Do you have intuition for "gravity losses"? I don't really, but if you do, it seems like it should translate.

The Oberth effect seems to be the same mechanism but with a sign flip - it's "gravity gains".

 at this point you've thought about this much more than me, so maybe they don't. But gravity losses refers to the fact that thrusting against gravity is wasteful, compared thrusting perpendicular to gravity if you're in a circular orbit already. So it seems to stand to reason that thrusting with gravity would be even more efficient. Maybe that's not true, it just seemed like it might be, sorry if that was actually a wild goose chase!

 Wait no, I'm still confused. How do gravity losses factor into this?

 Never mind, I just got it after reading Evan's comment. (Not an error in Wikipedia, was an error in my understanding.)

 Hmm, I thought I did, but the fact that I don't understand the bit flip framing makes me think I don't. Also, the Wikipedia page on the subject says that gravity losses are zero when the thrust is perpendicular to the gravitational field, whereas I thought it would be zero for any direction that isn't away from the gravitational field. (i.e. perpendicular or towards.) Unsure whether this is an error in Wikipedia or an error in my understanding.

From the rocket's reference frame, a large mass that passes by quickly has less impact than one that passes by slowly. Same (ish) force, less time, lower change in momentum. (Only "ish" because the trajectory follows a different curve.) This is hopefully intuitively true if you have a rocket at rest, a large mass approaching from far away, and no burn.

Performing an impulsive burn when the mass is close by changes from one of those scenarios to the other. The incoming mass has larger impact than the leaving mass; this is in addition to the impact of the impulsive burn.

The convenient reference frame for this is perhaps awkward to choose; the above might make intuitive sense the best if you look backwards in time from the rest frame of the rocket before the burn, then forward from the rest frame after the burn, then try to patch those two together to bridge the burn.

Hopefully that's the right level of handwavy; I assume if you wanted the calculus worked in the (initial) frame of reference of the rocket you'd have done that.

 This would infact work in the abscence of a gravitational field, as long as youre observing from an inertial frame of refference, I think.

 Thanks, this together with Chris's comment was the relevant insight I needed.

I'm still confused why Wikipedia talks about it in terms of speed, since this intuitive explanation has nothing to do with speed. And wouldn't work in the absence of a gravitational field, yet Wikipedia seems to think the Oberth effect would apply even there.

I don't know if this explanation will be any better than the typical online one, but here goes: A fixed amount of propellant being burned usually causes a fixed change in velocity. But kinetic energy is related to the square of the speed. So if I increase my speed by 1m/s when my speed is already high, that will do more to increase my kinetic energy than if I were to increase my speed by 1m/s when my speed was low.

 Your comment was just a restatement of the typical explanation online, it did not answer any of my questions.

 The rocket will remain at zero speed in its frame of refference afterwards as it was before. The same way, the speed and KE of exhaust will be the same in rockets refference frame independant of weather its firing in aphelion on perihelion. You must view this w.r.t. the gravity well, e.g. the planet to observe the Oberth effect. Imagine that you fire the rocket at aphelion (farthest), both rockets and exhausts PE remains the same, rocket gains some speed and KE in the direction it was going and exhaust leaves at exhaust speed (perhaps even changing orbit direction) and exhaust KE. Now when you do this in perihelion (closest), w.r.t. to the well, the exhaust basically stops dead, with 0 KE, and low PE, while the rocket gains all of the KE of the exhaust; all with respect to the well, which is most important. 

Edit. Forgot to mention that the rocket is not an inertial frame of reference because it is accelarating, firing the exhaust. The planet, for the purposes of the Oberth effect, is an inertial frame of refference.

 I am not saying that relativity doesn't apply here, I'm just saying you can't use it to get a 

 explanation of the Oberth effect or easily calculate how much extra speed it will give you.

 That doesn't make any sense. Someone inside the rocket should be able to perform the same physics calculations in their own frame of reference as someone on the planet performs in the planet's frame of reference and get the same ultimate behavior for the rocket. Relativity applies everywhere, there is no effect that cannot be explained in terms of relativity; if there were, it would be a counterexample to the theory, proving it wrong.

Yes I agree it's unsatisfying. I think the issue is that unlike a "slingshot" maneuver where you can reason about it by considering the relative frames of the planet and the sun, it doesn't really make sense to reason with relativity here because there is only one inertial frame - the body you are Oberthing around.

 Yeah that's the typical online explanation, doesn't make sense to me. It's presenting the math, but doesn't explain the actual physical principles at play. That's not understanding, just memorization of a formula. 

https://www.lesswrong.com/posts/N2pENnTPB75sfc9kb/outside-the-laboratory

In particular, I don't understand how this can be valid from the rocket's reference frame, where it's always at speed zero before firing, and therefore should always observe the same behavior during and after firing as per relativity.

None of the explanations I've found online have made sense to me. What's the explanation from the rocket's reference frame?

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