Latitude of Landing Sites and Return DV

[jsfalconero]

I'm fairly certain that this is the correct forum for this sort of thing (if not, feel free to move ).

I was wondering earlier what effect the latitude of your landing site (read: My insertion around the Mun was inclined) has on the required dV to return to Kerbin. For example, I've seen that the estimated dV to return from the Mun is 1km/s, but this is assuming an equatorial landing, correct? The idea being that from the equator you can burn nearly directly retrograde to the Mun's orbit for a minimum-dV return. From a higher (or lower) latitude though you can't because you are correcting for the inclined launch site (at least you are if you burn directly east or west from your landing site). So here's a question: is there a direction you can burn from a high/low latitude LZ that still takes roughly 1km/s dV to return to Kerbin?

Essentially, imagine that Jeb is (hypothetically, of course) landed on the Mun in a crater at approximately 45deg N latitude with just about 1km/s left in his lander. Burning due east would place him on an inclined orbit that does not apply all the thrust retrograde to his Kerbin-relative velocity. Should he instead burn to the south-east to correct for the inclination, or would that waste more dV on the inclination change than you pick up by applying the remaining thrust along the proper vector? Is there a best case scenario where Jebediah makes it home without having to get out and push?

Edited November 6, 2013 by jsfalconero

[Kasuha]

Inclined orbits have negligible effect on the minimum delta-v needed to return to Kerbin but have big effect on when it can be done. With equatorial orbit, you can burn for Kerbin with minimum dv anytime. With inclined orbit, you can do such burn only when part of that orbit is parallel to the mun's orbit, i.e. two times per mun's orbit around Kerbin (and you're exiting along this parallel part). Or you need to spend some dv to correct the angle.

If you start landed, there is no problem. As long as you liftoff at the right angle (which would lead to orbit with part parallel to Mun orbit), you can always get near minimum dv as if you started on equator.

But in KSP, delta-v to/from Mun is rarely an issue.

Edited November 5, 2013 by Kasuha

[Sirrobert]

From a landed state it takes very little deltaV to get the orbit you want.

If you are already in orbit: Burn out towards the retrograde vector of the Mun itself (but don't escape.) You can now adjust your inclination at apoaps very cheaply (obirth effect), and because the Periaps is still down at low Mun orbit, you can easey add a little bit of extra deltaV to turn your inclined orbit into an escape (for if you don't want to wait for the right moment like with Kasuha)

[Geschosskopf]

Essentially, imagine that Jeb is (hypothetically, of course) landed on the Mun in a crater at approximately 45deg N latitude with just about 1km/s left in his lander. Burning due east would place him on an inclined orbit that does not apply all the thrust retrograde to his Kerbin-relative velocity. Should he instead burn to the south-east to correct for the inclination, or would that waste more dV on the inclination change than you pick up by applying the remaining thrust along the proper vector? Is there a best case scenario where Jebediah makes it home without having to get out and push?

If you don't launch from the equator, you will always have an inclined orbit. The MINIMUM inclination is the latitude of your starting position and you can make this rather worse if you try hard enough. The higher your inclination, the more delta-V it takes to get back into the equatorial plane, so it's usually best not to mess with this until you're in space. So just take off heading east, which will give you close to the minimum inclination.

Now, usually it's no problem coming home from Mun from even landing on Mun's south pole. Sure, it takes a bit more delta-V to achieve Munar orbit because you don't have the rotation helping, but it's not enough to make a difference. Nor does this make an appreciable difference in getting back to Kerbin. Just burn prograde on the side of your orbit facing Kerbin until you get Munar escape and find yourself in a huge ellipse cocked at some crazy angle to Mun's orbit. All you have to do now is drop your Pe to about 30km in Kerbin's atmosphere and you're guaranteed to land. And you do this with a retrograde burn at your Ap. Because the Ap is out as far as Mun, this usually takes only a dozen or 2 m/s. And that's all the burning you need to do. It might take you 3 or 4 days to get home but you sure don't need much fuel to do it.

Now, if you want/need to get back into an equatorial orbit after an inclined launch, that's going to cost you delta-V. It's best to do plane changes as far from the planet as possible, but pushing your Ap out there also costs delta-V so it's something of a wash. Thus, if you're going to land at high latitudes and need to get equatorial again, you need to plan for this and pack some extra fuel.

NOTE: When transferring anywhere, it's very easy to change your inclination for very little delta-V either just before or right after the SOI change as you approach the target. You're way far from the planet so it usually takes less than 10m/s to give you any inclination you want. So, let's have no more accidental landings at high latitudes .

[Alistone]

The only additional deltaV added is from the slower "rotational speed" of the surface you are on. On the Earth it is nearly 1000 mph at the equator compared to 700mph at 45 degree latitude.

But in KSP; for the mun it is very small: maybe 10 m/s at the equator compared to 7 m/s at a 45 degree latitude. I don't have the exact number; but if you have landed you can toggle your velocity from "surface" to "orbit" to see what the actual value is, but as was said up above, on the mun it is essentially negligible.

It does however make plotting your optimal course harder...

[Kasuha]

...

just take off heading east, which will give you close to the minimum inclination.

...

burn prograde on the side of your orbit facing Kerbin until you get Munar escape

...

Simple, usually good enough but definitely not optimal unless you're sitting on the Mun longitude line directly facing or directly opposite to Kerbin.

Optimal, or at least better approach is:

Imagine there's a line of the Mun's orbit in space and it is going through the Mun's center. There are two points where it intersects Mun's surface. You want to lift off directly to/away from this point, i.e. establish an orbit which will go above both these points. You don't need to care about your inclination.

Establish an orbit and place a maneuver which will lead you to escape Mun's SOI. You want this escape to be against the Mun's orbital velocity and to be parallel with Mun's orbit at the place where it escapes the SOI. There you can adjust your burn to already lead to Kerbin's atmosphere. While adjusting you need to move the maneuver along the orbit to keep the escape parallel because the angle by which Mun's gravity will deflect you will depend on the amount of delta-v you apply. Notice that if you do it right, the orbit around Kerbin will be only very little inclined, the inclination will not be given by delta-v you applied but only by your offset above/below Mun's orbital plane where you exit its SOI.

Edited November 6, 2013 by Kasuha

[jsfalconero]

Blast! How do I mark this as answered? Thanks for all your input!

[Geschosskopf]

Optimal, or at least better approach is:

If I read your runes aright, what you say can be shown in this pic:

Is this correct? IOW, you burn at the highest or lowest latitude in your orbit, whichever happens to be on the inner side of Mun when you're heading in the opposite direction as Mun, so your vector is in a parallel plane with Mun's orbit. And you have to time this so your departure vector is parallel but in the opposite direction as Mun's orbit at the time, which occurs when the part of your orbit just in front of you crosses Mun's orbit.

So I can see why you want to get your orbit pointed in the right direction at launch. If you don't, you'll have to wait until the one point in Mun's orbit around Kerbin where your orbit crosses Mun's and you're heading in the right direction. By launching towards the intersection point, you're creating that point here and now.

Is this right? If so, I'll file this away as a very useful tip .

[Geschosskopf]

Blast! How do I mark this as answered? Thanks for all your input!

Edit the 1st post, select "Go Advanced", and you can change it.

[Kasuha]

If I read your runes aright, what you say can be shown in this pic:

Yes, you understand it right.