Determine orbit that requires least amount of �v from launch

[Bobe]

I'm attempting a sort of non-stop mission whereby I rendezvous with various planets and moons to refuel with a lander. I'm trying to figure out how to determine the optimum orbit around a planet to minimize the ÃŽâ€v required from my lander.

Obviously the closer to the surface I aim for the faster I have to go to maintain orbit, meaning I have to burn more. Conversely, the further away from the surface, the more I have to burn to get to that altitude and then circularize. Based on these extremes, it seems that there would be some sort of inverse bell curve that describes the required ÃŽâ€v against the target altitude.

Any ideas?

[KvickFlygarn87]

Well, lower is better since you get help from sir Oberth.

[nhnifong]

Having as small a lander as possible tends to save you the most ∆v, and that means you need to park in the lowest circular orbit possible.

If you parked in an eccentric orbit, your lander would need more fuel to complete the trip to the surface and back, and burning that fuel in the lander, is I believe less efficient than doing it in orbit, because of gravitational and steering losses.

You would also need to ensure that your argument of periapsis with respect to desired ejection angle was appropriate if you parked in an eccentric orbit. This depends on which direction you arrived from which in turn depends mostly on your timing of departure from the previous planet. You can see how that would require you to do some serious planning.

[Bobe]

Let me lay down some assumptions and constants.

First of all, I'm using MechJeb to do most of the heavy lifting. You can assume I'll always be in a circular equatorial orbit.

The ship I'm in weights in at just over 100 t, and the lander is an additional 42 t. Ironically, bringing the lander for the purpose of refueling seems to reduce my total ÃŽâ€v by more than a third, from 7191 m/s to 4467 m/s. So a direct trip to Jool is cutting it pretty close.

Considering the size of the ship, I can't really park it 5 km from the surface.

Obviously a smaller lander would be more efficient, but my lander doesn't have that luxury and it's really beside the point here as it pertains to determining a value based on many variables, among them the mass and thrust of the lander.

The lander, with it's single Poodle engine, has a total ÃŽâ€v of 922 m/s in a vacuum, though I typically afford myself a bit more by converting kethane as I ascend. For reference, a launch to a circular orbit of 60 km around Mun and maneuvers to rendezvous with the ship required my full starting fuel supply of 810 (litres I presume) plus the converted equivalent of almost 8000 kethane.

I actually didn't consider de-orbit burns, perhaps because I thought they would be closely correlated to the launch burns.

But I'm not searching for answers regarding my specific case, I'm just using it to help illustrate my query.

Edited January 3, 2014 by Bobe

[Kasuha]

You're coming in with your large heavy interplanetary ship and your tiny little lander. You're going to leave the ship at some orbit and put your lander to the surface, then get lander from the surface, pick up your interplanetary ship and leave the system.

The ship (without lander) goes only as low as to the parking orbit and then back up and away.

The lander goes all the way to the surface and then back up and away.

The lander weighs way less than the ship plus lander combo.

Conclusion: always park as high as possible - at the highest orbit which you are sure your lander will be able to reach after it returns from the surface. The higher you leave it, the less you need to brake its mass and the less you need to accelerate its mass again on ejection.

Edited January 3, 2014 by Kasuha

[Bobe]

Well that's more or less common sense. What I'm after however is a way to determine with a certain degree of accuracy the optimum orbit I should aim for, not just a superlative.

[Max Grant]

I do not know if there is a rule of thumb you can employ. Some bodies will be harder to get away from, and some will be harder to get to. Moho is an example of one that's hard to get to. Eve, hard to get away from.

You are going to have to just do the math. I use Engineer to design my landers, because it allows me to calculate TWR which is about the most important figure in determining whether you can get home (next to how much delta-v). So, I get a delta-v map, I get my lander, I make it what it has to be to get into orbit.

[maltesh]

An Earlier Version of this post included an incorrect graph. I'd loaded the wrong one I had saved in my Desmos account. It is corrected now, both in image and descriptive text.

Actually, the delta-V-to-deorbit function looks like this, assuming you're braking retrograde to do your deorbit:

Desmos graph that created the above image:https://www.desmos.com/calculator/ksjeuy8ast

The above graph is the delta-V necessary to put your periapsis at 70 km over Kerbin as a function of parking orbit altitude.

Horizontal axis is altitude in kilometers, Vertical Axis is delta-V in kilometers per second.

As it shows, it costs less to deorbit from a parking orbit up to about 350 km altitude than it does from a Parking orbit just inside the SOI.

You get a similarly-shaped function regardless of the world in question.

Ultimately, if you're doing an efficient ascent in your lander with the attendant gravity turn, during your ascent, you will almost certainly wind up at a point where, if you stopped burning, you would be in a low circular orbit above the body you're orbiting. It's only going to cost you more delta-V to push that apoapse higher if you wanted to, and still more to circularize.

So put your parking orbit as low as you can comfortably rendezvous with.

Edited January 3, 2014 by maltesh

[draeath]

You probably want to have a look at this tool.

[NikkyD]

Staging can do miracles. Lots of fuel (and thus boost) is needed only for the first 10 km to 20 km.

[Bobe]

You are going to have to just do the math..

That's why I'm asking for a method by which to do the maths.

You probably want to have a look at this tool.

That definitely looks like a useful tool, but it doesn't look like it does what I need it to.

Staging can do miracles. Lots of fuel (and thus boost) is needed only for the first 10 km to 20 km.

You didn't read my whole post did you?

[nhnifong]

Sorry I don't know the math involved. I just wing it in KSP and seem to get pretty close.

[Nao]

Having as small a lander as possible tends to save you the most ∆v, and that means you need to park in the lowest circular orbit possible.

If you parked in an eccentric orbit, your lander would need more fuel to complete the trip to the surface and back, and burning that fuel in the lander, is I believe less efficient than doing it in orbit, because of gravitational and steering losses.

You would also need to ensure that your argument of periapsis with respect to desired ejection angle was appropriate if you parked in an eccentric orbit. This depends on which direction you arrived from which in turn depends mostly on your timing of departure from the previous planet. You can see how that would require you to do some serious planning.

Actually depending on the entry and exit trajectories an highly eccentric orbit could be the best. If you think about it, most fuel will be burned by moving the heavy interplanetary stage. Even if the lander works on rockomaxes 48-7S and the interplanetary stage works on LV-N's if the lander weights more than 2.3 times less (confusing sentence ) it will be more fuel efficient to do any burns in orbit with it than the main stage. So the best parking orbit would be the one that takes the least amount of deltaV to achieve coming from interplanetary space (and we need to add the deltaV needed to exit the body in question, possibly in some other direction which complicate it even more).

So this is actually pretty interesting question. By hunch i would say that if the lander is much lighter than the main stage, either highly elliptical orbit (Pe just over surface and Ap just near SOI end) or high circular orbit (Ap/Pe near end of SOI) would be the best parking spot.

Finding that for bodies like Pol and Bop would be pretty easy, but let's think of Tylo for example. Lander weight for it would be more dependent on where the parking orbit is. So there could be some "in the middle" solution to the maximum efficiency parking orbit.

edit: Of course lowest possible Pe is the best spot for orbit insertion burn, so an high circular orbit would be silly. Thanks Kasuha!

Edited January 5, 2014 by Nao

[Kasuha]

Leaving aerobraking aside, the starting point is the mothership on a highly elliptical orbit with periapsis just safe above ground/atmosphere and apoapsis near the SOI boundary. This orbit is the base parking orbit as it costs least dv to put the mothership on it.

The other extreme is a circular orbit with the same periapsis as the base parking orbit and apoapsis at the same height.

Let's assume we're not limited by lander capabilities, i.e. the lander has no problem landing from our base parking orbit and returning to it from the surface. It must perform transfer between base parking orbit and minimum orbit twice.

The travel between the minimum orbit and surface - and back - is purely up to the lander. We can leave that out of the equation.

The travel from interplanetary space to base parking orbit - and back - is purely on mothership. We can leave that out of the equation as well.

So the only thing we need to find is the "real parking apoapsis" between base parking orbit's periapsis and apoapsis on which the mothership should park to minimize total fuel consumption to:

- decrease the apoapsis to the real parking apoapsis by mothership + lander

- decrease apoapsis to minimum orbit by the lander

- raise the apoapsis back to real parking apoapsis by lander with the fuel payload

- raise the apoapsis to base parking orbit by mothership + lander + fuel payload

I'm not going to do the whole math but I think this leads to relatively simple equation.

[Kerbart]

Would a three-stage approach make sense? The mothership stays in circular orbit near the SOI boundary. That takes the least amount of ÃŽâ€V to get into orbit and leave it.

A second ship, call it the "expedition cruiser" would then detach and get into a lowest-altitude parking orbit

A third ship, the lander, would then detach from the expedition cruiser and make touchdown. The lander doesn't need to carry the fuel needed for transfer from SOI boundary to low orbit and back. Of course that's only a few 100ÃŽâ€V but everything to keep the lander light helps, right?

Whether you're using a 2-ship or 3-ship approach you will likely also need to bring a variety of landers along. Something that can return from the surface of Duna needs to be a lot different than a Minmus lander.

[Epthelyn]

Would a three-stage approach make sense? The mothership stays in circular orbit near the SOI boundary. That takes the least amount of ÃŽâ€V to get into orbit and leave it.

A second ship, call it the "expedition cruiser" would then detach and get into a lowest-altitude parking orbit

A third ship, the lander, would then detach from the expedition cruiser and make touchdown. The lander doesn't need to carry the fuel needed for transfer from SOI boundary to low orbit and back. Of course that's only a few 100ÃŽâ€V but everything to keep the lander light helps, right?

Whether you're using a 2-ship or 3-ship approach you will likely also need to bring a variety of landers along. Something that can return from the surface of Duna needs to be a lot different than a Minmus lander.

By not bringing along the 2nd craft you'd probably be saving more or less the same dV you'd otherwise take to bring it to orbit in the first place. You might as well just put the mothership in Low Orbit.

In other words, I don't think that'd be any more efficient in terms of either overall fuel cost or lander fuel usage.

[cr055H41rz]

Conclusion: always park as high as possible - at the highest orbit which you are sure your lander will be able to reach after it returns from the surface. The higher you leave it, the less you need to brake its mass and the less you need to accelerate its mass again on ejection.

Wrong, parking low saves feul

coming into a system at 4 km/s and having to decelerate down to 2 km/s is less feul used than having to decelerate down to 500 m/s

vice versa on exit

[Kasuha]

Wrong, parking low saves feul

coming into a system at 4 km/s and having to decelerate down to 2 km/s is less feul used than having to decelerate down to 500 m/s

vice versa on exit

If you check my recent post here, you'll see that I agree that the cheapest parking orbit is a highly elliptical one. Because you brake low but still brake only necessary amount. Bringing your apoapsis all the way down to the planet is also fuel expensive, as well as pushing it back up before you escape.

High apoapsis also allows you to make cheap inclination changes before you escape the system.

[JiWint]

Thanks to uncle Oberth!

[technion]

If you check my recent post here, you'll see that I agree that the cheapest parking orbit is a highly elliptical one. Because you brake low but still brake only necessary amount. Bringing your apoapsis all the way down to the planet is also fuel expensive, as well as pushing it back up before you escape.

High apoapsis also allows you to make cheap inclination changes before you escape the system.

I saw a serious example of this last week. I was working on a Tylo landing. Turning a "just captured" elliptical orbit into a circular one was going to be hell, and doing it in my mothership scheduled a ridiculously long burn time. It was far more efficient to just drop the lander in that orbit, and let it deorbit from the pe. Once it got back into orbit, having a very lightweight lander meet that elliptical orbit on half of one small tank and rockomax engine was a lot more efficient than trying to circularise the mothership.