Transfers and inclined orbits

[Matt77]

Hullo. I couldn't find a thread that answers this question.

This is about Mun / Minmus really as my interplanetary stuff is more luck than judgement so far.

I hit a wall in the tech tree and decided the Mun biomes were the best plan of attack, as I can get to and from the Mun easily with a tried and tested lander/orbiter system.

The current mission involved an inclined orbit (about 50 degrees) which would allow me to get to almost any Mun biome. So far so good, I have just plotted my Mun escape burn having rendezvoused back with the orbiter surprisingly easily. And this is where the head-scratching begins.

Which direction should you burn to get home? On this occasion I am burning in the same plane as my orbit and keeping the inclination. I don't think this is right but I have the fuel and I can land in any Kerbin biome I choose when I get back... I expected to head back on a roughly equatorial plane as usual but not sure what's "right".

Edited December 1, 2013 by Matt77

[UmbralRaptor]

The minimum ÃŽâ€V approach would still be to burn so your orbit exits the Mün's SOI on the trailing side. It may be necessary to wait for up to half of the Mün's orbit to get things to line up properly, though.

This is likely to put you on a somewhat inclined Kerbin orbit, so if you care a great deal about where you land, some waiting in LKO (or additional maneuvers) may be necessary.

[Kasuha]

Honestly, whatever maneuver which gets you home is good enough. If you have enough fuel left, just put a maneuver on your orbit, pull prograde till it crosses Mun's SOI, then drag the maneuver along the orbit and watch what it does. Out of what you'll see, select the orbit with lowest Kerbin periapsis.

If you want the most efficient ejection, you may wait till your orbit crosses Mun's orbital line. Then burn towards Mun's retrograde.

The following image shows it for launch from a high latitude point. Since you're already in orbit you need to wait till your orbit is aligned the same way.

[Matt77]

Ahh, I see. I was about to ask Umbrella what he meant about getting things lined up meant, but yes I get it now. I don't really care about a specific landing spot for this mission but I did just land on a mountainside killing everybody on board. F9 try again

That diagram really helps with what seemed complicated at first.

Thanks both for the advice, I will be more prepared next mission.

[longhornchris]

Even from an incline Mun orbit, the easiest way home is to make sure you exit Mun orbit 'retrograde' to the Mun's orbit around Kerbin. You always want to burn 'prograde' for whatever orbit you are in in the Mun frame, but where you place you maneuver node will depend on the direction you are going when you takeoff.

From takeoff, you want to go either 90 or 270 from liftoff as this gets you the minimum inclination from your launch site (unless you are docking with an in-orbit transfer vehicle). Once you've got a good orbit (I like 8-10km just to be safe from mountains) drop a maneuver node, stretch it out till you get a mun escape, then pull the time around so the escape sends you out opposite Mun's orbit. This will minimize your Kerbin orbit PE. From here, increase the burn then adjust the time to keep your escape at the same angle an your Kerbin PE will shrink. Get it where you want and execute the node. If you do it right your Kerbin AP will be right around Mun's orbit, while the PE is wherever you want it. If the AP is well above Mun's orbit the your burn isn't optimally timed.

As a note, I've got a maneuver node editor to help with this process, much easier than using just the maneuver gizmo but the concept is still the same.

[Matt77]

Thanks longhornchris. Your well worded reply just improved all my Mun returns, not just the kind under discussion!

[LiveHereNow]

I understand this in theory but I'm still not sure how to do it right. I get an ejection from the Mun that seems to be retrograde to the Mun's orbit but I still end up with a largely circular orbit around Kerbin. In watching others' videos, it seems they more easily launch from the Mun and get a highly elliptical orbit back to Kerbin that requires less delta-v. I'm using far more fuel and can't figure out why. Let's assume I've landed equatorially and am leaving equatorially (seems simpler). I think once I "get it", I will get more mileage from my rockets. Any thoughts? Any good video links?

[FenrirWolf]

You probably aren't ejecting at the right angle. Basically you want your maneuver to start and end such that your ship leaves the Mun going exactly opposite to its own motion around Kerbin. You can try different angles if you like though, dragging your maneuver node around the orbit will let you see how the exact same delta-v expenditure can produce wildly different results depending on your departure time.

Edited December 2, 2013 by FenrirWolf

[Specialist290]

I don't have any handy videos to link to specifically on this subject, but hopefully this explanation and the accompanying roughly-sketched diagrams will prove enlightening. I apologize in advance for the relative low quality of the diagrams.

Let's say that you start out in an orbit around the Mun that looks something like this:

Thinking back to basic orbital mechanics, you may recall that the best way to raise or lower one part of an orbit is to make an appropriate burn on the opposite side of the orbit. We want to lower the periapsis of our orbit around Kerbin so that it's within Kerbin's atmosphere for an efficient return trajectory. However, there's a complicating factor: Our craft is not orbiting Kerbin directly, but is rather in orbit around the Mun, which is itself orbiting Kerbin. Thus, in order to lower our Kerbin-centered orbit's periapsis, we first have to raise our Mun-centered orbit's periapsis so that our craft escapes the Mun's sphere of influence in the correct direction.

It turns out that the best way to do that looks something like this:

You want to schedule a burn somewhere above the prograde-facing side of the Mun so that your Munar ejection is both parallel and retrograde to the Mun's direction of travel. The effect of that burn would be the same, relative to the craft's orbit around Kerbin, as a retrograde burn at the apoapsis of a Kerbin-centered orbit.

[Nao]

I understand this in theory but I'm still not sure how to do it right. I get an ejection from the Mun that seems to be retrograde to the Mun's orbit but I still end up with a largely circular orbit around Kerbin. In watching others' videos, it seems they more easily launch from the Mun and get a highly elliptical orbit back to Kerbin that requires less delta-v. I'm using far more fuel and can't figure out why. Let's assume I've landed equatorially and am leaving equatorially (seems simpler). I think once I "get it", I will get more mileage from my rockets. Any thoughts? Any good video links?

Are you doing one burn or two? Because the most efficient way is to get everything done in one burn. To get most efficient return, you bring Kerbin Pe into the atmosphere while still flying low over the Mun.

Also if you start from orbit and create maneuver node try to get the lowest Kerbin Ap, that will mean that you will exit the Mun properly retrograde and you will get on that Ap right after exiting Mun SOI.

[Kasuha]

I get an ejection from the Mun that seems to be retrograde to the Mun's orbit but I still end up with a largely circular orbit around Kerbin.

If you end your burn as soon as your elliptic trajectory around Mun stops being closed, you're burning too little. The trajectory is still elliptic, it just intersects SOI boundary by its end and you have almost zero speed (relative to Mun) at that point, which will put you on almost the same trajectory the Mun has when you exit the SOI. Add more prograde impulse to the maneuver and watch how the trajectory outside the SOI changes.

[Vanamonde]

I still end up with a largely circular orbit around Kerbin

When you are landed on or orbiting Mun, you have pretty much the same velocity it does (or else you wouldn't be able to linger in its vicinity). That means that any ejection trajectory starts off nearly matching Mun's orbital path, and only diverges from it in proportion to the length and direction of the burn.

[Frederf]

Indeed you should imagine leaving the Mun SOI "cage" is like jumping from a moving car. If you barely escape it you'll be going the same speed as it. To fall down for reentry you need to escape it rearward with some extra speed rearward. If you find what appoapsis speed for a 11400x30km orbit is and subtract the Mun's orbital speed, that's what you need to carry extra as you escape it's SOI.

[John FX]

Set up a maneuver point that has 278Dv prograde. Move the timing by grabbing the white ring until it looks like this

done. Should be the same every time.