Interplanetary burns from polar orbits.

[davidpsummers]

I find myself in an interesting quandry. I'm in an inclined orbit (73 degrees) about Duna and my guys want to get home. The orbit isn't as poorly lined up with the orbit (maybe 20-25 degrees?). So how long is my escape burn and in what direction? Fortunately, my reusable interplanetary booster has over 5000 m/s left (though I'm leaving a bit outside the window, so I will need @600 m/s to get home). I see three choices...

-Set up a maneuver node to burn and swing out over the poles, moving the maneuver node by eye until the escape looks as parallel with retrograde as I can (thought the orientation of the orbit will mean it has to be ~25 degrees off). Use the delta-V (if I just day dV, will people be confused?) from Alex Moon to determine the length of the burn, and adjust until I hit Kerbin.

-Do a minimum burn to escape from Kerbin dumping me out into Kerbols SOI as close to Duna's orbit as possible. Then use Alex Moon's tool (with as high an orbit it will accept) to calculate the burn. I've don't this before, but I guess I loose the Oberth effect.

-Spend about ~1000 m/s to get an equatorial orbit and then proceed as normal.

Edited November 6, 2014 by davidpsummers

[MarvinKitFox]

...

Of the three options you list, I advise using option number four.

Try this:

Shove your apogee out to almost the edge of Duna's SOI, with Apo being anywhere on the Duna equatorial plane in vaguely the direction you will want to exit.. Do not raise the peri..

Now, at extreme Apo, do your plane change. it should be very, very little dV indeed. (about 25m/s should do the job)

As this orbit takes a while, consider selecting your actual apo so that your next peri will match in time with the desired burn time for ejection.

As the next Perigee, now moving in a suitable plane, do your ejection burn.

You will actually use *less* total fuel this way, as your only loss is that 25m/s plane change (and a mountain of time, but time is free) while maximising Oberth due to splitting your burn into two seperate burns that can thus be closer to the planet.

[technion]

Of the three options you list, I advise using option number four.

73 degrees is an awful lot to shave off. Even done at the edge of the orbit, I'm interested in how expensive that would turn out to be.

[Yasmy]

Let's find out:

v^2 = mu (2/ r_SOI - 2/(r_SOI + r_LDO))

mu = 3e11 m^3/s^2

r_SOI = 4.8e7 m

r_LDO = 4e5 m

v = 10 m/s

So, 10 m/s * sqrt(2 - 2 cos(73)) = 24 m/s. (Law of cosines for a triangle with two length 10 sides with a 73 degree angle between them.)

Nailed it, MarvinKitFox!

Edited November 6, 2014 by Yasmy

[davidpsummers]

Of the three options you list, I advise using option number four.

Try this:

Shove your apogee out to almost the edge of Duna's SOI, with Apo being anywhere on the Duna equatorial plane in vaguely the direction you will want to exit.. Do not raise the peri..

Now, at extreme Apo, do your plane change. it should be very, very little dV indeed. (about 25m/s should do the job)

As this orbit takes a while, consider selecting your actual apo so that your next peri will match in time with the desired burn time for ejection.

As the next Perigee, now moving in a suitable plane, do your ejection burn.

You will actually use *less* total fuel this way, as your only loss is that 25m/s plane change (and a mountain of time, but time is free) while maximising Oberth due to splitting your burn into two seperate burns that can thus be closer to the planet.

I get some of what you are getting at, raising the apogee is worth it to get a cheap inclination change. 375 m/s to raise apogee and 30m/s to change inclination. Even if I just circularize back to a low orbit, it has cost me a few hundred less than a straight inclination change.

Of course circularizing back to a low orbit is a waste, I can, as you point out, just burn some more at perigee to get my transfer. However, I am now committed to escaping Duna in a specific direction (determined by the direction of my apogee relative to Duna's orbit). And my apogee was determined by the plain of the inclined orbit (since I have to burn as I cross the equator to get an apogee in the equatorial plane). And, in fact, I think that direction must be the same as it would in option #1. In both cases I end up leaving in a direction that lies in the plane of my original orbit.

So I think the point is the Oberth effect. I understand how it aids you when you burn in a lower orbit (or, to be honest, I understand the math but not why that conflicts with my understanding of other parts of physics). I've heard that splitting your burn up into multiple burns also helps? But, then why bother with the inclination change? It doesn't change the direction you leave the Duna sphere of influence unless you circularize.

[Brainlord Mesomorph]

Pardon me, but I'm wondering why fix the orbital inclination at all?

Given the distance of a Duna/Kerbin flight, why do you need an equatorial orbit? As long as your Duna escape is basically within the solar plane, you can do the plane change as a mid-course correction in solar orbit.

or am I missing something? (in this game I usually am)

[Laie]

Pardon me, but I'm wondering why fix the orbital inclination at all?

You're right: the 23° around Duna will become much less (perhaps 4°) in your orbit around the sun. Besides, even equatorial around Duna will become some inclination around the sun -- so starting at 23° doesn't mean you have to do anything special, only more of what you're doing anyway.

It is not necessary to fix the 23° degree inclination first. The OP proposals 1 & 3 would be far more expensive than dealing with it enroute (I don't get proposal #2).

However, MarvinKitFox's proposal #4 is by far the most energy-efficient, for two reasons: a) the Oberth effect only helps you with the prograde component of your burn while any normal component will have to come out of your own pocket; and plane changes at high apoapsis are dead cheap. If the OP is smart, he won't change into an equatorial orbit, but rather make the inclination so that he can leave with another prograde-only burn when the periapsis comes around again.

It takes three maneuvers and a lot of timewarp to leave a planet in this fashion; however, every single maneuver is dead simple and straightforward. Depending on how savy you are with maneuver nodes, you may find it easier to do it this way, rather than rolling everything into a single transfer burn.

And yes, plane changes at extreme apoapsis are dead cheap. Just plot a few maneuvers and see for yourself. It's an artefact of KSPs "you're always in one SOI until you change into another" model, btw. Don't try this in our universe.

Edited November 6, 2014 by Laie

[mhoram]

And yes, plane changes at extreme apoapsis are dead cheap. Just plot a few maneuvers and see for yourself. It's an artefact of KSPs "you're always in one SOI until you change into another" model, btw. Don't try this in our universe.

I have to disagree. In the real universe it is the same: plane changes get cheaper the farther away the apoapsis is.

[Laie]

I have to disagree. In the real universe it is the same: plane changes get cheaper the farther away the apoapsis is.

Yes, but KSP will let you take it to the point where you can do it on RCS, and have to be careful not to overshoot.

[ratchet freak]

You can just escape in polar orbit, the only difference from an ecliptic orbit is the amount of inclination the resulting solar orbit has

[moogoob]

You could burn straight from a polar orbit. Main issue - are you facing the right direction when you hit escape velocity? IE retrograde to Duna's orbit. In an equatorial orbit, all you have to do is wait until the right time.

[Yakky]

I get some of what you are getting at, raising the apogee is worth it to get a cheap inclination change. 375 m/s to raise apogee and 30m/s to change inclination. Even if I just circularize back to a low orbit, it has cost me a few hundred less than a straight inclination change.

Of course circularizing back to a low orbit is a waste, I can, as you point out, just burn some more at perigee to get my transfer. However, I am now committed to escaping Duna in a specific direction (determined by the direction of my apogee relative to Duna's orbit). And my apogee was determined by the plain of the inclined orbit (since I have to burn as I cross the equator to get an apogee in the equatorial plane). And, in fact, I think that direction must be the same as it would in option #1. In both cases I end up leaving in a direction that lies in the plane of my original orbit.

Once you are at the outer edge of Duna's SOI after you raise your apogee, it's also very cheap to circularize, probably less than 100 m/s of delta V or so, because the orbital velocity is slow at that altitude. So you could circularize, then un-circularize again at the appropriate point in order to fall in very close to Duna so that you maximize Oberth Effect for your escape burn in the proper direction. I bet this may be your overall cheapest strategy in terms of delta-V.

Edited November 6, 2014 by Yakky

[davidpsummers]

Pardon me, but I'm wondering why fix the orbital inclination at all?

Given the distance of a Duna/Kerbin flight, why do you need an equatorial orbit? As long as your Duna escape is basically within the solar plane, you can do the plane change as a mid-course correction in solar orbit.

or am I missing something? (in this game I usually am)

I guess mostly because the tools to calculate the transfer burn all assume you burn directly prograde or retrograde. Which means I will have try and arrange the actual intercept by trial and error which I find tricky in interplanetary trips.

[davidpsummers]

Once you are at the outer edge of Duna's SOI after you raise your apogee, it's also very cheap to circularize, probably less than 100 m/s of delta V or so, because the orbital velocity is slow at that altitude. So you could circularize, then un-circularize again at the appropriate point in order to fall in very close to Duna so that you maximize Oberth Effect for your escape burn in the proper direction. I bet this may be your overall cheapest strategy in terms of delta-V.

I see. When I "uncirculize" I would have to drop the periapsis down to where the right angle to retrograde is. I guess, with Kerbal Engineer, I find that and mark it with a maneuver node or something.

[davidpsummers]

Some have talked about leaving from a polar orbit producing an inclination. Since any orbit can produce and escape vector parallel to the Sun's equatorial plane, the only inclination resulting from leaving from polar orbit is the small one the results from the fact that any escape is offset a bit from the planets orbit (because you have to swing around the planet a bit).

The main restriction I can see is that you eave in a vector that lies along the plane of the original orbit, which will introduce a radial component to your burn.

[Yakky]

I see. When I "uncirculize" I would have to drop the periapsis down to where the right angle to retrograde is. I guess, with Kerbal Engineer, I find that and mark it with a maneuver node or something.

Yes, but I don't even think you need Kerbal Engineer. You can just create the "un-circularize" maneuver node, then create a second maneuver node off of that one, then just move them both around until they get you where you need to go. It can all be explored hypothetically with maneuver nodes before you do any actual burning.

[metaphor]

It doesn't take any more delta-v to get back to Kerbin from a polar Duna orbit as from an equatorial Duna orbit, as long as your polar orbit is lined up in the right way.

Wait until your polar orbit is lined up with the way Duna is moving around the Sun. That means that the prograde direction of Duna is in the plane of the polar orbit. Then you can burn straight prograde while in polar orbit, making sure that you exit Duna's SOI in the opposite direction that Duna is moving around the Sun (Duna's retrograde). This ideally would take you into a zero-inclination solar orbit with apoapsis at Duna and periapsis at Kerbin (i.e. a Duna-Kerbin Hohmann transfer orbit). If Kerbin is on the wrong side of the Sun at this time, you can tweak your solar orbit and wait a few orbits until you get a Kerbin encounter.

If you don't want to wait more than one solar orbit and you're unlucky with the way your polar orbit is lined up, there's another way to do it. Similar to MarvinKitFox's suggestion, but since you're not going to be facing in the right direction, after the plane change to equatorial orbit just aerobrake at Duna again until you have a low equatorial orbit. Proceed as usual from there.

In reality, interplanetary spacecraft are not first launched into equatorial orbit, but usually a high-inclination low orbit before burning to their destination.

[davidpsummers]

Not sure why I didn't think of aerobraking. Raise periapsis, change inclination, and then aerobrake for free. A 1000 m/s change for something like 400 m/s!

[Kryxal]

Isn't the OP time-limited for a good launch? I'd probably set up an escape trajectory, and use prograde/normal as needed to pretty well match the AP/PE from Alexmoon. Assuming you're a bit late for the window, you'll probably need to launch with a Kerbol-relative radial inwards component, though, and that could match up well or poorly with your current orbit.

[MarvinKitFox]

Let's find out:

He then proceeds to spout a stream of scary gibberish, something like

v^2 = mu (2/ r_SOI - 2/(r_SOI + r_LDO))

mu = 3e11 m^3/s^2

r_SOI = 4.8e7 m

r_LDO = 4e5 m

v = 10 m/s

Klathu berada nictu!

Ph’nglui mglw’nafh Cthulhu R’lyeh wgah-nagl ftaghn

So, 10 m/s * sqrt(2 - 2 cos(73)) = 24 m/s.

Nailed it, MarvinKitFox!

Sorry, but I could last follow that level of black mathemagic back in school. 1872, i think it was....

I "cheated", and flew a string of maneuver nodes to Duna, and *tested* the values needed!

[davidpsummers]

Isn't the OP time-limited for a good launch? I'd probably set up an escape trajectory, and use prograde/normal as needed to pretty well match the AP/PE from Alexmoon. Assuming you're a bit late for the window, you'll probably need to launch with a Kerbol-relative radial inwards component, though, and that could match up well or poorly with your current orbit.

Indeed. I think I have enough time to raise the apogee enough to get a cheap inclination burn. With aerobraking I only have to do one orbit. I definitely don't have enough time to wait until the orbitical plane lines up with Duna's orbit.