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Part of this answer was inspired if I remember correctly from a question about how to reverse the direction of a satellite in LEO with the minimum delta-v. Whether this was the original question or not, I'm still curious about the answer.

Apparently I would know this if I were an active KSP player, but I'm not.

So I would probably have to either do an internet search, or write a python script and do some hunting in maneuver space. If I did that I'd explore using 3-body effects, getting assistance from the Moon's or Sun's gravity by first rising to a higher, elliptical orbit, or maybe even an Apollo-style free-return.

However, it seems that the lowest delta-v maneuver is already known by many, just not me :-(

What is the lowest delta-v maneuver to reverse the direction of a satellite in LEO? For a satellite in a circular, equatorial LEO with an altitude of 600 km, and what would the value of that delta-v be?

The initial velocity is $\sqrt{GM_E/a}=$7558 m/s and so the worst case delta-v is double that.

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    $\begingroup$ The initial question included a rendezvous, so solutions that require precise timing (of the position of the Mun/Moon) might not work. Applying the answer that pericynthiongave is indeed what I'd do in game, with the altitude of apokee/apogee constrained to the Kerbin/Earth Sphere Of Influence size, and timed so the recircularization was also the rendezvous. I'll also add the single burn brute force method can fail in practice with your vessel falling into the atmosphere, something not obvious from pure numerical solutions. $\endgroup$
    – Slow Dog
    Commented Sep 1, 2017 at 7:24
  • $\begingroup$ @SlowDog you are welcome to leave an expanded answer. If I need to adjust the question a bit let me know, or I think you can just edit the question yourself. I'll see the flag and if it's reasonable I can accept your edit. I can't remember how the rendezvous was explained, so if you can help there it will be really appreciated. Thanks! $\endgroup$
    – uhoh
    Commented Sep 1, 2017 at 9:43
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    $\begingroup$ I don't think that's necessary. There's no reason to make changes that would affect Pericynthion's fine answer. $\endgroup$
    – Slow Dog
    Commented Sep 2, 2017 at 16:31

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Reversing direction is a special case of inclination- / plane-change.

Here's one way to find a better upper bound on the necessary delta-V for a 180 degree plane change:

Assume a 2-body system (i.e. ignoring the sun and moon). Burn once to increase your orbital energy to C3 = 0 (i.e. escape velocity), coast out to an infinite distance where your velocity is 0, make an infinitesimal velocity change, fall back to 600 km altitude and burn again to recircularize.

Escape velocity at 600 km altitude is 10690 m/s. Therefore the total delta-V required by this technique is $ 2*(10690 - 7558) = 6264 \mathrm{m}/\mathrm{s} $ - only 41% of that required by the single-burn "brute force" method.

Of course as given above it requires infinite distance and time, but you can still do pretty well by raising your apogee to something merely "very high", reversing direction and subsequently recircularizing at perigee. This is known as a bi-elliptic transfer with plane change.

As you suggest, you can do better still by making use of lunar flybys or Weak-Stability Boundary transfers. Those don't lend themselves to clean analytic solutions though. I'll also point out that the "lowest delta-V maneuver" may not be a practical one - in the real world there will definitely be time and distance constraints, and the optimal maneuver is also likely to depend on the starting conditions (e.g. the inclination of the 600 km circular orbit).

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    $\begingroup$ A KSP player did the math for in-game orbits (for which raising it to infinity is actually practical as it's the edge of the planet's Hill sphere) and it's about 60 degrees of plane change that make the trip to infinity the cheapest approach. $\endgroup$
    – Loren Pechtel
    Commented Sep 2, 2017 at 4:19
  • $\begingroup$ @LorenPechtel can you find a link for that - I'd like to read more and see the math. Thanks! $\endgroup$
    – uhoh
    Commented Sep 2, 2017 at 11:36
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    $\begingroup$ @uhoh forum.kerbalspaceprogram.com/index.php?/topic/… My memory was wrong, it's less than 60 degrees. Note that the years have done some damage to the math in the thread, some posts have been fixed, some haven't. $\endgroup$
    – Loren Pechtel
    Commented Sep 2, 2017 at 13:57
  • $\begingroup$ @LorenPechtel oh, that's good enough, I can understand what they are doing and I'll give it a go soon. Thanks! :-) $\endgroup$
    – uhoh
    Commented Sep 2, 2017 at 14:14
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    $\begingroup$ I'm surprised the optimal solution isn't gradually increase inclination via precession. $\endgroup$
    – user3528438
    Commented Apr 11, 2020 at 22:01
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I know this is quite an old post, reading the answers and thinking about it from a ksp perspective, I'm a gamer not a physicist lol, wouldnt the cheapest way to change orbital direction be to simply put your craft into a solar orbit, then burn in solar orbit to get back to the celestial body in question, but simply align your encounter so that your PE is on the opposite side of the planet, then burn retrograde at PE to put your self back into orbit in the opposite direction? Itd probably take 6 months or more to minimize dV required for an encounter once in solar orbit, but it seems like it might take a lot less dV then trying to normal burn and essentially brute force it, as well as less then going an infinite distance away and reversing direction

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    $\begingroup$ Welcome to Stack Exchange! +1 because you've made me stop and think but in general an answer should be more definitive than asking a question: "..wouldn't the cheapest way to change orbital direction be to...." Escape velocity from a circular orbit is $\sqrt{2}$ times the current orbital velocity, so to escape to heliocentric then drift away from Earth slowly, then later to drop back down would be $2(\sqrt{2}+ \delta) v_{orb}$ (where $\delta$ is small) which is about 42% larger than the worst case I give in the question $\endgroup$
    – uhoh
    Commented Apr 10, 2020 at 22:58
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A seasoned KSP player with foreknowledge of need of reversal of orbit direction would give the craft good heat shielding, lean aerodynamic profile and good hypersonic wings.

Perform a mild reentry burn. Reenter. When the air density is sufficient, pull up to level the flight, bank left or right and perform a 180 degrees turn slow enough that G-forces won't be too destructive. Then perform a burn that sets your apoapsis above the atmosphere and then circularize at the apoapsis.

Most of delta-V will be lost to atmospheric drag, and this number is difficult to determine, depending on the craft and its aerodynamics a lot; a ballpark of my KSP intuition tells me about 10-20% of delta-V needed to start. The obligatory reentry and circularization burns would be minuscule, of order of 100m/s.

Additionally, it might be possible to use scramjet engines for the most expensive part of the maneuver.

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  • $\begingroup$ I don't believe this will work with existing technology in Earth's orbit. I'm not asking for a game answer or what your intuition or inner space cadet believes. Please post enough data about this maneuver, including the value of the propulsive delta-v required so that I can check it numerically. What is the altitude, the lift/drag ratio, the time spent executing this huge momentum transfer to the atmosphere, etc. Answers in SXSE need to contain verifiable information. $\endgroup$
    – uhoh
    Commented Sep 2, 2017 at 11:32
  • $\begingroup$ btw "inner space cadet" isn't meant to be negative. When I think of orbital maneuvers I turn to python and odeint, I'm secretly jealous of KSP players with all this intuition I'm missing out on, but I will never admit it openly. Never! $\endgroup$
    – uhoh
    Commented Sep 2, 2017 at 14:17
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    $\begingroup$ @uhoh: I could give you the deorbit and circularization burns, but they are actually a minor part of the delta-V required. As I wrote, the bulk of delta-V (recovery of what was lost to drag) is beyond my ability. X-20 Dyna-soar was meant to be capable of maneuvering that way (not sure about 180 degrees, but some inclination change definitely), $\endgroup$
    – SF.
    Commented Sep 2, 2017 at 14:46
  • $\begingroup$ Please add the total propulsive delta-v to your answer. Otherwise I can't compare to the other values. If you don't have a viable spacecraft maneuver and the amount of propulsive delta-v used, then this doesn't qualify as an answer to the question. $\endgroup$
    – uhoh
    Commented Sep 2, 2017 at 15:01
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    $\begingroup$ No real wing could survive enough heat loading to make this approach useful. $\endgroup$
    – pericynthion
    Commented Sep 2, 2017 at 16:14
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I've done this on the moons, not kerbin. Burn down to a low orbit, maybe 75k for kerbin, then burn normal or anti-normal until you're done. I think you'd be looking at 2-5k delta v, excluding the altitude change to get you there. For kerbin, i've found it simpler to just launch another. Note that the delta V for the moons is significantly smaller, so it makes it more reasonable there. For instance it's only a few hundred around minimus.

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    $\begingroup$ The question is not about games. $\endgroup$
    – Organic Marble
    Commented May 12, 2018 at 17:50
  • $\begingroup$ @OrganicMarble: The question did mention KSP, but this answer is rather too focused on the minutia of KSP and less on the general principles. (Also, since it advises dropping altitude rather than gaining it, I'm not convinced it's correct.) $\endgroup$
    – Nathan Tuggy
    Commented May 12, 2018 at 17:58
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    $\begingroup$ Please note that the questioner stated " I'm not asking for a game answer" in a comment. $\endgroup$
    – Organic Marble
    Commented May 12, 2018 at 18:01
  • $\begingroup$ The question is looking for the lowest energy--you're giving about the highest. Not only do you pay the full cost this way, but by going lower you increase the cost. There is absolutely no redeeming feature of this approach. $\endgroup$
    – Loren Pechtel
    Commented Apr 11, 2020 at 4:11

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