Are the propellant mass and transfer time from orbit A to orbit B the same as from orbit B to orbit A, considering low-thrust propulsion? That is, in the case of dealing with Keplerian orbit parameters and having both points in the sphere of influence of Earth with any third body or perturbation effects neglected. In other words, can we say that the return transfer is a "mirror copy" of the forward one? If that is not generally the case, then under what conditions can we witness such symmetrical transfers?

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    $\begingroup$ There is a problem with your question as asked. What exactly to you mean by "points A and B"? If you mean orbits A and B the question is much more straightforward and answerable. $\endgroup$ – uhoh Jan 14 '17 at 9:36
  • $\begingroup$ To amplify on uhoh's comment, we can't even say whether points A and B are the same. For example, say point A is where I am, and point B is the location of the Apollo 11 landing. It's quite possible that I can get from A to B simply by waiting for the earth's motion to carry me there. In fact, there is guaranteed to be a frame of reference in which A, now, is the same as B, then. $\endgroup$ – Ben Crowell Jan 14 '17 at 22:25
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    $\begingroup$ You are right. I did not pose the question very clearly. I meant a point as a set of the six orbital parameters. $\endgroup$ – Aleksandar Petrov Jan 15 '17 at 2:32
  • $\begingroup$ @AleksandarPetrov: How about editing the question? If "low-thrust propulsion" means that expenditure of reaction mass is negligible, then I think the answer is yes, by the time-reversal symmetry of Newton's laws. But if you're using significant reaction mass, then it breaks the time-reversal symmetry. You can only expel reaction mass, not collect it. $\endgroup$ – Ben Crowell Jan 15 '17 at 3:48
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    $\begingroup$ Changed it to orbit a and b, as I think it's fairly clear that's what the intent of the question is. $\endgroup$ – PearsonArtPhoto Jan 15 '17 at 16:39

TL;DR: No, continuous thrust maneuvers introduce too many complexities for the "return" trajectory to be a mirror copy.

For continuous thrust maneuvers, we need to consider how many orbital parameters we want to change at once, and whether we plan on having a minimum efficiency during that transfer. Even if only one orbital parameter is changed at once, then we still can't have a mirror trajectory: this is because each orbital element will have an important effect on the optimum thrusting to change that element, as shown in Fig 1 below. Then, if attempting to change multiple orbital elements at once, we have to select which of the six "summation" control laws we wish to use (Petropoulos, Ruggiero, Naasz, and a few more). Each will emphasize the change of one orbital element more than another, and they all will lead to different "return" trajectories.

For example, changing semi major axis (a) from 42164km to 24396km (around Earth) requires 21kg of fuel (supposing a Snecma PPS1350 engine) and 45 days, whereas going from 24396km to 42164km requires 17kg and only 37 days.

Fig.1 (Ruggiero and Pergola - IEPC 2011-102) Fig.2 (same source)

NOTE: Both figures are from Ruggiero and Pergola - IEPC 2011-102.

Sources: those of my MSc. thesis on continuous thrust applied to return interplanetary mission designs. I also hope to open source my propagator in the coming weeks or months, which should allow anyone to try out a variety of missions and control laws for continuous thrust missions.

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    $\begingroup$ @uhoh not yet, no. I'm still finishing up the technical work and I'm defending mid March. $\endgroup$ – ChrisR Jan 16 '17 at 16:51
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    $\begingroup$ @ChrisR: In the example that I use, the initial mass and all the orbital elements are identical (apart from the semi major axis), so it's an idealized case for continuous thrusting. But it's not the idealized case in which zero reaction mass is consumed, because the roles of initial and final mass are swapped. That's the point of my answer. $\endgroup$ – Ben Crowell Jan 16 '17 at 23:02
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    $\begingroup$ @uhoh, I have open sourced my code for space mission design. It's not complete at all yet (one can only do patched conics for interplanetary mission design and the J2 pertubations don't yet affect the true anomaly), but it does have a few cool features like exporting the mission to Cosmographia. I've open sourced it because I'll use it in my interplanetary mission design class, where I might need to show the code to my professor. Anyway, here it is: github.com/ChristopherRabotin/smd (it's in golang by the way). $\endgroup$ – ChrisR Jan 21 '17 at 6:19
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    $\begingroup$ @ChrisR that's fantastic!!! I will definitely take a look. The open source approach has so many benefits, glad to see you've gone that direction. Also, I have been looking for an excuse to learn go for quite a while, this should make it a lot easier. Although I have an "allergy to curly braces", this seems too interesting to pass up! $\endgroup$ – uhoh Jan 21 '17 at 6:30
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    $\begingroup$ @uhoh I forgot to update the thread here, but my thesis is now online: scholar.colorado.edu/asen_gradetds/184 . $\endgroup$ – ChrisR Jul 6 '18 at 4:30

The answer is no except in the mathematically idealized case where the reaction mass used is negligible and the engine is an idealized device capable of making any force up to some maximum.

Let's consider a Hohmann transfer orbit from a lower orbit to a higher orbit. This requires a $\Delta \textbf{v}_1$ out of the lower orbit and a $\Delta \textbf{v}_2$ to insert into the higher orbit. The force exerted on the spacecraft during one of the burns is $\textbf{F}=m\textbf{a}$, where $m$ is the mass of the spacecraft, which is a function of time because we're expending reaction mass. The mass is large during the low-altitude burn and smaller during the high-altitude burn.

Now let's time-reverse the motion. Under time reversal, $\textbf{a}$ stays the same, which means that $\textbf{F}$ has the same direction. But $|\textbf{F}|$ doesn't stay the same, because $m(t)$ is now a different function. In this version, the mass is large during the high-altitude burn and smaller during the low-altitude burn. This breaks the symmetry.

However, when the reaction mass is negligible and the engine is an idealized device capable of making any force up to some maximum, $m(t)$ is a constant function, and time-reversing the orbit keeps $\textbf{F}$ the same.

This argument does not depend on the assumption of a Hohmann transfer orbit, which was just illustrative. The only assumption was that the gravitational field was static.

None of this actually depends in principle on whether it's low-thrust or high-thrust propulsion, continuous acceleration or acceleration in short burns. However, because of the high exhaust velocities used in low-thrust propulsion, it may be a better approximation to say that the reaction mass is negligible. I would assume that this is why, in the numerical example given by ChrisR's answer, there is at least approximate symmetry.

  • $\begingroup$ It was mentioned to me that there is also the question whether the case should consider Oberth effects and non-spherical primary objects. $\endgroup$ – kim holder Jan 16 '17 at 15:25
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    $\begingroup$ @kimholder: The Oberth effect isn't a separate effect that needs to be taken into account in addition to Newton's second law. Re a nonspherical primary, there is no assumption in the argument I've given that the gravitational field is of any particular form. I only assumed implicitly that it was static. The Hohmann transfer orbit is only an example. The argument doesn't require a Hohmann transfer orbit. I'll edit the answer to try to clarify those points. $\endgroup$ – Ben Crowell Jan 16 '17 at 15:29

Given no third body effects, atmospheric drag, etc, and an instantaneous burn, the amount of fuel required to switch to an orbit is the same as is required to return to the previous orbit. This assumes the following:

  1. The starting mass is the same.
  2. No fuel spent on station keeping.
  3. The exact reverse trajectory is kept.
  4. The profile of the thrust is the same.

For the low thrust case, however, the profile of the thrust changes. You have more thrust at the end of the maneuver, which will change the optimal trajectory slightly. I suspect moving to the closer orbit would take less fuel then moving to the higher one, but I haven't do a simulation to actually prove that. I suspect that except in extreme cases, where most of the mass of the spacecraft is fuel, then this is essentially the same.

  • $\begingroup$ I'm confused by your caveat for the "low thrust case". Does "...more thrust at the end of the maneuver..." mean more acceleration due to a lower mass a the end? A good ion engine will have roughly an order of magnitude higher specific impulse, so for a given delta-v the relative change in mass would be less than for a chemical engine. But maybe I am missing the point. $\endgroup$ – uhoh Jan 15 '17 at 18:32
  • $\begingroup$ Sorry, this answer is incorrect. Even a basic Hohmann transfer (supposing a perfect two-body computation) will lead to opposite departure and arrival deltaVs. Only the absolute value of the velocities will be the same. $\endgroup$ – ChrisR Jan 16 '17 at 5:28
  • $\begingroup$ I don't think this answer is right, for the reasons given in my answer. $\endgroup$ – Ben Crowell Jan 16 '17 at 15:21
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    $\begingroup$ @ChrisR there could be a commutation relation that involves a sign change, Considering the question includes "In other words, can we say that the return transfer is a 'mirror copy' of the forward one?" the OP has some flexibility here, and would still be interested in a less-than-purely-mathematically-rigorous use of the term. So do you think that "incorrect as written yet fixable with a caveat" might fly in this case? $\endgroup$ – uhoh Jan 16 '17 at 16:44
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    $\begingroup$ @uhoh yes, I agree. Let's go with "the answer is correct within the OP's constraints". Thanks for shining a light on that quote from the OP. $\endgroup$ – ChrisR Jan 16 '17 at 20:53

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