Reversing orbital direction?

[CrashyMcCrashFace]

So I get 2 separate contracts to rescue Kerbals from Mun orbit. Because I've done this a dozen times and think I'm getting hot-poop at it and I say to myself, "I'm super cool. I'll pick them both up at the same time and make some easy cash." Nope. I get out there and 1 is orbiting in the normal direction at about 10K but the other is 100K in the opposite direction. So I guess I'm making 2 trips. But this got me thinking there's almost always some trick to this space stuff. Is there a way to do a 180 without burning like 1,000+ dv worth of fuel? I've done just that around Minmus but not sure if it's reasonable for the Mun. I'm thinking, set a really high Ap where the speed will be slow to turn around but then gaining and then losing the Ap might be just the same as doing it low.

Edited April 13, 2018 by CrashyMcCrashFace

[Snark]

24 minutes ago, CrashyMcCrashFace said:

Is there a way to do a 180 without burning like 1,000+ dv worth of fuel?

Yes and no.

Let's consider the worst-case scenario for doing a reversal of direction:  let's say you're going from low Mun orbit in one direction to low Mun orbit in the other direction.  Your orbital speed is V.   Here are two things you could do:

• Just do a straight up "burn until you've reversed direction" maneuver.  Amount of dV needed:  2*V
• Burn to boost your Ap waaaay up as high as it can go.  Coast up to Ap.  Do a small burn to reverse direction.  Fall back to Pe and do a braking burn there.  Amount of dV needed, in the ideal case where the Mun's SoI goes out to infinity, would be that the first burn would be 0.414V (i.e. square root of 2, minus 1), then second burn would be negligibly tiny, then the third burn would be 0.414V again, for a total dV cost of 0.828V.  Now, you can't actually do that because the Mun has a finite SoI, so the actual dV requirement will be a bit higher.  However, overall, this would be considerably cheaper than the first option.

One other option to consider, if you're strapped for dV, is to use the fact that Kerbal EVA packs have a metric crapload of dV.  It's absolutely insane, like 600 m/s.  So, for example, if you wanted to, you could do something like this:

1. Rescue kerbal #1, the usual way.
2. For kerbal #2, all you need is a close encounter within 2.3 km to get within physics range.  Set that up.  Don't even try to match courses.  Just do a small burn that raises your Ap enough, at the right time that you go whizzing past kerbal #2 in the opposite direction.  All that matters is that you pass within 2.3 km when doing so.
3. Watch the target carefully.  As soon as it gets within 2.3 km, immediately hit the ] key to switch to it.
4. Congratulations!  You have now activated the kerbal, who can go EVA and has 600 m/s of dV to play with.
5. Use a big chunk of that kerbal's dV to put it in a friendlier orbit to match with your ship.

In this fashion, whatever dV your ship would have needed to do to match up with the 2nd kerbal... you can knock nearly 600 m/s off it, by letting that kerbal do a big chunk of the work for you.

[bewing]

But make sure you don't burn the Kerbal's EVA propellant until it's all gone. You will need a tiny bit left in order to actually board the rescue craft.

 

[Gargamel]

10 hours ago, Snark said:

However, overall, this would be considerably cheaper than the first option.

While still cheaper than a $2000 Hamburger, a $828 Hamburger is no bargain. 

But if you are deadset on one mission to nab both, flying out to the edge of the SOI is your best bet.  If you have the fuel to spare, go for it.  Just leave enough for at least the direct return, or you'll need an even bigger rescue vehicle. 

[Kerbal101]

Engines fight with gravitational influence of current body, in orbit or not. Less gravity influence = more efficient engines = more dV from fuel.

[Zeiss Ikon]

3 hours ago, Kerbal101 said:

Engines fight with gravitational influence of current body, in orbit or not. Less gravity influence = more efficient engines = more dV from fuel.

Actually, this is more or less backward.  You get the most dV from fuel by burning as far down in the gravity well as possible (just barely above 70 km, for Kerbin, just clear of mountain tops for the Mun, etc.), when the fuel you're burning is already going as fast as you can manage.  The reason you can change planes (an orbit reversal is nothing more or less than a 180 degree plane change) cheaply at the edge of the SOI is because orbital velocity is so low out there; a tiny bit of dV spent makes a huge difference in the orbit.  You can reverse a Kerbin orbit for around 4.7 km/s at 85 km, or you can do it for around 300 m/s at the high end of an orbit that grazes the atmosphere at the  bottom, but runs outside Minmus at the top.

Which leads toan important addition to @Snark's suggestion -- your orbit must be as eccentric as possible to maximize dV saving by changing plane at the top of the SOI.  A circular orbit that far out is significantly higher velocity than the Ap end of an orbit that barely clears the surface at Pe -- higher by the value of a circularizing burn.  Don't circularize, and you save up to three times the dV needed to circularize -- by not spending it in the first place, then not having to spend it twice more to reverse orbit.

Bottom line: to reverse your Mun orbit and get that second Kerbal in the most efficient way possible will cost you about the same dV as your return burn for direct reentry at 35 km Pe over Kerbin.  If you've got the dV, then go for it.

Edited April 14, 2018 by Zeiss Ikon

[Snark]

6 hours ago, Gargamel said:

While still cheaper than a $2000 Hamburger, a $828 Hamburger is no bargain.

Thus the comment,

16 hours ago, Snark said:

Yes and no.

 

[GoSlash27]

It'd take 392 m/sec DV to transition from a 100 km retrograde orbit to a 10 km prograde orbit.

158 m/sec to raise apoapsis to 2.4Mm
133 m/sec to reverse direction and raise periapsis to 10 km
201 m/sec to circularize at 10 km.

HTHs,
-Slashy
 

[blakemw]

The cheapest way is to leave the SOI then re-encounter the Mun again about a third of the way around Kerbin, now with the desired inclination.

This eliminates the 133m/s orbit-reversing burn, at the cost of ~2m/s to completely exit Mun's SOI.

 

Generally this can be done by just getting the minimal escaping ejection, then changing the timing (i.e. pulling the node around the orbit) so the ship re-encounters, then tweak the exact timing and the exact amount of prograde dV so it ends up retrograde instead of prograde, it's not nearly as hard as it may sound since it's really, really easy to re-encounter the Mun. In theory this can all be setup with the one burn, but a minuscule correction burn will probably be required because encounters at low relative speeds are extremely sensitive. 

Edited April 15, 2018 by blakemw

[Kryxal]

I'd probably run the rescue by first dropping your PE on the low, prograde orbit but keep your AP high up instead of bringing it down ... arrange it so next time around they have a close pass.  This allows you to EVA that Kerbal and catch up with your rescue craft, well within the EVA pack's budget.  After that, raise your AP if need be, then at AP do your plane-change (or reversal) and set up another close pass.