Talk:Tsiolkovsky rocket equation

message: I have a serious beef with the applicability section of this page. The second and third sentences of this paragraph:
>Also, the equation strictly applies only to a theoretical impulsive maneuver, in which the propellant is discharged and delta-v applied instantaneously. Orbital maneuvers involving significantly large delta-v (such as translunar injection) still are under the influence of gravity for the duration of the propellant discharge, which influences the vehicle's velocity. The equation is most accurately applied to relatively small delta-v maneuvers such as those involved in fine-tuning space rendezvous, or mid-course corrections in translunar or interplanetary flights where the gravity field is relatively weak.
The second and third sentences seem to imply that the rocket equation is just not valid for going to the moon, when it should specify that it does not work well with low T/W systems, and works just fine with high T/W systems (like in apollo). Unless, that is, I have a serious misunderstanding. The rocket equation works *just fine* for high-delta-V maneuvers, but not long duration maneuvers. If someone can show a citation proving otherwise I would like to see it. Otherwise I plan to revise that soon. LarsOsborne (talk) 02:51, 12 March 2015 (UTC)
replies: text: I believe your proposed edits are quite correct.  I'd say make the changes.  Be bold.  A reference for the low T/W limitation would be nice, but even without a ref, your correction is much better than what is currently there. MattDuchek (talk) 20:55, 25 June 2015 (UTC)

text: Actually, I'd say scrap all the text in that section and just make the distinction about non-applicability to low T/W.  Also, non-applicable to launch from planetary surface and landing through atmosphere. MattDuchek (talk) 20:44, 25 June 2015 (UTC)
replies: text:  I have made edits to the applicability section to clarify this and other points. Sanchazo (talk) 17:44, 26 June 2015 (UTC)
thread_title: Applicability

message: 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)
thread_title:  More about Isp 

message: 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)
replies: text: Speed of the rocket in which frame?  Stops dead according to who? 80.6.239.132 (talk) 14:01, 6 March 2012 (UTC)
replies: text: 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)
replies: text: 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)

text: 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)
thread_title:  energy gain? 

message: 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)
replies: text: 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)
replies: text: 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)
thread_title:  Why units? 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
thread_title:  Description of delta-v 

message: ::\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)
replies: text: I fixed it.--Patrick (talk) 23:35, 9 January 2010 (UTC)
thread_title:  Kinetic energy of accelerating the reaction mass 

message: I removed (temporarily?) the following excerpt:
"However a recently discovered pamphlet "A Treatise on the Motion of Rockets" by William Moore 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 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)
replies: text: I've undone your edit because there's absolutely no requirement that references be available online.WolfKeeper 22:12, 3 October 2007 (UTC)
thread_title: Moore's "derivation"

message: 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 (talk • contribs) 14:40, 30 November 2009 (UTC)
replies: text: ln is natural logarithm.--Patrick (talk) 22:50, 9 January 2010 (UTC)
thread_title:  Please explain ALL formula components 

message: 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)
replies: text: Ditto! I was about to write the same! When the exhaust speed (relative to the rocket) is equal to the rocket speed (v_e = V), then the speed relative to the observer should be 0, which you only get if  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)
replies: text:  Got the error! Indeed, the correct form is  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 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)
thread_title:  derivation 

message: 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)
thread_title:  Russian cosmism 

message: 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)
thread_title:  Specific impluse (relativistic; dimensional analysis) 

message: 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)
replies: text: Using "equivalent delta-v" is the standard practice I believe. Martijn Meijering (talk) 18:26, 15 March 2013 (UTC)

text: 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)
replies: text:  I believe the examples do contradict the applicability section, because that section specifically states that the equation is not applicable to launch from a planet with an atmosphere.  I do understand that it is standard practice to use delta-v allocations for drag and gravity losses, but there is nevertheless a contradiction here.  Perhaps the problem is that the equation is applicable to launch from a planet even with an atmosphere.  In fact, it is applicable; you just need additional delta-V in your rocket to overcome atmospheric drag than would be implied based on the orbital velocity that you are trying to reach.  I think a re-write of applicability is needed. Sanchazo (talk) 22:55, 25 June 2015 (UTC)

text:  I made edits to correct the discrepancy. Please discuss if you disagree. Sanchazo (talk) 17:46, 26 June 2015 (UTC)
thread_title:  Example section contradicts Applicability 

message: I do not understand why there is a need to have the misconception section. Newton's second law does work for all classical mechanical systems. The original statement of Newton's second law is the force applied is directly proportional to the rate of change of quantity of motion. Back in Newton's time, there was no concept or word for momentum. In modern terms, we write force as rate of change of momentum. So, on expanding the derivative, we get velocity multiplied by rate of change of mass plus mass multiplied by rate of change of velocity. Now, since we usually look at constant mass system, we have rate of change of mass equal to zero. That is why we normally write F = ma. But, force is rate of change of momentum, not mass times acceleration. 
What is written about relativity is true. Obviously, we need to consider the rate of change in mass term of force in a relativistic frame because the mass does change with the velocity of the frame. So, the complete form of the Newton's second law always works. Therefore, there needs to be a correction/deletion of this section of the article.  — Preceding unsigned comment added by 117.200.127.20 (talk) 12:02, 2 October 2014 (UTC)
replies: text: Have you read any of the three footnotes given in the "misconceptions" section (particularly the Plastino/Muzzio paper)? Could you please point out exactly where their analysis is incorrect? Then, could you please provide a similarly detailed reference which demonstrates that the "complete form of the Newton's second law" correctly handles ablation/accretion systems? After this, we can do as you wish and "correct/delete" the misconceptions section. Zueignung (talk) 15:48, 2 October 2014 (UTC)

text: I don't think the section should be removed as it appears to be well-sourced. Maybe the wording needs to be improved. Having a look at variable mass system might help. Martijn Meijering (talk) 16:46, 2 November 2014 (UTC)
replies: text:  I can see the point that perhaps this may not be a "common misconception", just a plain old misconception. We could just keep the first sentence (i.e., that you can't derive the Tsiolkovsky equation by setting F = 0 and then using the product rule on Newton's second law). Zueignung (talk) 17:55, 2 November 2014 (UTC)
replies: text: None of the sources though refer to it as a misconception, at least not in the quotes given. It just says you can't do it this way. Fine, the article shows the correct way, including how to use Newton's 2nd Law. We don't need to catalogue or even mention wrong methods, of which there are far too many to mention, unless they are especially notable. If sources remarked that this was a common misconception then it might be worth including. As it is though its some editors judgement and so original research and doesn't belong.--JohnBlackburnewordsdeeds 04:24, 3 November 2014 (UTC)
replies: text: It does look like a common misconception to me, but I see no great need to state that explicitly or to have a separate section. I think we should say something like "because the rocket is a variable mass system, we need to use the following form of Newton's second law" somewhere at the top of the derivation section. Of course, you then end up with something that's almosty identical to what you'd get if you applied Newton's law naively. It looks as if the sources we're citing now were copied from the other article, so we wouldn't lose any information this way. Martijn Meijering (talk) 11:53, 3 November 2014 (UTC)
thread_title:  Discussion on misconceptions 

message: Correct me if I'm wrong but shouldn't the graph at the top of the article be Change in Velocity as a function of the Mass Ratio, since the Mass ratio is the dependent variable in most practical cases. Also the scale of the graph is misleading since they are not scaled the same as well as the axis could be labeled much more clearly with units and what part of the equation it's referring back to. I suggest that a graph more like the one I provide in the link would make it much clearer to readers of the article. http://www.maththoughts.com/wp-content/uploads/2014/03/Graph.png  — Preceding unsigned comment added by 96.52.51.67 (talk) 04:37, 21 October 2014 (UTC)
thread_title:  Bad Graph 

message: The Victorian Era Minister William Leitch also independently derived many of the
principles of rocket flight.  
it's worth making a footnote to the article here
http://www.cgpublishing.com/prime/bookpages/9781926837369.html
--Patbahn (talk) 02:48, 13 October 2016 (UTC)
thread_title:  william leitch 

message: The belgian general Casimir Erasme Coquillart also derived the rocket equation in 1873, according to:
:Malcolm Macdonald, Viorel Badescu, The International Handbook of Space Technology, Springer, 2014, p. 2.
(accessible via Google.books). --Harlock81 (talk) 11:11, 13 November 2016 (UTC)
thread_title:  Casimir Erasme Coquillart 

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thread_title:  Accessibility 

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thread_title:  External links modified 

message: "It is a scalar that has the units of speed. As used in this context, it is not the same as the physical change in velocity of the vehicle." 
Two parts: 
I've never seen it written this way... "units of speed", it is speed. I'd recommend changing this to something like, "Its SI units are m/s". In the same vein a lot of these articles are sloppy with notation, sometimes bolding v when standardly that would be reserved for a velocity vector.
Second part: 
When used in what context? In the context of the rocket equation? No, it's exactly the physical change in the velocity of the vehicle. Simply because the equation fails to consider other forces that might be acting on your rocket, or silly things you might do, or engineering designs don't change that. It's my opinion that this caveat needs to be more explicit or done away with entirely. Halleffect (talk) 05:03, 20 December 2020 (UTC)
thread_title:  Delta V