Talk:Oberth effect

Unclear wording?

In the article under Example calculation it says then the specific kinetic energy per unit mass after the burn is. Should that be either just specific kinetic energy without the words per unit mass, or kinetic energy per unit mass without the word specific? I mean, isn't the per unit mass part already stated implicitly with the word specific? Doesn't the current wording mean, strictly speaking, kinetic energy per unit mass squared?130.234.5.138 (talk) 21:03, 18 December 2007 (UTC)

I cannot chime in on the Oberth Effect, as it is over my head. I'd like to propose A rewording of the introduction, that leaves out the efficiency-at-velocity point, because that is already discussed in the article on Thrust and in Rocket, and the wording here is ambiguous. The OE seems to be about firing in a gravity well, so better leave it at that. The thrust of a rocket remains the same, velocity notwithstanding, and this is not made clear enough in the text. —Preceding unsigned comment added by 134.2.125.243 (talk) 12:38, 1 April 2009 (UTC)

I am bookmarking this to review later, but let me flag that the first paragraph makes no sense on first reading. Out of the many possible interpretations of the effect, "usable energy" strikes me as the least accessible, and thus not a good lead-in to the topic. I suggest starting from the core idea that propellant imparts a certain amount of momentum, and thus delta-V as it is ejected. Then consider the effect of that delta-V on energy at low V vs high V scenario.

Something along the lines of:

A rocket works by transferring momentum to its propellant. At a fixed exhaust velocity, this will be a fixed amount of momentum per unit of propellant. For a given mass of rocket (including remaining propellant), this implies a fixed change in velocity. Because kinetic energy = mv^2, this change in velocity imparts a greater increase in kinetic energy at a high velocity than a low velocity. For example, if you have a 1kg rocket, at 1 m/s, adding 1 m/s increases KE from 1J to 4J, for a gain of 3J, but at 10 m/s, you start with a kinetic energy of 100J, and end up with 121J, for a net gain of 21J. This greater kinetic energy can then carry the rocket higher in the gravity well than if it were burned at a lower speed. </blockquote Bob Kerns (talk) 05:35, 11 November 2010 (UTC)

Disagreement over explantion of paradox

There is a disagreement between the explanation of the paradox, as would be given in the last paragraph of the description. It is (at the moment) a disagreement between User:Wolfkeeper and User:Wingedsubmariner. The current version of the page is by Wolfkeeper, the most recent by Wingedsubmariner is here.

The discussion to this point has been on user pages, here it is copy-and-pasted:

from User_talk:Wingedsubmariner

You're sort of half right. The kinetic energy of the propellant is transferred to the vehicle in the way you indicate, but the potential energy of the propellant is where the energy originates from if you trace it through.- (User) Wolfkeeper (Talk) 03:08, 25 February 2009 (UTC)

from User_talk:Wolfkeeper

I hope I didn't give offense: I only replaced the explanation because I was certain I was right. The paradox does need a resolution.

The potential energy difference does not explain the effect. Consider a modified version of the example I give on Oberth effect, where the difference in speed comes from the speed being invested into potential energy, that is, the spacecraft was moving away from some mass. We now have two different cases. One, the rocket at a certain distance from a mass moving away from it below escape velocity, two, the rocket after its speed has reached zero, converting all of the speed from case one into potential energy, 1/10 of which was invested into the propellant. It started with a speed of 9. We don't need to know the size of the mass or the exact distances. A speed of 9 gives initial kinetic/potential energy of 81, 8.1 of which is held in the propellant. The rocket fires in the same way as in my previous example, and kinetic energy is assigned in the same way. Potential energy does not change because nothing has changed its height. We still have an 89.1 difference in kinetic energy of the rocket between the two cases, and the difference in potential energy of the propellant is 8.1 between the two cases, insufficient to explain the effect.

--Wingedsubmariner (talk) 03:55, 25 February 2009 (UTC)

Well... the Oberth effect only happens when there is a change in potential energy. Your explanation doesn't cover potential energy, so is clearly incomplete. The kinetic energy of the propellant is generated from potential energy before it gives it up that energy to the body of the rocket, so the ultimate source of the energy is potential energy. In some cases the vehicle can have no significant initial kinetic energy until it falls close to the gravitating body.- (User) Wolfkeeper (Talk) 04:13, 25 February 2009 (UTC)

Sorry for not replying quickly, I've been away for awhile. The Oberth effect absolutely requires a change in kinetic energy, and this is almost always due to a change in potential energy, however it is not required: The kinetic energy could come from another supply of kinetic energy, such as after a collision, or a transfer from a skyhook-like device. My example above does in fact cover potential energy: I refer to the change in both the potential energy of the propellant and the rest of the craft, and it is not sufficient to explain the change in kinetic energy. The difference between the kinetic energy of the rocket between the two examples is 89.1, the difference in potential energy of both the propellant and the rocket is only 81, it is the difference between the initial kinetic energy between the two examples. The first example has a potential energy of 81, the second kinetic energy of 81 instead (it's the rocket if it had fallen and then used its engine).

I think I had already satisfied you that the potential energy of the propellant alome could not provide the energy, the equations are clear. I am surprised you reverted the article to your old version. Unless this argument convinces you, I think we need to ask for arbitration. Wingedsubmariner (talk) 05:12, 1 March 2009 (UTC)

Sorry, arbitration means something else...I mean "dispute resolution", as described at Wikipedia:DR. Wingedsubmariner (talk) 05:22, 1 March 2009 (UTC)

Sorry, you're simply talking nonsense. First you need to analyse everything in an inertial reference frame- we'll use Saturn. If the vehicle is travelling at 5 km/s and falls in towards Saturn and gains 20km/s, it then fires a thruster that emits exhaust at 3km/s relative to the rocket. It would leave the exhaust moving at 22km/s; whereas if it applies this burn in mid-space then the propellant would be left at 2 km/s; if it applies this close to saturn. The propellant has *more* kinetic energy when fired close to Saturn than when fired away; as does the rocket, and the rocket, when it has left the planet will be moving much faster than if it had applied it in deep space. The reason is that the potential energy of the propellant ends up much lower in the first case than the second, and the rocket was able to benefit from that.- (User) Wolfkeeper (Talk) 14:08, 1 March 2009 (UTC)

Yes, but it is the change in total energy that matters. 5 -> 2 means 25 - 4 = 19 change in potential energy, or 19, whereas 25 -> 22 means 625 - 484 = 141. Close to the planet the propellant loses more kinetic energy. It gained kinetic energy for its potential energy when falling into Saturn just as the rocket did, and still has this energy, less the 141 lost as a result of the rocket.

No. The Oberth effect the article discusses is the extra speed you get later, not simply the change in kinetic energy at the time of the burn.- (User) Wolfkeeper (Talk) 17:20, 1 March 2009 (UTC)

I don't disagree. However we are not arguing about what the Oberth effect is (a way to increase final speed) but its cause. You said it was due to the potential energy of the propellant, and I am showing that the change in the potential energy of the propellant isn't sufficient to explain the change in the kinetic energy of the rocket. Wingedsubmariner (talk) 17:38, 1 March 2009 (UTC)

The Oberth effect also works when slowing down a rocket. Firing the engines to slow down the rock in this case gives 5 -> 8 which means 25 - 64 = -39 as compared to 25 -> 28 which means 625 - 784 = -159. In this case, the propellant gains more energy despite being expended lower in the gravity well.

The Delta-v page gives a good description of the dependency on speed and direction of speed: "when applying delta-v in the direction of the velocity the specific orbital energy gained per unit delta-v is equal to the instantaneous speed." Wingedsubmariner (talk) 15:41, 1 March 2009 (UTC)

End copy-and-paste

Above by Wingedsubmariner (talk) 16:06, 1 March 2009 (UTC)

The thing is, I don't think the term Oberth effect is used to simply mean the fact that a rocket body gains more kinetic energy at high speed from its burn than at low speed (although that is true); I think it's used specifically with respect to a gravity field. In that case the vehicle can even give the propellant negative total energy, and the rest of the energy goes into the body of the rocket; when not close to a gravitating body it can't do that. It's a slightly more nuanced understanding.

However, I think that we simply need more good referenced POVs we can add here. If anybody has any we can add that to the article that would be very desirable.- (User) Wolfkeeper (Talk) 17:10, 1 March 2009 (UTC)

A brief survey of references on Google, excepting ones that are just copies of this page, do explain it as due to a gravity well. However, I think what Oberth actually said should be used as the definition, or perhaps another reference that explores the mathematics and proves it. Wingedsubmariner (talk) 17:38, 1 March 2009 (UTC)

Again, I am not disagreeing with you as to what the Oberth effect is. I have no problems with how the article at present defines it, and an explanation would be incomplete without discussing the importance of a gravity well. I only disagree over the portion of text we warred over, the explanation for the effect and where the energy comes from. Wingedsubmariner (talk) 17:50, 1 March 2009 (UTC)

Here, does this seem correct to you?

The resolution of the paradox is due to the propellant having less energy when expended lower in the gravitational field, which makes up for the gain to the rocket. As the vehicle falls into the gravitational field, the propellant (and the rest of the rocket) gains speed, and so during the burn the propellant loses more kinetic energy than if the burn had occurred higher in the gravitational field. From conservation of energy there must be an equal and opposite gain to the rocket's energy.

It covers the Oberth effect operating with a gravitational field, and explains where the energy for the rocket comes from; the paradox disappears once you include changes in the propellants energy. Wingedsubmariner (talk) 18:13, 1 March 2009 (UTC)

I'm here because of the post in WikiProject Physics; here's my take on this. There's a larger increase in total energy of the rocket when you fire closer to the planet, which had better be balanced by a larger decrease in total energy of the propellant. How you split that up into components depends on the details of the problem setup. If the propellant escapes to infinity in both firing scenarios then the difference must ultimately be entirely kinetic. If it falls into the planet in both scenarios then I suppose the difference is ultimately mostly thermal. If it falls into the planet in one scenario and escapes to infinity in the other then the difference is partially gravitational potential, but not entirely. If you're considering the change at the time of firing and the distance from the planet stays roughly constant then the difference must be mostly kinetic. So in short I think Wingedsubmariner is right, if I'm understanding correctly what you've both written.

The other thing, though, is that you seem to be debating what "explains" the Oberth effect, but that's inherently a subjective thing—if an increase in one quantity is balanced by a decrease in another it's a matter of choice whether you say the increase explains the decrease or the decrease explains the increase or neither. It seems to me that the most useful description of the problem is the one that talks only about the kinetic energy of the rocket, in part because it extends to non-rocket forms of propulsion. You can then go on to say "this might appear to violate conservation of energy in the case of self-propulsion, but if you carefully consider all forms of energy then you'll find that it doesn't". But neither of those is the explanation of the effect. In fact the second seems more like an explanation of a common mistake made by people first encountering the effect than an explanation of the effect itself (assuming for the sake of argument that people do commonly make that mistake). -- BenRG (talk) 15:19, 3 March 2009 (UTC)

That's an important point, I hadn't though about non-rocket forms of propulsion. In the case of a solar sail the Oberth effect would still operate, but the energy would come from increased red-shifting of the light striking the sail, rather than either decreased potential or kinetic energy of a massive propellant.

Thanks for taking the time to help resolve this. Wingedsubmariner (talk) 18:00, 3 March 2009 (UTC)

Gee thanks. In any case I'm continuing to find this edit[1] both offensive and inaccurate, as well as offensively inaccurate. The text that is currently in the article describes the situation better. I also find that discussing only kinetic energy is formally and logically incorrect. The resolution of the paradox is because the total energy of the propellant is lower when emitted in a gravity well, and the fact that it has a positive or imaginary velocity at infinity is not important. That this is not simply about kinetic energy is not negotiable or debatable.- (User) Wolfkeeper (Talk) 19:04, 3 March 2009 (UTC)

I've also not ever seen use of the 'Oberth effect' in non rocketry scenarios. Are you aware of any references that use it in that way?- (User) Wolfkeeper (Talk) 19:04, 3 March 2009 (UTC)

Just in case you're still not getting it- there are cases where the propellant is emitted at the same speed within the gravitational field (with respect to the rest frame) as when it is emitted outside, and the rocket still gains more energy.- (User) Wolfkeeper (Talk) 19:10, 3 March 2009 (UTC)

Could you give a concrete example? The speed the propellant is emitted at is a factor of the exhaust velocity and the speed the rocket was going at before the burn. The exhaust velocity isn't different in the two cases, and so you would be implying the speed of the rocket is the same. Potential energy would be different at the two heights in the gravitational field, and so the starting energy would be different. This would not a valid comparison. Wingedsubmariner (talk) 20:09, 3 March 2009 (UTC)

Careful here, the exhaust velocity is relative to the rocket, whereas energy is best analysed in center of mass frames. If the speed of the rocket is 1km/s and the exhaust velocity is 2km/s, then the exhaust velocity is 1km/s in the gravity free case. If the vehicle falls down into a gravity well, so that the vehicle is moving 3km/s before emitting the exhaust, and then it emits the exhaust then the exhaust velocity is still 1km/s, but the vehicle velocity will be greater. That particular example is only true for an infinitesimal point in time though, but it's possible to emit the same amount of exhaust with the same total kinetic energy in and out of a gravity well, and the effect is still true; the vehicle gains more energy.- (User) Wolfkeeper (Talk) 00:42, 4 March 2009 (UTC)

In fact, in some cases the exhaust can end up travelling faster down the gravity well than outside, as well as the rocket having more energy.- (User) Wolfkeeper (Talk) 00:42, 4 March 2009 (UTC)

Right, my mistake. I was think of velocity, which isn't equal in the two cases, though the speed is. Wingedsubmariner (talk) 16:33, 4 March 2009 (UTC)

I agree that Wingedsubmariner's resolution in the revision you linked is wrong. There are cases where the speed of the propellant is the same in both scenarios, and that's fine because the initial-final energy delta is different, so everything still evens out. I'm not aware of uses of the Oberth effect outside rocketry or for that matter inside rocketry; I'm just applying basic principles of Newtonian physics here. I didn't realize I'd introduced the idea of a non-rocket Oberth effect, I thought I was repeating something that Wingedsubmariner had already said (the bit about the skyhook). There are probably various other things I misunderstood. At any rate, the main problem here seems to be the whole idea of a paradox that needs resolving. There are various errors you can make in analyzing this problem, such as the one Wingedsubmariner appear to have just made, and for every such error you could explain in the article what's wrong with it. But which errors are paradoxical enough to include? Newtonian physics isn't going to tell you that, it's purely subjective. None of the ways of misanalyzing this problem seem especially interesting to me. It's a high school physics problem. There's nothing here that deserves the exalted name of "paradox" in my opinion. -- BenRG (talk) 11:44, 4 March 2009 (UTC)

Well, it never says that it is a true paradox, and there are no true paradoxes in science anyway. If there ever are we come up with new science until there aren't!- (User) Wolfkeeper (Talk) 17:01, 4 March 2009 (UTC)

And I think it's more complicated than it seems. Notably, different bits of the propellant end up moving at different speeds, even within a single impulsive burn; an impulsive rocket burn is not the same as firing a cannon.- (User) Wolfkeeper (Talk) 17:01, 4 March 2009 (UTC)

On rereading it, I agree, that revision is wrong. As Wolfkeeper shows above, the propellant can end with the same kinetic energy, but the rocket can still be moving faster. The picture is incomplete without considering the potential energy of the propellant. However, the magnitude of the potential energy difference may not be great enough to explain the difference in energy of the rocket, so I still dislike Wolfkeeper's revision. Wingedsubmariner (talk) 16:33, 4 March 2009 (UTC)

Well, conservation of energy must still apply. You have KE and PE (both gravitational and chemical) going in and KE and PE coming out, and the sums must be the same in all cases.- (User) Wolfkeeper (Talk) 17:01, 4 March 2009 (UTC)

I agree, the whole paradox is just how we explain how conservation of energy holds with the Oberth effect. What do you think of my revision below? I think it presents this clearly. Wingedsubmariner (talk) 05:13, 5 March 2009 (UTC)

I suggest this paragraph to replace the current last two paragraphs in the description:

It may seem that the rocket is getting energy for free, which would violate conservation of energy. However, any gain to the rocket's energy is balanced by an equal decrease in the energy the propellant is left with. When expended lower in the gravitational field, the propellant is left with less total energy.

It should cover every single situation any of us have come up with to this point. Wingedsubmariner (talk) 16:49, 4 March 2009 (UTC)

Since no one objected, I've put the paragraph above on to the page. Wingedsubmariner (talk) 14:48, 7 March 2009 (UTC)

It doesn't seem to me to have even as much clarity, and even the original wasn't as clear as I'd like; as this discussion showed.- (User) Wolfkeeper (Talk) 15:05, 7 March 2009 (UTC)

There is a lot it leaves unsaid, but I assumed more advanced readers would work out the math themselves. The preceding paragraph really describes the effect, this paragraph is just there for people who can't reconcile it with conservation of energy.

The original doesn't cover the situation you gave above where the kinetic energy of the propellant does not change. Wingedsubmariner (talk) 15:18, 7 March 2009 (UTC)

Yes, it does.- (User) Wolfkeeper (Talk) 16:08, 7 March 2009 (UTC)

Well, alright, maybe more like a modification of your example. If the exhaust velocity is 2 km/s and the rocket is traveling 1 km/s per second in the gravity well, then no kinetic energy is lost from the propellant when the burn is made. The original revision used "loses more kinetic energy", which isn't broad enough to cover this case. Wingedsubmariner (talk) 17:01, 7 March 2009 (UTC)

3PO (this was at the top of this list, but...) I'm not sure how much the 3PO can help here. This seems to be a nuanced issue in astronautical engineering, so it may make sense to email a professor in the field. The reference desk may also be able to help (or at least find a good professor). Also, since the 3PO was requested I see a third person has entered the conversation. Although the tone of discussion here is very congenial, if you do find that an agreement is not forthcoming, consider an RFC. (removing 3PO request) NJGW (talk) 03:24, 6 March 2009 (UTC)

Math check for my reverted edit

I'm finally coming back to my edit that got immediately reverted. I offer a more detailed explanation of the correct mathematics as I see it.

The residual speed at infinity is always $v_\infty=\sqrt{2e}=\sqrt{v^2+2u}$, where e is the specific total energy and $u=-\frac{GM}r$ is the specific gravitational potential energy. For an instantaneous change in speed, the return in the form of $v_\infty$ per investment of delta-v is $\frac{dv_\infty}{dv}=\frac{2v}{2\sqrt{v^2+2u}}=\frac v{v_\infty}$. Certainly this only holds for infinitessimal burns, since if the delta-v is much greater than the escape velocity, not much of it will be lost (so $v_\infty\approx v$) even if that ratio is huge.

This is the linearity that I mentioned: the ratio is $v/v_\infty$, not its square root, and it only applies for small burns. If there are no objections or refinements, I'll restore my edit. --Tardis (talk) 14:50, 10 April 2009 (UTC)

The math looks good to me. In any case, its very easy to construct counterexamples to the square root rule given in the current version, so it definitely needs to go. As far as I can tell, that is the main change your revision would make. Wingedsubmariner (talk) 04:12, 13 April 2009 (UTC)

This bothered me both times I read it, and I'm really glad to see your explanation here. I think it should be somehow referenced in or inserted into the page itself; I had the impulse to edit the paragraph to change final velocity to say the speed goes up as the square root as well. So it should probably be explained better.66.30.14.175 (talk) 19:02, 5 September 2009 (UTC)

I haven't checked it carefully, but there's probably nothing wrong with your maths, but the article is about non infinitesimal burns. The article is about impulsive burns (to which many real burns approximate fairly well). This is very different.- (User) Wolfkeeper (Talk) 20:26, 5 September 2009 (UTC)

Note that small burns under strong gravity gives a very big Oberth effect; from the equation in the article a 1m/s burn close to the Earth where the escape velocity is say, 10.5 km/s gives a delta-v at infinity of... 145 m/s; over 100x better delta-v!- (User) Wolfkeeper (Talk) 20:31, 5 September 2009 (UTC)

3.3?

"It can be easily shown that the impulse is multiplied by a factor of:

$\sqrt{1 +\frac{2 V_\text{esc}}{\Delta v}}$

Plugging in 50 km/s escape velocity and 5 km/s burn we get a multiplier of 3.3."

When I plug in those values, I get

$\sqrt{1 +\frac{2 \times 50}{5}}$

$=\sqrt{1 +\frac{100}{5}}$

$=\sqrt{1 + 20}$

$=\sqrt{21}$

which is 4.6 not 3.3. Am I missing something? Art LaPella (talk) 15:43, 7 April 2010 (UTC)

no ;-)- Wolfkeeper 16:12, 7 April 2010 (UTC)

dE/dt is not F v

In a discussion on the Talk:Conservation of energy page, I made the point that the time rate of change of the kinetic energy of the rocket engine is not simply Fv where F is the thrust and v is the engine velocity. A second term which constitutes an energy loss due to the mass of propellant being lost to the exhaust must be included. The correct statement is:

$dE/dt=Fv-\frac{1}{2}\mu v^2$

where $\mu$ is the rate at which propellant mass is converted to exhaust mass. As noted by 95.134.115.203, the total power P and thrust F produced by the engine can be assumed constant, so, assuming no heating effects, if Po=Fv is the rate at which the engine increases its kinetic energy, and Px is the rate of energy increase of the exhaust train, then P=Fv+Px or Px=P-Fv and at a high enough velocity, the rate of change of energy of the exhaust train is negative. Thats impossible.

I would edit the article myself, but I am not familiar enough with it to trace the implications of this throughout the article. PAR (talk) 21:52, 9 April 2010 (UTC)

The energy gained by the mass m1 of the rocket that isn't going to be released during the burn is still proportional to v; it's absorbing a fraction of the kinetic energy being produced that depends only on where you are in the burn in terms of mass ratios. The fact that you're losing energy in the exhaust doesn't seem to matter.- Wolfkeeper 23:21, 9 April 2010 (UTC)

But I'll think about ways to clarify this.- Wolfkeeper 23:21, 9 April 2010 (UTC)

Introduction

In astronautics, the Oberth effect is an effect where the use of a rocket engine at high speed generates much more useful energy than one at low speed.

This does not seem to be a good description of the Oberth effect.Ordinary Person (talk) 00:29, 14 April 2010 (UTC)

Do you have a good reference?- Wolfkeeper 00:39, 14 April 2010 (UTC)

Parabolic example

We need to do a bit of work on the parabolic example - the example given is not strictly a parabolic example, but rather an example which starts from a parabolic orbit (orbital speed at periapsis equals V_esc) and transitions to a hyperbolic orbit (orbital speed at periapsis after acceleration now exceeds V_esc). Let me dig out my old copy of Stearns' "Navigation and Guidance in Space" and see what he has to say about it; then I'll take a shot at rewriting. Skål - Williamborg (Bill) 22:34, 22 May 2010 (UTC)

If you have a good reference that would be very useful, we're desperately short of refs for this.- Wolfkeeper 16:47, 24 May 2010 (UTC)

Still short of good references. The Oberth effect is not mentioned in Stearns' or in about a half dozen other similar texts. So I looked for it in the literature. Oberth shows up, of course, but no mention is made of the "Oberth effect". Have to look further next time I'm down at Powell's Books. Williamborg (Bill) 03:11, 1 June 2010 (UTC)

I've found Oberth talking about this kind of stuff, but obviously he doesn't use the term 'Oberth effect'. See: [2] around page 200 (pdf page 209) he talks about energy, and includes a powered slingshot example.- Wolfkeeper 03:59, 1 June 2010 (UTC)

Homeland-issue

Oberth was born in Nagyszeben (Hermannstadt), which was the part of Hungary that time, not Romania. http://en.wikipedia.org/wiki/Hermann_Oberth I've corrected it.GergelyGergely (talk) 21:44, 28 November 2010 (UTC)