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Which of these three is the more efficient trajectory for orbiting Mun?


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Can any of you kerbalnaughts tell me which is the most fuel efficient way of reaching Mun out of the three options below? The DeltaV of all approaches is apparently 815m/s. However, getting into orbit around the mun with those three approaches will require different amounts of retrograde burn, i.e. fuel. So which approach will require the least fuel cost? Is having a longer "purple line" / longer amount of time in the Mun's SoI the better way?

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I would say go with a path where you will pass the mun before it on its orbital trajectory, not after it. That way the gravity will work in your favour as you slow down and getting captured in its orbit.

Also, while the longer the purple line meaning the longer time you have in its SOI, it doesn't always mean the path is favourable to you. Some path will slingshots you to a completely different direction, which may require even more fuel to correct.

I think you may find a better trajectory playing with the angle and orbit (wait a few orbit till optimal moment, for example.) However, out of the 3, I wouuld either go with 2 or 3. Less turn and twist on the path means easier to handle, at least in my experience.

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Ah thanks!

Does escaping from Kerbin's orbit use more fuel than circularising my orbit at the Mun; i.e. if I have a high delta V escape from Kerbin but a low Pe at the Mun, is that more fuel efficient than a low Delta V escape from Kerbin with a high Pe?

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Does escaping from Kerbin's orbit use more fuel than circularising my orbit at the Mun; i.e. if I have a high delta V escape from Kerbin but a low Pe at the Mun, is that more fuel efficient than a low Delta V escape from Kerbin with a high Pe?

There's not a simple answer to that, I'm afraid. It depends on how much more velocity you're carrying when you encounter the Mun and how different the periapses are. Why not quicksave and try a few different approaches? That will give you a better feel for how to do it efficiently.

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Ah thanks!

Does escaping from Kerbin's orbit use more fuel than circularising my orbit at the Mun; i.e. if I have a high delta V escape from Kerbin but a low Pe at the Mun, is that more fuel efficient than a low Delta V escape from Kerbin with a high Pe?

It depends on exactly what you mean by "high/low delta-V" and "high/low Pe".

In general, there are two things going on. #1: the Oberth effect favors burns at high velocity, because you can better change your orbital energy at high velocity*. This favors a low Munar periapsis, so you can do your braking burn at a higher Munar orbit velocity.

*The Oberth effect comes out of the fact that kinetic energy is proportional to the square of velocity. Thus, braking from 100 to 0 m/s eliminates 5,000 J/kg of kinetic energy (KE = 1/2 m*v^2), while braking from 1100 m/s to 1000 m/s eliminates 105,000 J/kg of kinetic energy. Same expenditure of delta-V, but the second changed your kinetic energy much more. This also works in the reverse direction, and explains why Kerbin escapes are more efficient with low periapsis.

#2: you want to spend the absolute minimum amount of dV necessary to get into the Mun's SOI. At periapsis, your orbital velocity will be, at absolute minimum, escape velocity (as your trajectory in is, at best, Munar escape in reverse). The less velocity you have remaining at Munar capture, the less you will have to cancel when you hit periapsis.

The conclusion from this is that you want a transfer which just barely puts you in the Mun's SOI long enough for its gravity to pull you in, and gives you a projected periapsis as close to the surface as you dare. The only exception I can think of is if you are planning a Munar flyby only: for that, you want to do a free-return trajectory, which is a different story.

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One thing you can do is set a second node at the lunar periapsis, and have it burn retrograde until you have an orbit. Then you add up the deltaV from both maneuvers. My gut says that whichever has the lowest periapsis will win, because of the Oberth effect -- you want to do your orbit injection burn as low as possible on Mun.

You're implicitly saying that you're doing a Hohmann transfer and are expecting to do exactly two burns to set up an orbit. This is not the cheapest overall option for getting to the Mun, but it's close enough that most of us don't care to optimize much more than that.

What is cheapest without gravity assist is a bi-elliptic transfer. That makes you start at Kerbin 70km, fly all the way out past Mun and Minmus to the edge of Kerbin's SoI, then raise periapsis there. You burn more overall to get to that point, but the savings come when you enter Mun's SoI: instead of entering at 366 m/s with a Hohmann, you enter with 175 m/s. Overall you save about 40 m/s by doing the bi-elliptic. But it's more complicated, and it's just 40 m/s, so most people don't bother (I don't). When you're really big into optimizing, you can use multiple gravity assists off Mun to raise your Kerbin periapsis and get close to matching Mun's orbit. The closer you are to Mun's orbit, the less excess speed you have to burn off in order to establish an orbit. IIRC you can save a total of about 80 m/s.

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I hope you have used 'Hohmann Transfers' to raise and lower your orbit around Kerbin (or anywhere else) already.

Then consider raising your orbit to the same as Mun's (without worrying about an encounter): your 'transfer' burn should raise the apoapsis to the required altitude, then you drift there and 'circularise'. Note that you are not 'escaping' Kerbin's orbit, you and Mun are simply orbiting it at the same altitude. You might like to try this to see how much deltaV the circularisation burn requires without Mun's gravity to influence it.

Introducing the requirement for an encounter and orbit, you obviously need to do the transfer burn at the right point around your (comparatively low) Kerbin orbit. When you orbit around Mun it, and therefore you, are still orbiting Kerbin. You will, in fact, be orbiting Kerbin with the exact same position and altitude as Mun on average (your Mun-orbit representing only insignificant wobbles relative to Kerbin-orbit). After the transfer burn and transfer drift to apoapsis, however, before you do the 'injection' burn, you are orbiting slower than Mun (even if your Ap is higher), which is why you fall back to Kerbin. The injection burn therefore 'circularises' your Kerbin-orbit and, if Mun were a massless-point you'd therefore just want to time and scale the transfer burn to get you as close to it as possible. In practice, of course, Mun represents a bloody great rock with gravity and the more obvious purpose of the injection burn is to establish an orbit around it - so miss, but only just.

What all this means is that the 'ideal' transfer burn would put you on a direct collision-course with Mun except that actually hitting it would be what aerospace professionals call "a bad thing". Missing by a little bit - smallest divergence from Mun's own orbit - gives you the lowest periapsis and hence the greatest benefit from Mun's gravity. It also means you take the greatest advantage of the Oberth effect, but as you've seen that's a bit arcane. Missing by more just means you get less help from those and you have a greater difference between your orbital velocity and Mun's to make up.

(A prograde orbit around Mun is generally best so you probably want to miss behind it, rather than in front. That rather depends on what you intend to do next though so it doesn't greatly matter).

...bi-elliptic transfer ... it's more complicated, and it's just 40 m/s, so most people don't bother (I don't)...

And it takes a lot longer (if you care about things like life support)

Edited by Pecan
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There is a fourth alternative. A little less efficient but a guaranteed return to a landing on Kerbal if you fail to perform the Mun capture burn. It is the figure eight maneuver. Difficult to set up but not impossible.

rbOVDrA.jpg

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There is a fourth alternative.

('free return trajectory')

That requires a little more d-v initially, but might save more on the capture burn and end up being the most fuel efficient. Getting it just right with a low Pe is rather tricky.

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