AstrodynamicsOrbital mechanics

Kepler's laws

Tsiolkovsky rocket equation

Vis-viva equation

Payload fraction

Propellant mass fraction

Mass ratio

Rocket mass ratios versus final

velocity calculated from the rocket

equation

Tsiolkovsky rocket equationFrom Wikipedia, the free encyclopedia

The Tsiolkovsky rocket equation, or ideal rocketequation, is a mathematical equation that relates the delta-v(the maximum change of speed of the rocket if no otherexternal forces act) with the effective exhaust velocity and theinitial and final mass of a rocket or other reaction engine.

The equation is named after Konstantin Tsiolkovsky who

independently derived it and published it in his 1903 work.[1]

It considers the principle of a rocket: a device that can applyacceleration to itself (a thrust) by expelling part of its masswith high speed in the opposite direction due to theconservation of momentum.

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

where:

m0 is the initial total mass, including propellant.

m1 is the final total mass.

ve is the effective exhaust velocity. ( where Isp is the

specific impulse expressed as a time period)

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.

Contents

1 History

2 Stages

3 Energy

4 Derivation

5 Examples

6 Applicability

7 See also

8 References

9 External links

Equations

Efficiency measures

History

This equation was independently derived by Konstantin Tsiolkovsky towards the end of the 19th century and iswidely known under this name and 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.

Stages

In the case of sequentially thrusting rocket stages, the equation applies for each stage, where for each stage theinitial mass in the equation is the total mass of the rocket after discarding the previous stage, and the final mass in theequation is the total mass of the rocket just before discarding the stage concerned. For each stage the specificimpulse 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

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

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, and88.9% for fuel. This would give

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

Energy

Main article: Spacecraft propulsion

In the ideal case m1 is useful payload and m0 − m1 is reaction mass (this corresponds to empty tanks having no

mass, etc.). The energy required can simply be computed as

This corresponds to the kinetic energy the expelled reaction mass would have at a speed equal to the exhaustspeed. If the reaction mass would have to be accelerated from zero speed to the exhaust speed all energyproduced would go into the reaction mass and nothing would be left for kinetic energy gain by the rocket andpayload. However, if the rocket already moves and accelerates (the reaction mass is expelled in the directionopposite to the direction in which the rocket moves) less kinetic energy is added to the reaction mass. To see this,if, for example, ve=10 km/s and the speed of the rocket is 3 km/s, then the speed of a small amount of expended

reaction mass changes from 3 forwards to 7 km/s rearwards. Thus, while the energy required is 50 MJ per kgreaction mass, only 20 MJ is used for the increase in speed of the reaction mass. The remaining 30 MJ is theincrease of the kinetic energy of the rocket and payload.

In general:

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

We have

where ε is the specific energy of the rocket (potential plus kinetic energy) and Δv is a separate variable, not just thechange in v. In the case of using the rocket for deceleration, i.e. expelling reaction mass in the direction of thevelocity, 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 itis 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 , where for the fuelvalue also the mass of the oxidizer has to be taken into account. A typical value is ve = 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 theexhaust that the nozzle was unable to extract.

The required energy E is

Conclusions:

for we have

for a given Δv, the minimum energy is needed if ve = 0.6275Δv, requiring an energy of

E = 0.772m1(Δv)2.

Starting from zero speed this is 54.4 % more than just the kinetic energy of the payload. In this optimal case the

initial mass is 4.92 times the final mass.

These results apply for a fixed exhaust speed.

Due to the Oberth effect, and starting from a nonzero speed the required potential energy needed from thepropellant may be less than the increase in energy in the vehicle and payload. This can be the case when thereaction mass has a lower speed after being expelled than before- rockets are able to liberate some or all of theinitial kinetic energy of the propellant.

Also, for a given objective such as moving from one orbit to another, the required Δv may depend greatly on therate at which the engine can produce Δv and maneuvers may even be impossible if that rate is too low. Forexample, a launch to LEO normally requires a Δv of ca. 9.5 km/s (mostly for the speed to be acquired), but if theengine could produce Δv at a rate of only slightly more than g, it would be a slow launch requiring altogether a verylarge Δv (think of hovering without making any progress in speed or altitude, it would cost a Δv of 9.8 m/s eachsecond). 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

where F is the thrust and a the acceleration due to it. Thus the theoretically possible thrust per unit power is 2divided 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 ve the possible acceleration is inversely proportional

to it, hence the time to reach a required delta-v is proportional to ve; with 100% efficiency:

for we have

Examples:

power 1000 W, mass 100 kg, Δv= 5 km/s, ve= 16 km/s, takes 1.5 months.

power 1000 W, mass 100 kg, Δv= 5 km/s, ve= 50 km/s, takes 5 months.

Thus ve should not be too large.

Derivation

Consider the following system:

In the following derivation, "the rocket" is taken to mean "the rocket and all of its unburned propellant". Newton'ssecond 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 Ve in the observer frame is related to the velocity of the exhaust in the rocket frame veby

Ve = V + ve

Solving yields:

and

If there are no external forces then and

Assuming is constant, this may be integrated to yield:

or equivalently

or or

where m0 is the initial total mass including propellant, m1 the final total mass, and ve the velocity of the rocket

exhaust with respect to the rocket (the specific impulse, or, if measured in time, that multiplied by gravity-on-Earthacceleration). The value m0 − m1 is the total mass of propellant expended.

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 inthe 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 velocityexperienced by the vehicle. Hence delta-v is not usually the actual change in speed or velocity of the vehicle.

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

Clearly, to achieve a large delta-v, either m0 must be huge (growing exponentially as delta-v rises), or m1 must be

tiny, or ve must be very high, or some combination of all of these.

In practice, this has been achieved by using very large rockets (increasing m0), with multiple stages (decreasing

m1), 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 m1, not as a function of time, but as a function of

delta-v produced. The delta-v that is the corresponding "half-life" is

Examples

Assume an exhaust velocity of 4.5 km/s and a Δv 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 spaceshuttle, it would also include the orbiter.

Two stage to orbit: suppose that the first stage should provide a Δv 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 massis 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 Δv of 4.7 km/s; 1 − e − 4.7 / 4.5 = 0.648, therefore 64.8% of theremaining 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.

Applicability

The rocket equation 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.

However, it does not apply to other technologies such as gun launches, space elevators, launch loops, tetherpropulsion and air-breathing engines.

See also

Delta-v

Delta-v budget

Oberth effect applying delta-v in a gravity well increases the final velocity

Specific impulseSpacecraft propulsion

Mass ratio

Working mass

Relativistic rocket

Reversibility of orbits

References

1. ^ К. Э. Циолковский, Исследование мировых пространств реактивными приборами, 1903. It is available onlinehere (http://epizodsspace.airbase.ru/bibl/dorev-knigi/ciolkovskiy/sm.rar) in a RARed PDF

2. ^ Moore, William; of the Military Academy at Woolwich (1813). A Treatise on the Motion of Rockets. To which isadded, An Essay on Naval Gunnery. London: G. and S. Robinson.

3. ^ Johnson, W. (1995). "Contents and commentary on William Moore's a treatise on the motion of rockets and anessay on naval gunnery" (http://www.sciencedirect.com/science/article/B6V3K-3Y5FP5P-

11/2/c3e98a6cec8f083c93dc4e4e157282bb) . International Journal of Impact Engineering 16 (3): 499–521.doi:10.1016/0734-743X(94)00052-X (http://dx.doi.org/10.1016%2F0734-743X%2894%2900052-X) . ISSN 0734-743X (http://www.worldcat.org/issn/0734-743X) . http://www.sciencedirect.com/science/article/B6V3K-3Y5FP5P-11/2/c3e98a6cec8f083c93dc4e4e157282bb.

External links

How to derive the rocket equation (http://ed-thelen.org/rocket-eq.html)

Relativity Calculator - Learn Tsiolkovsky's rocket equations

Relativity Calculator - Learn Tsiolkovsky's rocket equations(http://www.relativitycalculator.com/rocket_equations.shtml)

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Categories: Astrodynamics | Equations

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