentum equation

m d v = v e d m

for a simple rocket that emits mass at a constant velocity ( $d m$ is here the reaction mass; if it is the change of the rocket mass then there is a minus sign in the latter equation).

Although an extreme simplification, the rocket equation captures the essentials of rocket flight physics in a single short equation. It also happens that delta-v is one of the most important quantities in orbital mechanics, that quantifies how difficult it is to perform a given orbital maneuver.

Clearly, 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, this has been achieved by using very large rockets (increasing $m 0$ ), with multiple stages (decreasing $m 1$ ), and rockets with very high exhaust velocities. The Saturn V rockets used in the Apollo space program and the ion thrusters used in long-distance unmanned probes are good examples of this.

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 $v_e \ln 2 \approx 0.693 v_e$

[edit] History

This equation was independently derived by Konstantin Tsiolkovsky towards the end of the 19th century and is widely known under this name. 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 in Royal Military Academy at Woolwich in England in 1813, and was used for weapons research.

[edit] Stages

In the case of subsequently 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

$\Delta v \ = v_e \ln 5 \ = 1.61 v_e$

With three similar, subsequently smaller stages we have

$\Delta v \ = 3 v_e \ln 5 \ = 4.83 v_e$

and the payload is 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

$\Delta v \ = v_e \ln(100/11.1) \ = 2.20 v_e$

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

[edit] Energy

In the ideal case $m 1$ is useful payload and $m 0 - m 1$ is reaction mass (this corresponds to empty tanks having no mass, etc.). The energy required can simply be computed as

$\frac{1}{2}(m_0-m_1)v_e^2$

Seemingly this is just the kinetic energy of the reaction mass and not the kinetic energy required for the payload, but if, for example, $v e$ =10 km/s and the speed of the rocket is 3 km/s, then the speed of the reaction mass changes only from 3 to 7 km/s; the energy thus "saved" corresponds to the increase of the specific kinetic energy (kinetic energy per kg) for the rocket. In general:

$d\left(\frac{1}{2}v^2\right)=vdv=vv_edm/m=\frac{1}{2}\left(v_e^2-(v-v_e)^2+v^2\right)dm/m$

Thus the specific energy gain of the rocket in any small time interval is the energy gain of the rocket including the remaining fuel, divided by its mass, where the energy gain is equal to the energy produced by the fuel minus the energy gain of the reaction mass. The larger the speed of the rocket, the smaller the energy gain of the reaction mass; if the rocket speed is more than half of the exhaust speed the reaction mass even loses energy on being expelled, to the benefit of the energy gain of the rocket; the larger the speed of the rocket, the larger the energy loss of the reaction mass.

We have

$\Delta \epsilon = \int v\, d (\Delta v)$

where $ε$ is the specific energy of the rocket (potential plus kinetic energy) and $Δ v$ is a separate variable, not just the change in $v$ . In the case of using the rocket for deceleration, i.e. expelling reaction mass in the direction of the velocity, $v$ should be taken negative.

The formula is for the ideal case again, with no energy lost on heat, etc. The latter causes a reduction of thrust, so it is a disadvantage even when the objective is to lose energy (deceleration).

If the energy is produced by the mass itself, as in a chemical rocket, the fuel value has to be : $\frac{1}{2}v_e^2$ , where for the fuel value also the mass of the oxidizer has to be taken into account. A typical value is $v e$ = 4.5 km/s, corresponding to a fuel value of 10.1 MJ/kg. The actual fuel value is higher, but much of the energy is lost as waste heat in the exhaust that the nozzle was unable to extract.

The required energy is

$E = \frac{1}{2}m_1\left(e^{\Delta v\ / v_e}-1\right)v_e^2$

Conclusions:

for $\Delta v \ll v_e$ we have $E\approx \frac{1}{2}m_1 v_e \Delta v$
for a given $Δ v$ , the minimum energy is needed if $v e = 0.6275Δ v$ , requiring an energy of

E = 0.772 m 1 (Δ v) 2

Starting from zero speed this is 54.4 % more than just the kinetic energy of the payload. Starting from a nonzero speed the required energy may be less than the increase in energy in the payload. This can be the case when the reaction mass has a lower speed after being expelled than before. For example, from a LEO of 300 km altitude to an escape orbit is an increase of 29.8 MJ/kg, which, using a specific impulse of 4.5 km/s, has a net cost of 20.6 MJ/kg (

Δ v

= 3.20 km/s; the energies are per kg payload).

This optimization does not take into account the masses of various kinds of engines.

Also, for a given objective such as moving from one orbit to another, the required $Δ v$ may depend greatly on the rate at which the engine can produce $Δ v$ and maneuvers may even be impossible if that rate is too low. For example, a launch to LEO normally requires a $Δ v$ of ca. 9.5 km/s (mostly for the speed to be acquired), but if the engine could produce $Δ v$ at a rate of only slightly more than g, it would be a slow launch requiring altogether a very large $Δ v$ (think of hovering without making any progress in speed or altitude, it would cost a $Δ v$ of 9.8 m/s each second). If the possible rate is only $g$ or less, the maneuver can not be carried out at all with this engine.

The power is given by

$P= \frac{1}{2} m a v_e = \frac{1}{2}F v_e$

where $F$ is the thrust and $a$ the acceleration due to it. Thus the theoretically possible thrust per unit power is 2 divided by the specific impulse in m/s. The thrust efficiency is the actual thrust as percentage of this.

If e.g. solar power is used this restricts $a$ ; in the case of a large $v e$ the possible acceleration is inversely proportional to it, hence the time to reach a required delta-v is proportional to $v e$ ; with 100% efficiency:

for $\Delta v \ll v_e$ we have $t\approx \frac{m v_e \Delta v}{2P}$

Examples:

power 1000 W, mass 100 kg, $Δ v$ = 5 km/s, $v e$ = 16 km/s, takes 1.5 months.
power 1000 W, mass 100 kg, $Δ v$ = 5 km/s, $v e$ = 50 km/s, takes 5 months.

Thus $v e$ should not be too large.

[edit] Examples

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

SSTO 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 $Δ v$ 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. The remaining mass is 32.9 %. After deposing 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 $Δ v$ 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.

[edit] See also

[edit] References

^ К. Э. Циолковский, Исследование мировых пространств реактивными приборами, 1903. It is available online here
^ Johnson W., "Contents and commentary on William Moore's a treatise on the motion of rockets and an essay on naval gunnery", International Journal of Impact Engineering, Volume 16, Number 3, June 1995, pp. 499-521