Moon exit trajectory question

[McKermack]

So I've learned how exiting a moon's SOI to return to the parent body is best done by setting a course for it's retrograde vector(if I'm saying that right). I'm not clear on why this is so however as I don't understand the mechanics and I haven't read anything that articulates it well enough.

Can someone explain it succinctly if you please?

Also in relation to this, if my lander is on a moon and my coordinates are at the very trailing edge of the moon in relation to it's vector, would it not be feasible to just burn straight up since my heading is already as the moon's retrograde?

Edited May 25, 2015 by McKermack

[cybersol]

The simple explanation is that you burn prograde to raise your orbit and retrograde to lower it. In effect a higher orbit is a higher energy state; yet, most of that energy is potential energy and thus your orbital speed is actually lower in higher orbits.

Yes, you could launch straight up in that instance, modulo the fact that you and the moon already have some eastward velocity due to the fact that it is spinning.

[Starwhip]

What you really heard was that you should burn prograde in your own orbit, but retrograde in relation to the parent body.

If the moon is orbiting 'clockwise' and you are orbiting 'clockwise', if you burn prograde when you are between the moon and planet, you will slow yourself down in relation to the parent body.

If the moon is orbiting as above, but you are going in the opposite direction, burning prograde when the moon is between you and the planet will have the same effect as above.

[McKermack]

http://images.akamai.steamusercontent.com/ugc/29616984106954889/50E52E096CF82FEA53259CB989337E60DB65E911/

Is this not optimal? Considering I was "behind" the moon, I set my heading for 90° at a 70° angle. Just trying to build a better reference.

[Starwhip]

http://images.akamai.steamusercontent.com/ugc/29616984106954889/50E52E096CF82FEA53259CB989337E60DB65E911/

Is this not optimal? Considering I was "behind" the moon, I set my heading for 90° at a 70° angle. Just trying to build a better reference.

Not really. It is much more effective to burn exactly prograde exactly between the planet and moon in this case. Try it, it seems counterintuitive, bit it works.

[MathmoRichard]

So I've learned how exiting a moon's SOI to return to the parent body is best done by setting a course for it's retrograde vector(if I'm saying that right). I'm not clear on why this is so however as I don't understand the mechanics and I haven't read anything that articulates it well enough.

Can someone explain it succinctly if you please?

Sure I can try, so I hope this helps anyway.

So the first thing to understand (and this is really important in general) is that in general* to lower an orbit you must lower the the energy of the orbit (and vice versa). The physical reason for this is that energy [= kinetic energy + potential energy] is conserved, and the lower your potential energy the lower your altitude. Beyond those two sentences it's a little Maths heavy, but I'll happily explain later if you want.

The only way to lower your total energy is to lower the kinetic energy of the orbit, which equals 0.5mv^2 which means your speed needs to decrease. Now we need to think about how speed changes and what it is. Speed is the magnitude of velocity, and if you change the direction of your velocity then your speed doesn't change, this means that any addition to your velocity in a perpendicular direction will change direction not speed. Thus to change our speed we must accelerate directly along our velocity vector direction. To decrease the speed, we do this in the opposite (retrograde) direction (same line but 180° away).

That was all ignoring the Mun. Now imagine you've just magically escaped the Mun's SOI and haven't changed velocity much, then you're travelling in the Mun's orbit so you have the same retrograde direction that the Mun has. Then to get back to Kerbin we need to accelerate in the direction opposite to how the Mun is travelling.

Now let's think about escaping the Mun. The position you burn can vary so let's just consider the direction you're travelling as you are leaving the SOI. If this was along the Mun's retrograde vector, we have basically put all of our extra escape speed along the retrograde direction so we more or less got the maximum beneficial effect of the escape. If we hadn't been going in that direction then (with Pythagoras in mind) you will have a smaller component of velocity going retrograde, and the other component of the velocity went towards or away from Kerbin (or possibly perpendicular to the plane) and that part is wasted because it will just make the orbit change shape just as we said perpendicular meant a change of direction.

The practical point here is that it's about the direction you leave the SOI, not the direction you burn. But the time you burn will be more or less on the prograde side of the Mun, and you should burn along the prograde direction of Munar orbit.

Also in relation to this, if my lander is on a moon and my coordinates are at the very trailing edge of the moon in relation to it's vector, would it not be feasible to just burn straight up since my heading is already as the moon's retrograde?

Entirely because of gravity losses. Put simply, this is the loss of fuel efficiency which is the same as burning on the launchpad without enough thrust to go anywhere.

Put precisely: the portion of thrust fighting gravity (i.e. along the direction of gravity's pull) is completely wasted.

This is why to escape you should gain merely ground clearance and then burn horizontally.

*Disclaimer: again this is complicated and on some occasions radial burning can do more, but only for much more complicated situations, and only when considering the periapsis since when ever you burn perpendicular an increase in the periapsis will decrease the apoapsis and vice versa.

Edited May 25, 2015 by MathmoRichard

[McKermack]

Not really. It is much more effective to burn exactly prograde exactly between the planet and moon in this case. Try it, it seems counterintuitive, bit it works.

What approximate heading and angle would that be? I can't relate what you mean by exactly between in this case, not sure if my understanding spatially is on par with most other peoples'.

Do you mean to point straight to the parent body from my landed position?

Thanks for helping by the way!

[MathmoRichard]

The best way to work out the angle McKermack is with manoeuvre nodes. It is actually the best way to understand all this. Try setting one up in a circular Munar orbit with a set delta V of 300 m/s say, and then slide the node about to see which gives you the smallest Kerbin periapsis. I use this technique every single time, it's the only way to be most efficient without mods.

[Laie]

Can someone explain it succinctly if you please?

First rule of spaceflight: an orbiting vessel is always falling down and constantly missing the ground. That's why you have to turn your rocket and move sideways: the goal is to be over the horizon before you hit the ground.

As you made it to the Mun, you have already learned that accelerating prodgrade raises your orbit, going retrograde lowers it. If you slow down enough, you will not make it over the horizon; in other words: return to Kerbin.

The smart space traveller leaves the Mun in such a direction & velocity that he slows down relative to Kerbin. Next time you return from the Mun, take the time to focus the view on Kerbin and watch your ship's movement from there: it will look as if the Mun moves on while you stay behind. After a while, the Mun will be so far away that you're no longer in it's SOI; your (nearly) stationary ship now starts falling towards Kerbin.

[cybersol]

Do you mean to point straight to the parent body from my landed position?

Until you know and experiment a lot more, you are likely better off establishing a local orbit around the body you are landed on first. So first launch into a low prograde (eastward) orbit and circularize. Then you can set up a maneuver node and play around with it as MathmoRichard suggests.

[MathmoRichard]

And if you ever wonder why people tell you to do any launch eastward it is purely because of the small boost given by the fact that the body is rotating eastward (counterclockwise).

[McKermack]

I think some of this is starting to click. You might think by my line of questioning I haven't achieved much but I have done many landings and returns, even to other planets. It's just my grasp on the why is behind, not the how.

Correct me if I'm wrong:

Heading east in a sense is moving with the Mun's rotation and so maximizes your velocity gain, which saves fuel. And going straight out is never optimal. Eh?

Again thanks for the help everyone!

[cybersol]

Yes, heading east lets you gain the rotational speed of the body you are leaving for free. Going straight up is actually probably optimal at the exact right time with perfect piloting. Its just that in practice even the tiniest errors in either timing or piloting will likely cost you more than circularizing and making an optimal maneuver node before your ejection burn.

[McKermack]

Yes, heading east lets you gain the rotational speed of the body you are leaving for free. Going straight up is actually probably optimal at the exact right time with perfect piloting. Its just that in practice even the tiniest errors in either timing or piloting will likely cost you more than circularizing and making an optimal maneuver node before your ejection burn.

That reply I think cleared it up for me. So there is a point nonetheless when and where you can burn straight out but it's very slim so not practical.

Also if a moon has an elliptical orbit what point of the orbit(the moon's) would be most optimal to eject on course to the parent?

[Zyffyr]

Also if a moon has an elliptical orbit what point of the orbit(the moon's) would be most optimal to eject on course to the parent?

The ejection itself is going to be the same regardless of when you do it. What will differ is the resulting orbit of the parent world. The closer in the moon is, the tighter the resulting orbit.

[McKermack]

I alright then I think I'll call this one answered. Thanks for all the replies!

[Wanderfound]

The simplest way to do it: look up before you launch.

If Kerbin is straight above you, burn eastwards until you hit escape velocity. If Kerbin is on the western horizon, burn straight up until escape velocity.

If Kerbin is on the eastern horizon, burn eastwards as if you were setting up a 10km prograde orbit, coast to apoapsis, then burn prograde until escape velocity. If Kerbin isn't visible at all, do the same as that, but with a westwards launch for a retrograde orbit.

Basically, you're just trying to aim your launch burn in the same direction that you would point a retrograde burn if the Mun itself was the ship. Get it right and you can go from the Munar surface to Kerbin splashdown in a single burn.

But, with all of these, it is simpler and almost as efficient to just circularise in LMO first.

[Snark]

There's another reason to circularize before the homeward burn, besides the advantage of controllability that folks have already pointed out: taking maximum advantage of the Oberth effect.

You want as much of your burn as possible to be at as low an altitude as possible. If you just burn straight up, then unless you have a ridiculously high TWR to keep the burn short, you're going to be doing a lot of the burn at high altitude, since you're rapidly putting distance between you and the surface.

If, on the other hand, you immediately turn due east upon takeoff and circularize at a really low altitude, then when the time comes to do your homeward burn, you're doing it horizontally, thus finishing the whole burn at low altitude. This conserves your dV.

[Laie]

Heading east in a sense is moving with the Mun's rotation and so maximizes your velocity gain, which saves fuel.

Yes. Next time you put a vessel on the launch pad, switch your navball to orbital velocity and find that you're already moving at 174m/s. If you want to launch into a polar otbit by going straight north or south, you'll find that the orbit will still be inclined a little because you inherit some sideways momentum from the planet. If you want to go into a retrograde orbit (launch west), you have to spend as much dV as you always do, plus 2x 174m/s to make up for the launchpad moving in the wrong direction.

It's the same on the Mun, of course, though the figure will probably be much lower than 174m/s.

[Khaur]

Entirely because of gravity losses. Put simply, this is the loss of fuel efficiency which is the same as burning on the launchpad without enough thrust to go anywhere.

Put precisely: the portion of thrust fighting gravity (i.e. along the direction of gravity's pull) is completely wasted.

This is why to escape you should gain merely ground clearance and then burn horizontally.

This is just wrong.

Thrust is not wasted because it is against the direction of gravity, it is wasted because you could have netted a better Oberth effect by going sideways.

Take this simple case: moving radially away from Kerbin at 1km/s, with a TWR of 1 (adjust throttle to stay at no acceleration as you move away and gravity gets weaker). If as you say "the portion of thrust fighting gravity (i.e. along the direction of gravity's pull) is completely wasted", then you'd never escape Kerbin, even though you're moving steadily at 1km/s away from it. That's an obvious contradiction.

How much energy you get per dv is dictated by the average speed at which you spend dv. At equal thrust (along the prograde), the lower you stay the faster you'll go on average, therefore the less dv you need to spend to reach escape energy.

Now just to be clear, I'm not saying burning radially is an efficient way to escape (it's not in most conditions), I'm saying that your justification of why is based on a misconception.

[Laie]

This is just wrong.

Thrust is not wasted because it is against the direction of gravity, it is wasted because you could have netted a better Oberth effect by going sideways.

Nope. By just going up you may go to space, but you will not stay in space. Unless you go up so fast that you leave the SOI entirely, of course. But as long as you stay in the SOI, mereley going up will end with you coming down again (cue Spinning Wheel). What keeps you in space is going sideways, always speeding over the edge of the world.

All vertical velocity will eventually be eaten by gravity, but horizontal velocity is yours forever. It therefore makes sense to turn sideways as soon and as aggressively as possible.

Oberth has nothing to do with this, it's something else entirely. If you're doing it right, you're getting a considerable Oberth Effect -- but moving sideways would still be best practice even without Oberth.

[Juicy]

Basically, you want to depart along the blue arrow, to get into the blue orbit. If you angle more towards Kerbin, you get a much tighter (direct) orbit.

[Khaur]

All vertical velocity will eventually be eaten by gravity, but horizontal velocity is yours forever.

It doesn't disappear into thin air. You're getting altitude (potential energy) in exchange for that. And that altitude will get converted back to velocity (kinetic energy) when you're on the way down.

Horizontal velocity isn't yours forever: look at your velocities at periapsis an apoapsis: unless you're on a circular orbit, they'll be different, despite both being completely horizontal.

What is yours forever is orbital energy (kinetic+potential). Well, as long as you don't collide with anything, but at high altitude, even a tiny nudge sideways is enough to avoid collision with the orbited body.

[MathmoRichard]

It doesn't disappear into thin air. You're getting altitude (potential energy) in exchange for that. And that altitude will get converted back to velocity (kinetic energy) when you're on the way down.

Horizontal velocity isn't yours forever: look at your velocities at periapsis an apoapsis: unless you're on a circular orbit, they'll be different, despite both being completely horizontal.

What is yours forever is orbital energy (kinetic+potential). Well, as long as you don't collide with anything, but at high altitude, even a tiny nudge sideways is enough to avoid collision with the orbited body.

After a whole days thought, I think we were all a little wrong and a little right. With a certain interpretation of "Oberth effect", you are right Khaur, in that our gain in energy is directly proportional to our speed. Many (not you) people think it's an altitude/gravitational thing, but it isn't, except than on free orbit the highest speed is at periapsis and lowering the periapsis counterintuitively gives you a bigger maximum speed and it turns out more efficient that way.

I was wrong and right I think because unless I've got it all backwards, (the Oberth effect is irrelevant as long as your circular orbit is low) in fact the delta-v required is the same, ​which is really nice if true don't you think?

That's all based on high acceleration and low thrust time and low orbits as I said. If you need a large circular orbit, you will lose significant delta-v by Oberth. If you don't have the thrust to get up the escape velocity quickly going straight up will always lose because of gravity losses, and that is exactly as I stated before. The gravity losses will be the integral of the gravitational strength with time, or simply gt. Basically the t needs to be really small.

Edited May 25, 2015 by MathmoRichard

[Khaur]

I think a lot of the confusion comes from that the term "effect" is a bit of a misnomer, at least in how it's used. It's not like there's a basic price in m/s to pay for a given amount of orbital energy, and the Oberth effect comes on top of that to apply a discount or a surcharge. Discarding the Oberth "effect" is akin to discarding Newton's second law of motion: one gives the rate to convert dv into orbital energy, the other to convert thrust into acceleration.

Back to the Mun escape case: what we want is a given escape velocity. Up to patched conics approximation errors, that's only tied to the orbital energy around the Mun. How much dv is required depends exclusively on the average speed when performing the burn(s) (weighted to compensate for variable thrust over time). In terms of energy, as long as the burns are along the (orbital) prograde vector, it doesn't matter which way it is pointing. To maximise this we want to stay as close to the centre as we can.

Enters constraint number 1, the lithosphere: we really don't want to go through it (atmosphere isn't as hard a constraint and makes things messy, but there's none in our case). That means we have to stay above a certain altitude, so we have to settle for as low a gravity turn as ground clearance will allow.

Constraint 2 is on the escape vector. If we want to return to Kerbin we want it close to retrograde, but the reasoning applies to all cases. Unless we're really lucky, that means we have to give up on our grazing single-burn trajectory.

If we're on the right hemisphere (actually a little more), we can still try a direct ascent, but it will get less and less efficient the closer to the zenith our desired escape vector is.

The other (and only if we're on the wrong hemisphere) option is to transition through an orbit (or a quasi-orbit). This means going a little bit higher to steer clear of additional obstacles, and possibly coasting to apoapsis to circularise, both of which lower efficiency slightly. The upside is that we'll get to perform the rest of the escape burn from a rather low altitude and we have fine control over the escape vector.

Which one is more efficient depends on your craft, and how quickly it can reach the desired orbital energy. If it can get up to speed before going higher than the required clearing altitude for the orbit, go for a direct ascent. Otherwise, use an orbit. Additionally, you may want to consider the "free" velocity from the surface rotation, and higher piloting error in the case of direct ascent.

If your desired escape vector is straight up, it would requires insane amounts of thrust for option 1 to be better.

TL;DR: Save yourself the trouble and just get into a low orbit first if you care about dv.