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Although my asking this question now was inspired by this one, it's a question that's occurred to me before in the context of orbital capturing. Another way to put this might be: is there an equivalent Oberth effect for a retrograde burn?

The wiki is ambiguous on this point. I also reviewed Hop David's excellent answer to this question, but I can't seem to get my old brain to process it in reverse.

If the answer is yes, are there any instances where this has been used for orbital capture around some object?

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    $\begingroup$ To change apses, fire at periapsis. To change inclination fire at the node further away from the planet. A mantra a day keeps the ballistician away. $\endgroup$
    – Deer Hunter
    Dec 4, 2014 at 22:26
  • $\begingroup$ note that burning prograde at periapsis you are increasing the strength of Oberth effect (while benefitting from it). Burning retrograde, you're decreasing it. But still, getting down from a very high orbit, one burn to bring periapsis as low as possible, then burn(s) at periapsis to lower the apoapsis as desired will cost way less than spiralling down to the planet. Although in most cases it's all moot due to aerobraking. $\endgroup$
    – SF.
    Jun 30, 2016 at 22:30
  • $\begingroup$ Related (duplicate?) space.stackexchange.com/questions/9554/… $\endgroup$
    – SF.
    Mar 1, 2017 at 12:49

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Yes, it works exactly the same in reverse. Every orbit insertion about a planet has taken advantage of this effect, by virtue of targeting each insertion to take place as close to the body as reasonable. The Moon, Mars, Venus, Jupiter, Saturn, and Mercury. Low-thrust missions like SMART-1 and DAWN don't get this Oberth advantage, since they can't generate a large impulse in a short time, so they have to start slowing down long before they can feel the gravity of the object. However they get a high Isp advantage.

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Oberth works both retrograde and prograde. The equivalent Oberth effect for retrograde is in fact the Oberth effect =) Pretend we are at periapsis and are about to get an infinitesimal but nonzero impulse from thrust. Let's look at the prograde impulse case first.

Prograde, we have some orbital velocity in the same direction as thrust. The exhaust is pushed retrograde and therefore into a lower orbit. It loses energy. Conservation of energy says it has to go somewhere - it goes into propelling the rocket itself, faster/into a higher orbit than it would get simply from the impulse.

Retrograde, we have some orbital velocity in the opposite direction as thrust. The exhaust is thus pushed into a higher orbit because it is moving prograde. Conservation of energy says it had to get that energy from somewhere - it gets it from the rocket itself propelling the thrust, so that the rocket gets a slower/lower orbit than it would get simply from the impulse.

Because normal, antinormal, radial, and antiradial are all orthogonal to the velocity vector, no Oberth effect is obtained from these components of any impulse. Hope this helps!

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  • $\begingroup$ Very helpful answer. Is there an equivalent way to frame the Oberth effect in terms of conservation of momentum? CoE has always been hard for me to work into my intuition. $\endgroup$
    – Russell Borogove
    May 14, 2015 at 14:09
  • $\begingroup$ Framing it without energy is turning out to be harder than I thought - I think without using energy methods you can't explain why it isn't just what you'd expect out of the rocket equation. If you think about it you get different Oberth effects at different orbital heights because it's intrinsically conserving orbital energy. Can you think about it in terms of work? Consider that you're doing the same work with a force F whether the ship is aligned prograde or retrograde - and this holds well with understanding that the four other axes do no work - they're orthogonal to the distance traveled. $\endgroup$
    – Paul L.
    May 14, 2015 at 14:47
  • $\begingroup$ I have the same intuitive block about work as I do energy. On the bright side, I'm starting to think that when I unblock on it, I'll understand relativity as well for free. Thanks! $\endgroup$
    – Russell Borogove
    May 14, 2015 at 21:45
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I think the Oberth effect is easier to understand without referring to conservation of energy or momentum. I don't know the etiquette of repeating answers here but if anyone objects I'll try to figure out how to post a link to another question where I gave essentially the same answer, and remove this one:

The Oberth effect can be explained this way. Every second you are falling in a gravitational field, it is altering your velocity--perhaps helping you, perhaps hurting you. If you do your burn when you're deeper in the gravity well, you're either increasing the time gravity will help you, or decreasing the time it will hurt you. For example:

  • If you're moving away and you want to slow down, burn now. By slowing down, you spend more time near the planet with gravity pulling you back.
  • If you're moving toward the planet and you want to speed up, wait until you're closer. If you burn now, you spend less time falling toward the planet getting accelerated by gravity.

...and so on for the other cases.

The equations and laws tell you that the Oberth effect will happen. They don't tell how it works. If your capsule is going faster, something had to push or pull on it. In this case, that thing was gravity. Conservation of energy doesn't speed things up or slow them down, it just tells you that something will do so.

(Related: is there a tag for soapbox? Tired rant, maybe?)

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