Question about the Oberth effect

[Right]

[quote name='GoSlash27']Right,
I'm with RIC on this one. the Oberth effect has nothing to do with gravity wells. It's simply the result of the exponential relationship between velocity and energy. It holds true even without gravity wells.

Best,
-Slashy[/QUOTE]

Okay I've got my terminology confused. I'm describing powered gravity assists.

Strictly speaking, the Oberth effect has nothing to do with gravity. However, isn't it fair to say that the Oberth effect is not the sole reason periapsis burns are the most effective?

[GoSlash27]

[quote name='Right']Okay I've got my terminology confused. I'm describing powered gravity assists.

Strictly speaking, the Oberth effect has nothing to do with gravity. However, isn't it fair to say that the Oberth effect is not the sole reason periapsis burns are the most effective?[/QUOTE]

Right,
Actually, I would say that it is *not* fair to say that. In fact, periapsis burns are not necessarily the most effective even with Oberth effect. Without the Oberth effect, it would make more sense to do your burns as far out of the gravity well as possible.

As it stands, it makes sense to do your transfers from LKO because that's where you happen to be. It would be more efficient to do your transfer at a higher altitude (depending on where you're going), but it always takes more energy to get up there than it saves.
OhioBob did a really good write- up on this a couple months back.

Best,
-Slashy

[Right]

[quote name='Red Iron Crown']Have a look at a rocket stage's speed vs. time graph:[/QUOTE]

Not that it invalidates your point, but that graph would happen even if you were wrong due to decreasing fuel mass and atmospheric drag.

[Red Iron Crown]

[quote name='Right']Not that it invalidates your point, but that graph would happen even if you were wrong due to decreasing fuel mass and atmospheric drag.[/QUOTE]
If mass were not changing significantly and there was no drag, it would be linear. F=ma, so if F and m are constant a must be, too. :)

[Yasmy]

[quote name='Tatonf']Thanks for all the replies, but I still don't understand why I gain more "kinetic energy" from burning in LKO rather than in Sun's orbit.
You say that I can't compare my velocity from LKO and the velocity from my Sun orbit, but what does matter at the end ? My velocity relatively to the Sun ?
That would mean that, if I go orbiting the Sun at an altitude where my velocity is greater than 9284.5 + 3431.0 = 12 715.5 m/s, the Oberth effect I'll get from the Sun will be better than the one I get from LKO ?[/QUOTE]

Short answer: Yes.

Long answer: It is your sun-relative energy that matters for orbits in or crossing the sun's SOI. And for any particular solar orbit, your total sun-relative energy is the same no matter how you got on that orbit. (Not strictly true in KSP due to Kerbin's SOI system, which is not a conservative system, but close enough.)

Setup: in LKO burn so you push your orbit to just barely escape Kerbin's SOI.
Case 1: Wait until you exit Kerbin's SOI, then burn dv.
Case 2: Burn dv immediately.
Arrange in either case to exit Kerbin's SOI along Kerbin's solar prograde.

Note, since F = m*a, and F = -G*M*m/r^2, the acceleration due to gravity (a = -G*M/r^2) is independent of mass, i.e. objects of different mass fall at the same rate. Thus we can ignore mass. Instead of energy, I'll use "specific" energy, which is just e = E/m.

Case 1:
Initially near Kerbin, e = 1/2 v^2 - GM/r^2 = 0. Your Kerbin-relative energy is zero. This is what it means to have just enough energy to escape Kerbin.
Exiting Kerbin's soi, relative to the sun, you have Kerbin's velocity V, and are at Kerbin's orbital radius R. Now add dv to your velocity, and add in the sun's potential:
e = 1/2 (V + dv)^2 - G Ms / R^2, with Ms equal to the mass of the sun.

Case 2:
Initially near Kerbin, e = 1/2 v^2 - GM/r^2 = 0. Your Kerbin-relative energy is zero. Add dv to your velocity:
e = 1/2 (v+dv)^2 - GM/r^2 = [1/2 v^2 - GM/r^2] + v dv + 1/2 dv^2.
Note the bracketed term is zero.
e = v dv + 1/2 dv^2. (Aside: This is the Oberth effect: de/dv = v)
When you exit Kerbin's SOI, you retain that excess energy as kinetic energy:
e = 1/2 Ve^2 = v dv + 1/2 dv^2.
Ve = sqrt(2 v dv + dv^2),
which gets added to Kerbin's solar velocity V:
e = 1/2 (V + Ve)^2 - G Ms/R^2
Now Ve is always greater than dv, so right now you know that you got more bang (Ve) for your buck (dv) at LKO that in solar orbit (dv).

Now to make it clear which case is better, take the difference of the total sun-relative energies, case 2 - case 1.
de = (1/2 (V + Ve)^2 - G Ms/R^2) - (1/2 (V + dv)^2 - G Ms / R^2)
de = 1/2 V^2 + V Ve + 1/2 Ve^2 - 1/2 V^2 - V dv - dv^2
de = V Ve + 1/2 Ve^2 - V dv - 1/2 dv^2
de = V Ve + 1/2 (2 v dv + dv^2) - V dv - 1/2 dv^2
de = V (Ve-dv) + v dv
de = V dv (sqrt(1 + 2v/dv)-1) + v dv

Sure it is a bit ugly, but it is strictly non-negative. It's never better to burn in solar orbit when coming from LKO, from a dv efficiency standpoint.
In the limit of dv -> 0, de -> 0.
In the limit of dv >> v, then we can make the good approximation that sqrt(1+2v/dv) ~= 1 + v/dv, giving de ~= v(V + dv).
But for most transfers from LKO you can't simplify the expression. None the less we know it is positive and significant. Edited November 18, 2015 by Yasmy

[quyxkh]

For another way of looking at the elephant, try this:

Whatever the rocket's doing, it gets kinetic energy out and then splits the kinetic energy so some goes ← that way with the exhaust and the rest of the rocket goes → this way (split so the momentum of the whole system stays constant too). But what the rocket's doing to split kinetic energy applies to [I]all[/I] the kinetic energy -- including any that's already there. That gets carried along with the rest. More energy getting thrown around, more acceleration. Edited November 18, 2015 by quyxkh
Gaarst corrected my physics, I had the split wrong, Thanks for that, Gaarst!

[GoSlash27]

[quote name='Right']Not that it invalidates your point, but that graph would happen even if you were wrong due to decreasing fuel mass and atmospheric drag.[/QUOTE]

Right,
That's kinda the point. You've got to be careful with car analogies because cars don't work like rockets.
A more accurate car analogy would go like this:
If you're in a parked car and accelerate to 10 kph into a crowd, you're not going to seriously injure anyone. If, OTOH, you're in a car doing 100 kph and you accelerate to 110, the difference that 10 kph makes is much more catastrophic.
the difference is nonlinear and that's what the Oberth effect is about.

Best,
-Slashy

[Gaarst]

[quote name='quyxkh']For another way of looking at the elephant, try this:

Whatever the rocket's doing, it gets kinetic energy out and then splits the kinetic energy so half goes ← that way with the exhaust and the rest of the rocket goes → this way. But what the rocket's doing to split kinetic energy applies to [I]all[/I] the kinetic energy -- including any that's already there. That gets carried along with the rest. More energy getting thrown around, more acceleration.[/QUOTE]

Be careful there, the energy is not evenly splitted across the ejected gases and the moving rocket. Momentum is, due to conservation of momentum laws.
If you take m[SUB]gas[/SUB]*v[SUB]gas[/SUB] = m[SUB]rocket[/SUB]*v[SUB]rocket[/SUB], then the energy is not necessarily the same.
Actually, as the ejected gas mass is often smaller than the rocket mass, its speed is higher, and therefore the kinetic energy of the ejected gas is greater than the kinetic energy of the rocket. That is in a reference frame where m[SUB]gas[/SUB]*v[SUB]gas[/SUB] = m[SUB]rocket[/SUB]*v[SUB]rocket[/SUB] at all points and times (or when both were static before burning the fuel).

The conclusion of your post is correct, but I'm not so sure about the way to get to this conclusion. Edited November 18, 2015 by Gaarst

[Right]

GoSlash,

Could you give me an example? I'm having trouble thinking of one outside of a few minor things.

For conceptual simplicity, I think of it this way: You're passing through a friendly black hole (practically linear hyperbolic orbit). If the singularity is stationary, and you do nothing, you emerge from its SOI at the same speed you entered. Gravitational energy gained = gravitational energy lost

If you burn 10d/V at the periapsis, your exit journey is a bit faster and shorter than your entrance. You emerge from the SOI with >10d/V. Gravitation energy gained > gravitation energy lost

RIC,

I agree. But most of my rockets' masses change significantly while burning.
If that graph was in vaccum and with infinite fuel flipped on, I'd be sold.

[FancyMouse]

[quote name='Right']
I agree. But most of my rockets' masses change significantly while burning.
If that graph was in vaccum and with infinite fuel flipped on, I'd be sold.[/QUOTE]
Think of everything (mass/energy/...) with respect to payload instead of the whole rocket.

[Red Iron Crown]

Familiar examples are hard to come by, and aren't very intuitive. The best I've come up with is:

Imagine a skateboarder on a zero-friction halfpipe with a bowling ball in his hands. He's oscillating up and down, reaching the rim exactly on each side before falling back down. He decides to throw the bowling ball away from himself to push himself higher. If he throws the ball at the highest point, he won't go as far up the other side than if he waited and threw the ball at the lowest, fastest part of the pipe. (I wonder if anyone has made a video of this?)

[Empiro]

The Oberth effect is true whether or not you're in the presence of a gravity field, but using gravity gives you another way to look at things.

Gravity wells are considered to be conservative fields. In particular, when you fall toward a body, you lose gravitational potential energy but gain the same amount of kinetic energy. The Oberth effect says that the faster you're moving (i.e. more kinetic energy you have), the more energy you gain from spent delta-V (it comes from the square in the kinetic energy equation: 0.5mv^2). Therefore, burning at periapsis or deep within the gravity well works best: you've traded all your gravitational energy for kinetic energy.

[GoSlash27]

[quote name='Right']GoSlash,

Could you give me an example? I'm having trouble thinking of one outside of a few minor things..[/QUOTE]

Right,
It holds true for every interplanetary journey, but I'll give Duna as an example.

To eject from Kerbin to Duna at 70 km altitude on a Hohmann transfer, you would need 1,078 m/sec. You would think that this is the best you can do, *but* if you hyperedit to an altitude of 8 Mm you will find that it only requires 649 m/sec. If you raise the altitude to 10Mm, the required DV again increases to 652Mm.
The advantage of the Oberth effect is offset by the disadvantage of climbing out of a gravity well and there is an altitude where the total losses are at a minimum.

This is also true at Duna's end. You don't necessarily want to retroburn at the lowest periapsis for the lowest DV expended. Closing your retroburn orbit will cost 616 m/sec if you do it at 60 km, but only 584 m/sec if you do it at 500 km. Above that, it gets more expensive.
Of course, it costs more to orbit your vehicle at 8Mm than you save by transferring from up there, but it illustrates the point:
Gravity wells work [I]against[/I] the Oberth effect, not for it (and are certainly not the cause of it). Gravity wells are valleys you have to climb out of. The Oberth effect, OTOH, is a slingshot effect that boosts your efficiency with velocity. They counteract each other.
The Oberth effect is about velocity, not gravity.
best,
-Slashy Edited November 18, 2015 by GoSlash27

[RocketPropelledGiraffe]

Well, let me expand on this discussion and add a question of reference frame:
The gain in kinetic energy is proportional to the velocity, that's the basic idea and that's why you usually want to do pro/retro burns at periaps. So far so good.

Now we think about going interplanetary, to the outer solar system. This is easy mode, because you burn prograde in LKO, which is also prograde relative to the sun. Ideally my orbital velocity relative to Kerbin adds to the orbital velocity of Kerbin around the sun (~9200 + 2200 m/s). Do everything in one burn and you are better off, simple.

Now let's go to the inner solar system. We burn retrograde in LKO to escape Kerbin's gravity. This is done retrograde to the sun as well, which means the velocity relative to the sun is minimal, because we have to substract our velocity relative to Kerbin ( ~9200 - 2200 m/s).
What is the tradeoff here? Is it better to burn to escape Kerbin in LKO, with good relative velocity to Kerbin, and then do another burn afterwards, because we travel at the full 9200 m/s in solar orbit then?

Think about appliying 2000 m/s delta-v in two instantaneous burns of 1000 m/s each (One to escape Kerbin, one to get to the destination):
If we do everything in LKO we get 1000 m/s applied at 9200-2200 = 7000 m/s relative solar velocity, and the second one at 9200-2200-1000= 6000 m/s, which is even slower relative to the sun.
If we do 1000 m/s in LKO just to escape we apply it at the same 7000 m/s. The second 1000 m/s burn occurs after escape from Kerbin, with 9200 m/s relative solar velocity.
With regard to the Oberth effect the second one sounds better. And, at least after three wheat beers tonight, it also seems very reasonable. :huh:

[GoSlash27]

RPG,
Same as prograde burns, retrograde burns add (or subtract) the orbital velocity to Kerbin's velocity about the sun. If you barely eject from Kerbin, your velocity relative the sun is Kerbin's velocity itself.
Better to start your burn at 9300 m/sec prograde or 7000 m/sec prograde?

The Oberth effect still works in your favor for retrograde ejections, but launching out of a gravity well works against you. There is a happy medium in there.

Best,
-Slashy Edited November 18, 2015 by GoSlash27

[RocketPropelledGiraffe]

Hey Slash,
I did add the relative velocity in my example, no? Where did I go wrong?

Are you saying when going to the inner solar system I am better off to burn prograde in LKO to escape (1000 m/s delta-v at 9200+2200 m/s relative solar velocity), then wait to escape Kerbin and then burn retrograde the rest of the way at 9200 m/s in solar orbit? That means always do two burns, not one, to get to the inner planets? I never tried it like that...

[GoSlash27]

[quote name='RocketPropelledGiraffe']Hey Slash,
I did add the relative velocity in my example, no? Where did I go wrong?

Are you saying when going to the inner solar system I am better off to burn prograde in LKO to escape (1000 m/s delta-v at 9200+2200 m/s relative solar velocity), then wait to escape Kerbin and then burn retrograde the rest of the way at 9200 m/s in solar orbit? That means always do two burns, not one, to get to the inner planets? I never tried it like that...[/QUOTE]

RPG,
You would burn prograde relative Kerbin's surface, but retrograde relative Kerbin's orbit about the sun. It's a net subtraction.
Forget about different effects of "velocity"; the velocity winds up the same no matter what. This is why the Oberth effect is so confusing; it's not about a change in velocity, but rather a change in kinetic energy.
What matters is how much change in kinetic energy you create in the process vs how much you lose by climbing out of the gravity well.

Best,
-Slashy Edited November 18, 2015 by GoSlash27

[Goody1981]

Hey everyone, massively long time lurker but finally thought I should join the forum when I saw this thread. I know it's flagged as answered now, had to get home from work...hopefully this helps! I read - maybe a year ago - a response in a similar thread. I wish I could remember who it was because it was a really great, simple example of the oberth effect - and I'd like to give credit.

Anyway, it went something like this: imagine you have a ball and you throw it up in the air with an initial velocity of 5ms-1. Using simple Newtonian equations (specifically, v^2 = u^2 + 2as) you can work out it gets to a height of 1.28m. If you catch it right there at the top, then do the same again (another 5ms-1 throw), it will go another 1.28m for a final height of 2.56m.

BUT! If you throw it from the original position at a speed of 10ms-1 it will reach a height of 5.1m.

The oberth effect :)

[Right]

[quote name='Red Iron Crown']Familiar examples are hard to come by, and aren't very intuitive. The best I've come up with is...[/QUOTE]

Sorry RIC, that was meant for GoSlash in looking for examples of when burning at the periapsis was not ideal. If that was what your example was intended to be however, I would say that is in fact a counter example.

I like the illustration though - While driving, I used to speed up at the top of a hill to save gas getting over the next hill, but since I found KSP I know thats backwards. I now speed up at the bottom.


[quote name='GoSlash27']This is also true at Duna's end. You don't necessarily want to retroburn at the lowest periapsis for the lowest DV expended. Closing your retroburn orbit will cost 616 m/sec if you do it at 60 km, but only 584 m/sec if you do it at 500 km. Above that, it gets more expensive.
Of course, it costs more to orbit your vehicle at 8Mm than you save by transferring from up there, but it illustrates the point:
Gravity wells work [I]against[/I] the Oberth effect, not for it (and are certainly not the cause of it). Gravity wells are valleys you have to climb out of. The Oberth effect, OTOH, is a slingshot effect that boosts your efficiency with velocity. They counteract each other.
The Oberth effect is about velocity, not gravity.
best,
-Slashy[/QUOTE]

Hmm, I don't think that example works. In both situations you named, burning at the periapsis is still optimal. I could be taking you too literally, but you said "In fact, periapsis burns are not necessarily the most effective even with Oberth effect."

Also, climbing out of gravity wells is hard, yes - but it the energy lost during the climb goes down the faster you do it. And these valleys work both ways, so if you gather a bunch of energy on the down swing, and then push out and lose less energy on the way out, you've got a net gain.

I concur, Oberth effect has nothing to do with gravity, but there is a different effect that - in addition to Oberth - makes periapsis burning more efficient. This one [I]is [/I]directly related to gravity. I tried to illustrate this with my black hole scenario, but that may have been a bad explanation. What did you think of it?

[RocketPropelledGiraffe]

Hmm, I guess I have to think about it some more and do some actual calculations. [URL="https://xkcd.com/1356/"]Orbital Mechanics[/URL] are difficult...

[Red Iron Crown]

[quote name='Right']Sorry RIC, that was meant for GoSlash in looking for examples of when burning at the periapsis was not ideal. If that was what your example was intended to be however, I would say that is in fact a counter example.

I like the illustration though - While driving, I used to speed up at the top of a hill to save gas getting over the next hill, but since I found KSP I know thats backwards. I now speed up at the bottom.[/QUOTE]
It doesn't work with cars for the reason I mentioned earlier: Cars have fixed power and variable "thrust" when at full throttle. Rockets have fixed thrust and variable power when at full throttle. Oberth [i]only[/i] works for reaction engines that carry their own reaction mass, it does not work for traction-propelled vehicles or reaction engines that are using the ambient medium for reaction mass (jets, propellers, etc). More specifically, it works [i]because[/i] they carry their reaction mass with them.

[Right]

[quote name='Red Iron Crown']It doesn't work with cars for the reason I mentioned earlier: Cars have fixed power and variable "thrust" when at full throttle.[/QUOTE]

If you're saying that one in fact doesn't save on fuel going from one hill to another by accelerating at the troughs versus the peaks, then I'll have to take issue.

By power you mean?

If the Oberth effect is as you described in you're linked post, then I see no reason why it would be limited to reaction engines. It is merely a side effect of kinetic energy having a quadratic growth with speed. I can't see the source of the velocity change impacting this.

[GoSlash27]

Right,
[quote]Hmm, I don't think that example works. In both situations you named, burning at the periapsis is still optimal. I could be taking you too literally, but you said "In fact, periapsis burns are not necessarily the most effective even with Oberth effect."[/quote]
Oh, I can assure you it *does* work. If you hyperedit into the orbits I outlined and try it, you'll see that it is indeed the case.
The problem is that you have to be at these altitudes in the first place in order to see the savings. The DV required to attain a 10Mm orbit is a lot more than the transfer from there would save.

[quote]Also, climbing out of gravity wells is hard, yes - but it the energy lost during the climb goes down the faster you do it. And these valleys work both ways, so if you gather a bunch of energy on the down swing, and then push out and lose less energy on the way out, you've got a net gain.[/quote]
You don't actually lose or gain any more energy going downhill vs. going uphill. All energy is conserved, it's just converted from potential to kinetic and back. You do gain or lose energy by entering a gravity well by using the velocity of the parent to your advantage. That's how slingshots work.
[quote]I concur, Oberth effect has nothing to do with gravity, but there is a different effect that - in addition to Oberth - makes periapsis burning more efficient. This one [I]is [/I]directly related to gravity. I tried to illustrate this with my black hole scenario, but that may have been a bad explanation. What did you think of it?[/quote]
Honestly, I couldn't make heads or tails of it. Apologies...
There is not an effect (at least in KSP) that makes periapsis burns most efficient other than the Oberth effect. You have 3 factors at work:
-The Oberth effect, which is completely a result of velocity. This tends to make low periapsis burns more efficient because your orbital velocity is higher.
-Gravity itself, which is (in KSP) a result of the parent body's mass. This tends to make low periapsis burns [I]less[/I] efficient.
-KSC's location, which means that (for better or worse) you start out at the bottom of the gravity well. Being that it's always free to burn from where you are and it always costs DV to change your orbit to something else, it tends to make low periapsis maneuvers more efficient.

Best,
-Slashy

[Red Iron Crown]

"Power" in the physics sense, work per unit time (in SI terms, joules per second or Watts).

Carrying the propellant is important because the reaction mass needs to have the higher kinetic energy from additional speed just as the vessel itself does. It is the greater exchange of kinetic energy between propellant and vessel at higher speeds that is the heart of the Oberth effect. These are the only type of engine that apply the same force no matter the speed. Engines that do not carry their propellant do not apply the same force, they apply the same power; so their force decreases with speed.

It is really best to stop trying to think of it in terms of cars or physics class frictionless carts, the intuitive sense of how these things work that we've all built over decades of experience does not apply to this problem.

Another simple way to think of it is: To change the energy of an orbit, work must be done. Recall that work is determined by the equation: Work = Force * Distance. Since a given burn will apply the same force for the same time, if the burn is done at higher speed more distance will be covered, thus more work is done and the orbit's energy is changed by a greater amount.

The real key here is thinking in terms of energy, not speed. When a vessel is in an eccentric orbit its speed is constantly changing but its total energy is constant. Oberth is about applying force when the division of that energy between kinetic and potential energy favors kinetic as much as possible (i.e. lowest altitude/highest speed), to maximize the energy change from a given change in speed.

[Lukaszenko]

[quote name='Right']If you're saying that one in fact doesn't save on fuel going from one hill to another by accelerating at the troughs versus the peaks, then I'll have to take issue.

By power you mean?

If the Oberth effect is as you described in you're linked post, then I see no reason why it would be limited to reaction engines. It is merely a side effect of kinetic energy having a quadratic growth with speed. I can't see the source of the velocity change impacting this.[/QUOTE]

In a car you have to burn more fuel to accelerate at the bottom of the hill, than you do to accelerate the same amount at the top of the hill. So in the end, you don't actually gain anything. In fact, you lost some energy because you put kinetic energy into your fuel...and then you just threw that kinetic energy away. A car engine only harnesses the fuel's chemical energy.

In a rocket, you will burn the same amount of fuel in both situations. So, you DO gain something. A rocket harnesses the chemical AND kinetic energy of its fuel. Edited November 18, 2015 by Lukaszenko