Return from the Mun: Lies, lies and 276 dV!

[Benno]

So here's the thing. I was building a 3-crew Mun lander to collect a big packet of science (well more of a Mun hopper really), and after doing all my dV calcs etc, realised I'd neglected to plan for any means of getting my three intrepid explorers back to Kerbin. Whilst my Mun hopper should have arrived at it's destination with a healthy 3500 odd dV to go exploring with, a quick look at WAC's excellent dV map suggested I was going to lose a good portion of this trying get back home. So, I was debating whether to place some sort of station in low Mun orbit that could manage the return trip, or accept that my Mun hopper was going to have a somewhat shortened trip.

Looking at the chart, we can see that (from Kerbin take-off to Mun landing), it requires around 6260 dV, which is about right.

Now common wisdom would have it that getting home would take about the same, less some of the 4550 dV required for the ascent from Kerbin. That leaves around say 1710 dV, which is quite a bit still.

As I was considering docking with a station anyhow though, let's just look at the dV to get home from a low Mun orbit (14km). According to the chart (860 + 210), that's still 1070 dV, and probably a little more to deorbit from Kerbin. Another chart I looked at gave a similar estimate of around 1100 - 1200 dV.

So, that's around 1/3 of my dV. Admittedly I wouldn't need to take the whole lander back, but it was definately a bit of a downer and causing me to lean towards putting a transfer station in orbit around the Mun.

But, 1100 to 1200 just didn't feel right, so I did a little experiment and came up with this:

Yup, that's right, you can get back to the surface of Kerbin from a 14km Mun orbit using just 276 dV! Incidentally, for a basic 3-person capsule and 410 ISP engine, that's just 45 units of liquid fuel! Or, from the Mun's surface, that would be 916 dV. Where did all the free fuely goodness come from? I think there are two main factors that aren't obvious in the charts:

1) There's no need to circularise your orbit around Kerbin. You may be able to a direct descent (although I exploded from either heat or g-forces as I entered at around 3.5Km/s), but at the very least you can come in with a roughly 35km periapsis and do around 3 (automatic) aerobraking orbits before landing.

2) Launching from the Mun so that you are in a clockwise orbit around the Mun (remember, the Mun is orbiting counterclockwise around Kerbin, so when you do your Mun exit burn you want to be burning opposite to the direction of the Mun's orbit).

Needless to say, I'm now reconsidering my need to launch a transfer station prior to the main mission (although there would be other benefits too).

Does anyone have any similar revelations?

Edited February 22, 2014 by Benno
typo

[Dave Kerbin]

The chart numbers are for non-aerobraking transfers. Places where aerobraking is possible are labeled on the chart with a white arrow - in those cases you can either burn the delta-v or save some or all of it by aerobraking. You'll see the 860 on the path to the Mun there is a white arrow pointing to Kerbin and that means you can use aerobraking if you are going in that direction. If you wanted to return to a Kerbin 70km circular orbit without passing through the atmosphere it would use around the advertised 860 m/s, depending on how much of an assist you got coming out from the Mun.

Edited February 22, 2014 by Dave Kerbin

[Yasmy]

Right. On a delta-v map, from low orbit around one body to another body, you see an intercept/transfer burn and a circularization burn.

Going in reverse, undoing the circularization burn is the new intercept/transfer burn, and reversing the intercept/transfer burn makes it a circularization burn.

Edited February 22, 2014 by Yasmy
Nice reply Dave Kerbin! I never bothered to read the legend on that map.

[Starstrider42]

2) Launching from the Mun so that you are in a clockwise orbit around the Mun (remember, the Mun is orbiting counterclockwise around Kerbin, so when you do your Mun exit burn you want to be burning opposite to the direction of the Mun's orbit).

This part shouldn't matter. Yes, you want to escape the Mun in a retrograde direction relative to the Mun's orbit around Kerbin, but the delta-V required to do that shouldn't depend on whether your orbit around the Mun is prograde or retrograde -- you just start your burn on the opposite side of the Mun in one case vs. the other.

[Benno]

This part shouldn't matter. Yes, you want to escape the Mun in a retrograde direction relative to the Mun's orbit around Kerbin, but the delta-V required to do that shouldn't depend on whether your orbit around the Mun is prograde or retrograde -- you just start your burn on the opposite side of the Mun in one case vs. the other.

I think it does matter (correct me if I'm wrong), because wouldn't you always want to burn on the far side of the Mun (relative to Kerbin)? If you burn on the near side, once you escape the Mun you will be further out from Kerbin than had you burned on the far side?

Edit: Actually, you might be right. Looking at the tragectory on my image, it's not like there is much of an arc before you escape. If I flipped the burn to the other side of the Mun, the exit point would be pretty similar.

Edited February 23, 2014 by Benno

[Benno]

If you wanted to return to a Kerbin 70km circular orbit without passing through the atmosphere it would use around the advertised 860 m/s, depending on how much of an assist you got coming out from the Mun.

Yes, I think this is the essence of it. You don't need to do most of the 860 to go from Mun Intercept to LKO unless you plan to circularise. So (ignoring the 4550 because aerobraking takes care of that) that just leaves the 210 to go from Low Mun Orbit to to the Mun intercept/escape, plus a little bit more to bring your periapsis down in to Kerbin's atmosphere.

[Claw]

I think it does matter (correct me if I'm wrong), because wouldn't you always want to burn on the far side of the Mun (relative to Kerbin)? If you burn on the near side, once you escape the Mun you will be further out from Kerbin than had you burned on the far side?

You burn on the far side when going clockwise, and on the near side when counterclockwise (when viewed from above). This has the result that you are burning prograde in your orbit when traveling retro compared to the Mun's orbit. (Effectively slowing down your orbit around Kerbin when compared to the Mun.)

Edited February 23, 2014 by Claw

[Claw]

Yes, I think this is the essence of it. You don't need to do most of the 860 to go from Mun Intercept to LKO unless you plan to circularise. So (ignoring the 4550 because aerobraking takes care of that) that just leaves the 210 to go from Low Mun Orbit to to the Mun intercept/escape, plus a little bit more to bring your periapsis down in to Kerbin's atmosphere.

You don't need to do most of the 860 to go from Mun intercept to LKO unless you plan to circularize and don't want to aerobrake at all.

Ignoring the 4550 can be done because that's if you want to land. Reentry and parachutes/landing devices take care of that.

[sjwt]

With good luck/planning and multiple burns, you can get just sub 200 IIRC.

[Captain Sierra]

With good luck/planning and multiple burns, you can get just sub 200 IIRC.

If I'm good, I can do a minmus return for under 450, but 200 from the Mun? Are you sure? I think it takes at least 180 just to hit escape velocity.

Citation needed.

[Superfluous J]

If I'm good, I can do a minmus return for under 450, but 200 from the Mun? Are you sure? I think it takes at least 180 just to hit escape velocity.

Citation needed.

I've been giving it some thought. If you make sure your orbit is a multiple of Mun's, you'll come back into its SOI on a regular basis. Seems you could plan an at-periapsis burn for a dozen dV or so to aim your next Mun encounter in such a way that you'd end up coming back to Kerbin.

However, rather than do all that I'd rather just bring the extra liter or two of fuel