# Talk:Tsiolkovsky rocket equation

## Derivations

Could please someone change the direction of the vector V_e in the right picture? It contradicts the equation for P_2. Thank you! --193.170.243.211 (talk) 08:17, 24 October 2013 (UTC)

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)

In the Special Relativity paragraph, where is the given formula taken from? The use of Isp/c does not seem to work properly. It seems to have the wrong units, and in an example I calculated for a rocket with relativistic exhaust at 2/3 c and a small mass of propellant, the given formula produced a nonsensical result. Should it be Effective Exhaust Velocity instead? Would that perhaps be calculated as ve*ExhaustGammaFactor? That method of calculating Effective Exhaust Velocity matches one made from the (Exhaust) Relativistic Momentum given by the Wolfram Alpha Calculator for momentum of 1 gram at 2/3 c YodaWhat (talk) 04:45, 19 June 2012 (UTC)

## 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.

## 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]

## 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

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)

## 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)
Speed of the rocket in which frame? Stops dead according to who? 80.6.239.132 (talk) 14:01, 6 March 2012 (UTC)
That would be the Center of mass frame/zero momentum frame which is the frame with the lowest total energy and very probably the one you build the rocket in, or at least launch from, and the one you're most interested in (leaving!)Teapeat (talk) 16:32, 6 March 2012 (UTC)
How do you build a rocket "in" a reference frame, or launch a rocket "from" a reference frame?" MarcusMaximus (talk) 16:07, 19 June 2012 (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.) --Doradus 02:13, 18 September 2006 (UTC)

## 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? --Doradus 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. --Doradus 15:17, 18 September 2007 (UTC)

## 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. --Doradus 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. --Doradus 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. --Doradus 14:31, 30 September 2007 (UTC)

## Kinetic energy of accelerating the reaction mass

${\displaystyle {\frac {1}{2}}(m_{0}-m_{1})v_{e}^{2}}$
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)

I fixed it.--Patrick (talk) 23:35, 9 January 2010 (UTC)

## 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)

It is proper to honor Tsiolkovsky by naming the equation after him, because of the context and conclusions he drew from it: that spaceflight was possible and that it required a new kind of rocket more energetic than the black-powder artillery rockets of his time. That is, it led him to propose the liquid fuel rocket. He did much more than just integrate a = F/M, which any high school calculus student could do. DonPMitchell (talk) 04:19, 12 November 2010 (UTC)

## Please explain ALL formula components

I can find no reference to 'ln' in this document that explain what those two characters are representative of. —Preceding unsigned comment added by Battleshield (talkcontribs) 14:40, 30 November 2009 (UTC)

ln is natural logarithm.--Patrick (talk) 22:50, 9 January 2010 (UTC)

chutiya ho tum —Preceding unsigned comment added by 117.201.70.175 (talk) 06:36, 19 February 2010 (UTC)

Me too! I just came here to leave a note and find one already here - since 2009! Any chance of putting that explanation in the article? Robotics1 (talk) 12:09, 20 August 2012 (UTC)

## derivation

Hallo, I'm wondering, why do you take Ve = V + ve and not Ve = V-ve? Does Ve not have to be zero when the rocket is at speed ve?

Thanks, — Preceding unsigned comment added by 84.192.164.31 (talk) 15:17, 4 June 2011 (UTC)

Ditto! I was about to write the same! When the exhaust speed (relative to the rocket) is equal to the rocket speed (${\displaystyle v_{e}=V}$), then the speed relative to the observer should be 0, which you only get if ${\displaystyle V_{e}=V-v_{e}}$. I think the derivation is wrong, but I don't really know how to fix it. -- 140.181.115.42 (talk) 12:07, 5 August 2011 (UTC)
Got the error! Indeed, the correct form is ${\displaystyle V_{e}=V-v_{e}}$. To get the correct sign in the equation, one has to keep in mind that the mass differential is actually subtracted from the rocket, i.e. that ${\displaystyle dm(propellant)=-dm(rocket)}$. Since we are calculating the change in mass of the rocket, we have to enter the minus sign, and the final expression is correct again. I'll change it in some time, the derivation needs some rephrasing. Of course any other kind person could do it too... :-) - 140.181.115.42 (talk) 16:32, 20 September 2011 (UTC)

## Russian cosmism

Please explain the relevance of Russian cosmism before adding it back. I'll be happy to clean up the English if necessary, but only if it is worthy of inclusion in this article. Martijn Meijering (talk) 18:33, 13 August 2011 (UTC)

I see the edit in question has been restored again, without discussing it first. In its present form this change isn't helpful. The claim that the Tsiolkovsky equation only makes sense in the context of Russian cosmism is simply false: aeronautics textbooks tend not to even mention Russian cosmism. The statement about simulacrum is incomprehensible to me. I have no idea what it is supposed to mean. Please observe Wikipedia guidelines and discuss this before making controversial edits. Martijn Meijering (talk) 11:37, 14 August 2011 (UTC)

I looked up simulacrum and it refers to when something is a copy of something else and has a similar image, but is not at all the same, such as an idol of a god. Perhaps Tsiolkovsky is the god of rocketry, but since Moore was first he can hardly be a copy. Also, Moore's military application does not invalidate his work. After all, many rockets still are military. The history of English analysis of rocket propulsion for the military is quite substantial. I suppose in the history section one could mention that Tsiolkovsky's work and derivation of this equation was foundational in astronautics, but I wonder why all the rest of this is in the article.69.231.117.173 (talk) 04:06, 20 August 2011 (UTC)

The rocket equation is named after Tsiolkovsky because he was first to consider the wider problem of spaceflight, whether it was possible to attain orbital velocity with chemical fuel, the necessity of high-energy liquid fuel rockets, etc. It is relevant to note William Moore's work, but not very interesting to bring up Meshersky. The comment about Meshersky doesn't make sense to me, is someone claiming that Tsiolovsky was a follow of him? I've never hear that anywhere. DonPMitchell (talk) 21:31, 25 August 2011 (UTC)

I (temporarily) withdraw my comment. Sorry about that.

## Specific impluse (relativistic; dimensional analysis)

In the relativistic case, the power in the form m0/m1 = [(1+deltaV/c)/(1-deltaV/c)]^(c/2Isp) should surely be dimensionless.

From the definition give of Isp in the introduction, Isp = ve/g0, so has time units (indeed "expressed as a time period"); but then c/2Isp is not dimensionless (it distance/time^2).

Can someone explain what I'm missing? Ridcully Jack (talk) 20:42, 16 September 2012 (UTC)

• I suspect the formulas are in error (That's why I came to the talk page). I think the relativistic versions of the equation are using Isp when they should be using Ve. Think about it, where would we have gotten a "multiply by little g" step in there? If you substitute the velocity Ve instead of the time unit Isp, the units work out. — Preceding unsigned comment added by Sephalon1 (talkcontribs) 04:38, 1 October 2012 (UTC)
• That's what I thought too, but I've forgotten too much physics to edit this page. Ridcully Jack (talk) 07:19, 1 October 2012 (UTC)

## "Incorrect derivation"?

The final section contains these words:

"Using this formula with ${\displaystyle m(t)}$ as the varying mass of the rocket is mathematically equivalent to the derived Tsiolkovsky rocket equation, but this derivation is not correct.{{ safesubst:#invoke:Unsubst||date=__DATE__ |\$B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }}

A simple counter example is to consider a rocket travelling with a constant velocity ${\displaystyle v}$ 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 ${\displaystyle F=dp/dt}$ would result in a non-zero but non-accelerating force, leading to nonsensical answers."

The second paragraph is a nonsense. Momentum is a vector quantity, as is velocity, so it's perfectly plain that there is zero net force on the rocket. Is there any merit to the above or should it be deleted? 131.111.184.8 (talk) 12:49, 7 October 2012 (UTC)

I agree. The counter example puts rocket's velocity in the place of v. Why wouldn't there be the vrel which is, in this case, zero? For me it's hard to see why one must not use Newton's general second law (F = dp/dt) to solve variable-mass problems but instead one should use "same-looking" formulas which are derived using the same general NII and momentum (in an unnecessarily complicated way) only to get there the essential subscripts like vrelative and Fext and Fnet. Where's my blind spot? What's the difference? I'd like to keep it simple and use only F = dp/dt = vdm(t)/dt + mdv(t)/dt. I see no contradiction there. What this simple equation gives, for example, when there are no external forces (F=0) and mass is losed with constant relative velocity v? --- It gives a result similar to Tsiolkovsky's (write: -vrel * (1/m) * dm = dv => integrate over time => get Tsiolkovsky). Oulal (talk) 13:04, 24 October 2012 (UTC)
Yes, there is merit to the paragraph and no, it should not be deleted. Set up coordinates so that the rocket travels with constant velocity vx along the x-axis, and so that its thrusters are ejecting mass in the +y and –y directions. Now take the equation
${\displaystyle \mathbf {F} =m{\frac {d\mathbf {v} }{dt}}+\mathbf {v} {\frac {dm}{dt}}}$
and examine its x component:
${\displaystyle F_{x}=m{\frac {dv_{x}}{dt}}+v_{x}{\frac {dm}{dt}}.}$
Since the velocity vx is constant, this reduces to
${\displaystyle F_{x}=v_{x}{\frac {dm}{dt}}.}$
Now, since vx and dm/dt are both nonzero, this equation says that there is a nonzero force in the x direction. Hence, the rocket is accelerating, which contradicts the assumption that the velocity is constant. So you are forced to accept one of the two statements: (1) a rocket which ejects mass in some direction (the y direction, in this case) must experience acceleration in a different direction (the x axis, in this case), or (2) the formula F = m dv/dt + v dm/dt is incorrect. My money is on the second of these. Zueignung (talk) 22:22, 23 November 2012 (UTC)
Your counterexample is confused. If the mass to be emitted in the +/- y direction is to have no x momentum component in the original frame, it has to be given a push in the -x direction by the moving rocket, which means it will accelerate. — Preceding unsigned comment added by 217.162.141.172 (talk) 17:13, 20 December 2012 (UTC)
You are correct, I removed the section — Preceding unsigned comment added by 192.5.110.4 (talk) 05:56, 24 February 2013 (UTC)
No, it is you who is confused. If you attempt to use this putative variable-mass form of Newton's second law to compute the trajectory of the rocket, you do not get to say anything about what the ejected gas is doing (notice that I made no claim about the momentum of the ejected gas; I only indicated the orientation of the thrusters). This putative variable-mass equation is a relationship between the force on the rocket, the mass of the rocket, the velocity of the rocket, and nothing else. Notice that neither the mass nor the velocity of the ejecta occurs anywhere in the equation. As soon as you attempt to discuss the behavior of the ejecta, you no longer get to claim that you're applying Newton's second law to a variable-mass system. Every "derivation" of the Tsiolkovsky equation that uses this putative variable-mass second law ends up cheating somewhere, often by surreptitiously swapping the rocket velocity v for the ejecta velocity vex, by claiming mvex is the force exerted by the ejecta on the rocket (which it isn't), or by invoking the correct, constant-mass second law (i.e., a bait-and-switch). You are invited to read the discussion in Kleppner and Kolenkow's mechanics book or Halliday and Resnick's introductory physics book, or else to look at the relevant sections on the pages variable-mass system (which is even linked multiple times the deleted paragraph, for Christ's sake), Newton's laws of motion and Momentum. I've reinserted the paragraph and added a number of references which demonstrate the incorrect nature of the putative variable-mass second law. Zueignung (talk) 15:57, 2 April 2013 (UTC)
In your example, the momentum of the ejecta is given, it just happens to be zero. The force on the rocket is mdotvex. In your example mdot in the positive x direction equals mdot in the negative direction, so the x vector of mdotvex must be 0. In the sources you cite, they specifically state that for a closed system, newtons second law holds. In this case, it is a closed system because the propellant is being counted. — Preceding unsigned comment added by 192.5.110.4 (talk) 07:38, 3 April 2013 (UTC)
changed so that it was clear that the derivation is only invalid when the control area is taken as the rocket alone. — Preceding unsigned comment added by 192.5.110.4 (talk) 07:47, 3 April 2013 (UTC)
In my example, the momentum of the ejecta is not considered, which is precisely why I'm able to arrive at a nonsensical answer as to the motion of the rocket. People in this discussion have been claiming the exact opposite; i.e., that one can ignore the behavior of the ejecta and still get the Tsiolkovsky equation by applying Newton's second law to a variable mass system. I'm glad that you're able to insert a ${\displaystyle {\dot {m}}v_{\mathrm {ex} }}$ term to make the derivation correct, but the point is that people on this talk page have been claiming that one can get the Tsiolkovsky equation without including such a term to account for the momentum of the ejecta (see Oulal's comment, for example). The central question that is being discussed here is Does the equation F = m dv/dt + v dm/dt correctly describe the motion of a rocket of mass m(t) and velocity v(t)?, and the answer is No. The "common misconceptions" section exists to bang the reader over the head with the fact that you must always include the momentum of the ejecta, or, equivalently, you must never attempt to use Newton's second law on a variable-mass system (or a variable-mass "control area", as you call it). Zueignung (talk) 16:20, 3 April 2013 (UTC)
The section claims one is dealing with the "relativistic force equation". Relativity is the physics of going fast. For relativistic forces, one changed from a derivative with respect to time to one where there is a change with respect to the interval tau. The equation shown is a direct application of the product rule of calculus. That rule is never wrong as one can do a formal proof that the product rule is true. It cannot be applied in this case because there are 2 velocities, not one. Like an earlier commenter, I think pointing out there is not just one velocity makes this issue on confusion less confusing. I am not going to edit the page myself since I don't like to actively engage in wikipedia issues. Dougsweetser (talk) 12:02, 11 May 2013 (UTC)

Since this equation neglects non-rocket forces (aerodynamic and gravity), it cannot be used to accurately calculate mass ratio required for a launch vehicle, without pulling "equivalent" delta v estimates out of the air (or somewhere else.) Therefore, I think the two examples listed (single-stage-to-orbit and two-stage-to-orbit) are really poor ones to use. I think a single example, such as a translunar injection, would be better. JustinTime55 (talk) 18:22, 15 March 2013 (UTC)

Using "equivalent delta-v" is the standard practice I believe. Martijn Meijering (talk) 18:26, 15 March 2013 (UTC)
While I agree the examples might not be the best, they are not contradictory to the rest of the article. The 9700m/s stated in the example assumes almost 2000m/s of gravity and drag losses, and is used as a general benchmark for launch vehicles that need to reach LEO. I don't see how it matters (as far as an example of how to use the equation goes) whether we propose a hypothetical 9700m/s launcher or a 9700m/s hohmann transfer. I've changed the example to reflect the assumption of drag and gravity losses. Arthree (talk) 19:57, 21 October 2013 (UTC)

## Applicability half-section way off topic

The last two paragraphs seem to be out of scope here:

The Saturn V rocket used in the Apollo space program is an example of a large, serially staged rocket. The Space Shuttle is an example of parallel staging, where all of its engines are ignited on the ground and some (the solid rocket boosters) are jettisoned to lose weight before reaching orbit.
The ion thruster is an example of a high exhaust velocity rocket. Instead of storing energy in the propellant itself as in a chemical rocket, ion and other electric rockets separate energy storage from the reaction (propellant) mass storage. Not only does this allow very large (and in principle unlimited) amounts of energy to be applied to small amounts of ejected mass to achieve very high exhaust velocities, but energy sources far more compact than chemical propellants can be used, such as nuclear reactors. In the inner solar system solar power can be used, entirely eliminating the need for a large internal primary energy storage system.