Talk:Tsiolkovsky rocket equation

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[edit] Derivations

Perhaps we could have more mathematically rigorous derivations for the various formulae? -- The Anome 10:51, 10 Oct 2004 (UTC)

I would welcome that. But is this a request or an offer?--Patrick 21:32, 2004 Oct 10 (UTC)
The derivation is simple. Force is assumed constant, and a = F/m. The mass decreases as fuel is consumed, so you are integrating 1/m, which is ln(m). Tsiolkovsky did this calculation in 1897, according to his notebooks. DonPMitchell 07:42, 26 August 2006 (UTC)

[edit] historical note

The original form of the rocket equation, and the form which is still used in practice, corresponds to the original definition of specific impulse (given as an alternative definition in Wikipedia): m1/m0 = exp(-dv/g/Isp). This has the distinct advantage that Isp is the same in both English and metric units; furthermore, delta-V expressed in various units (e.g., m/s or km/s) can be accommodated by converting g to the same units.

[edit] historical note (correction)

On further reflection I am almost certain that Tsiolkovsky's work predates the development of the concept of specific impulse (Isp) by many decades, so the original form of the rocket equation must have used exhaust velocity. But I can state pretty authoritatively that the form generally used today includes g*Isp to replace exhaust velocity, since propellants are almost universally characterized by Isp in seconds (i.e., the alternative definition of specific impulse given in Wikipedia). [note and correction by Ted Sweetser, tsweetser@aol.com]

[edit] More concerning Isp

Specific impulse is indeed in seconds. Specific impulse is not equivalent to exaust velocity, so describing Ve as Isp is false. [correction by T.Cooper, the.centipede[at]gmail[dot]com

[edit] More about Isp

It's a measure of impulse per mass of fuel consumed. Impulse equals momentum, incidently. The units are often given as "seconds", but it is more accurate to give them as "kgf-sec/kg", kilogram-force seconds per kilogram, or pound seconds per mass-pound. The number is the same in both cases. DonPMitchell 07:40, 26 August 2006 (UTC)

[edit] energy gain?

I'm not an expert, so forgive me if I am making a silly question: If at some point in the future we get to make a rocket propulsion system in which the propellant mass is negligible and its potential energy goes entirely to the rocket acceleration itself (so that the percentage of that energy taken by the propellant being expelled is not inversally proportional to its mass), wouldn't we get a kinetic energy gain in the rocket that would increase for the same amount of fuel spent in a given time frame as velocity got higher? If the rocket was at 20.000 m/s, for instance, and we used a given energy amount to accelerate it to 20.001 m/s, wouldn't we get a greater increase in energy than if we were just accelerating from 100m/s to 101m/s, using the same quantity of fuel? If it is so, couldn't it mean a kinetic energy gain to the rocket greater than the energy contained in the fuel that was spent to obtain it, if we got our rocket at a sufficiently high speed?

It's not a silly question, it's a good question, but if you do the maths it turns out that energy is conserved (of course). The reason is that the kinetic energy of the propellant in a rocket going 20,000 is much higher than one going at 100 m/s. So, when emitted by the rocket, the propellant loses much more energy at high speed, than at low speed, and that's why the high speed rocket gains more energy than at low speed. Rockets are actually optimally efficient when the exhaust speed and the speed of the rocket match- in that case 100% of the energy ends up in the rocket and the exhaust just stops dead. Above and below that speed, efficiency drops. Hope this helps.WolfKeeper 22:03, 17 September 2006 (UTC)
Yep, as Wolfkeeper says, you have to include the rocket plus all propellant (including spent propellant) to get this right. You'll find, of course, that energy and momentum are conserved, since these are the principles from which the rocket equation itself was derived. As the mass of propellant gets smaller, its velocity becomes correspondingly greater, and so you can never neglect the spent propellant no matter how quickly it exits the rockets. (Of course there are also relativistic effects to consider.) --P3d0 02:13, 18 September 2006 (UTC)

[edit] Why units?

This edit added units to all the bullet points. The equation is correct regardless of the units used (as long as they are consistent with each other) so I think the units just confuse things. Can we remove them? --P3d0 18:16, 16 September 2007 (UTC)

I think they simplify things because they make the objects more concrete; people know that if something is in m/s that it is a speed, or has speed-like properties. Furthermore we are really supposed to be using SI units in the wikipedia, and I see no reason to encourage the use of non SI units.WolfKeeper 00:01, 17 September 2007 (UTC)
But if we say it's a speed, we don't need the units to convey that it has speed-like properties, right? And you're the one who added the non-SI units, not me; I'd rather just remove all units. I think adding units to this kind of equation is an unusual thing to do, looking around at articles like Newton's law of universal gravitation or heat conduction. A counterexample is ideal gas law, because there's a constant in that equation whose magnitude depends on the units chosen; this is not the case in Tsiolkovsky's equation. --P3d0 15:17, 18 September 2007 (UTC)

[edit] Description of delta-v

When thrust is applied in a fixed direction, delta-v is precisely the magnitude of the change in velocity, is it not? I'd like to reinstate this explanation that was removed by this edit if that's ok with everyone. --P3d0 18:18, 16 September 2007 (UTC)

Trouble is, the rocket equation still applies when the thrusts are not in the same direction or when gravity is acting. Delta-v is a bit more subtle that that; on balance I decided that having people click if they didn't already know what it was, was probably a good idea.WolfKeeper 00:04, 17 September 2007 (UTC)
Ok. I think we address sufficiently that by allowing for "a journey involving multiple such maneuvers". (However, the current phrasing with "such maneuvers" currently doesn't make any sense, because the context has been deleted.) I think we lose more by removing the succinct definition than we gain by being completely correct. Some people (like me) occasionally print out Wikipedia pages to read later, and though I know we're not supposed to cater to this, it seems kind of mean to make the text intentionally terse to encourage people to click the link and read another article. --P3d0 15:47, 18 September 2007 (UTC)
This is only the introduction though, I think having a short introduction is highly desirable. If you want to add other text to expand on delta-v elsewhere, I certainly wouldn't stop you.WolfKeeper 16:39, 18 September 2007 (UTC)
Ok. I think I'd still lean toward having at least some kind of summary of delta-V in the intro, but I can see your point. --P3d0 14:31, 30 September 2007 (UTC)

[edit] Kinetic energy of accelerating the reaction mass

\frac{1}{2}(m_0-m_1)v_e^2
This is the kinetic energy of accelerating the reaction mass.

This is either vague or contradicting what follows.--Patrick 22:25, 16 September 2007 (UTC)

[edit] Moore's "derivation"

I removed (temporarily?) the following excerpt:

"However a recently discovered pamphlet "A Treatise on the Motion of Rockets" by William Moore [1] 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."

Because now, when one more link to faximile of Tsiolkovsky's manuscript, explicitly showing that he derived the formula on May 10, 1897, is added, a real supporting evidence to a claim of William Moore's "earlier derivation" should be provided. It's not enough to point to "pages 499-521" of an article in a journal, which is unavailable online and hence is difficult to verify. The place, where William Moore derived this formula in his work should be cited precisely and this citation should be easily verifiable. Until this is done a "claim" of William Moore's superiority is nothing more than weasel words and WP:OR. Cmapm 19:33, 3 October 2007 (UTC)

I've undone your edit because there's absolutely no requirement that references be available online.WolfKeeper 22:12, 3 October 2007 (UTC)