What would be the best way to reverse orbital direction?

[Brainlord Mesomorph]

After some sloppy interplanetary piloting I have arrived at my target planet, but I’m in retrograde orbit instead of prograde orbit. Now I may, or may not actually, try to fix that. But it has me wondering:

What would be the best way to reverse orbital direction?

It would be great if we could just do a U-turn, like they do in Star Trek. (LOL) But obviously that won’t work. For the purposes of this discussion let’s assume the following nice round number: you are in low orbit of a body, at 1000 m/s.

Method 1:

A simple 2000 m/s retrograde burn. Problem, unless you have a ridiculous amount of thrust, you’re probably going to deorbit and crash before you finish the burn.

Method 2:

A series of perigee kicks to raise the apogee into an elliptical orbit (so the vehicle will be much slower at apogee) theoretically we could reduce speed at apogee to 500 m/s. Then it would only take a 1000 m/s to reverse direction, and you are much higher so you have much more time. But something tells me it’s going to cost about 500 m/s to raise the apogee , and another 500 m/s to get back down, and we’re still at a total of 2000 m/s. (I’m not a physicist but that just sounds right to me). And I right about that?

Method 3:

A series of plane-change burns? Change to a 45 degree retrograde orbit, then a polar orbit, and 45° to prograde, and finally prograde. At least then we are not falling out of orbit. Let me guess, that’ll be a 2000 m/s total, right? Or more?

Method 4:

The one I haven’t thought of. That you are about to tell me.

But my guess is that if you’re going to 1000 m/s in the wrong direction no matter what we do its going to cost 2000 m/s to turn around. Right?

Opinions? Physics?

Please discuss.

Edited August 17, 2015 by Brainlord Mesomorph

[Signo]

I think (and that is my usual habit too) that the best place where to reverse your orbit is as soon as you enter the target's SOI burning radial/antiradial. So this should fall under your category 2.

[KerikBalm]

Stick your Apoapsis way out there, reverse direction, then take it back down again.

Consider Kerbin orbit, since we actually have a rough idea of the numbers involved without doing much math.

Orbital velocity: ~2,300 m/s

Reversing direction would thus take 4,600 m/s

Or you can stick your Apoapsis out past minmus for ~900 m/s

Minmus orbits at ~250 m/s, but we won't even be going that fast because our PE is not that far out... lets say velocity at apoapsis is ~200 m/s

A direct reversal then is 400 m/s

Then to lower AP again is another 900 m/s

Total: 900+400+900 = 2200 m/s

That is less than what it takes to even do a 90 degree plane change (which will require cancelling your forward velocity, and adding velocity normal to your current orbit, best done with a 45 degree burn, which will require ~3250 m/s)

If there is a body like a moon to help reverse direction/gravity assist you into a prograde orbit, even better.

If the body has an atmosphere and you can aerobrake your apoapsis back down, even better.

In the example of kerbin, just aerobraking should get a complete reversal down to about 900+400 = 1,300 m/s

If you can manage to get mun or minmus to reverse your direction, you're looking at <1,000 m/s

the PE kicking and plane change at apoapsis is definitely superior.

I'm not sure at what point it becomes better, its certainly not worth it to adjust your inclination by 1 degree, but it is certainly worth it for a 90 degree plane change.

If you are retrograde, then you should always do this.

If you just captured into a retrograde orbit, you can do a short burn to just barely capture (don't circularize), and then you've already got a really high apoapsis, and you can plane change and complete the circularization for not much more than if you just captured prograde to start with.

I recently made a mistake like this with my eve probe... but I had intended to have it visit gilly and eve anyway... and it had an ISRU scanner and I wanted to go polar anyway... so it didn't really matter except that I didn't insert into a polar orbit in the first place... so I kicked my apoapsis way out there, to visit and scan gilly - and then from there went to a polar eve orbit.

---Edit---

Yes, as the one above me says, a radial/antiradial burn far out is best, but that happens before capture. If you find your last quicksave is right before perapsis and the insertion burn, then its far too late, and you want to insert into a highly eliptical orbit.

If you've saved after the insertion burn, then raise your PE.

As to method 3... a series of burns will take more dV than doing it all at once, because you waste dV adding velocity normal to the planets rotation, only to cancel it again.

We can also take the example of Duna, instead of your guess numbers.

~1300 orbital speed -> 2,600 for direct reversal

~600 for a very eliptical orbit-> ~1200 to raise and then lower Apoapsis (but you can aerobrake it down, for only 600)

~ 50m/s if that from a very eliptical duna orbit (assuming you went right to the SOI edge, which ~610 will take you to)

~1250 for a reversal, ~650 with aerobraking.

It wins again

Edited August 17, 2015 by KerikBalm

[Sharpy]

If the body has a moon, use it for a gravitational assist and reverse.

If it doesn't, increase apoapsis to nearly escape speed and reverse there. Edge of Kerbin SOI is 912m/s, and reversal there is 52m/s. Airbrake the rest of the way back.

[GoSlash27]

Method 2: raise your apoapsis to near- escape (using gravity assist if possible) and reverse at apoapsis. Use atmosphere to brake if possible.

That's the cheapest way you're going to pull it off.

Best,

-Slashy

[Brainlord Mesomorph]

Method two is it.

Thanks to all. (and a minor pat on my own back for having thought of it, even if i didn't know it was better.)

[Red Iron Crown]

You can generalize it a bit: For a greater than 60 degree change in inclination it is worth raising your Ap as high as possible while staying in the SoI; for plane changes between 38.4 and 60 degrees it is worth raising Ap by a lesser degree, given by this formula:

Apoapsis = r * sin(θ/2) / (1 - 2 * sin(θ/2))

(r being orbital radius)

See more discussion here.

[Nich]

No one has mentioned leaving the SOI. I am pretty sure if you leave at the right speed/angle you wont even need an adjustment burn saving 50-100 dv and costing only 5-10 dv more then going highly elliptical. Although the extra time would be extremely long.

[Brainlord Mesomorph]

No one has mentioned leaving the SOI. I am pretty sure if you leave at the right speed/angle you wont even need an adjustment burn saving 50-100 dv and costing only 5-10 dv more then going highly elliptical. Although the extra time would be extremely long.

You're Right!

using KerikBalm's Kerbin/Minmus analogy, it would cost 930 m/s to reach Minmus orbit, and 400 m/s to reverse direction. Leaving SOI costs 950 m/s with almost zero to reverse, SAVING 360 m/s!

THAT'S METHOD FOUR!

That's funny. The best way to reverse orbital direction is to leave orbit!

Edited August 17, 2015 by Brainlord Mesomorph

[tomf]

My experience is that if you leave the SOI you normally need a fairly significant burn just to encounter Kerbin again though.

[Red Iron Crown]

You're Right!

using KerikBalm's Kerbin/Minmus analogy, it would cost 930 m/s to reach Minmus orbit, and 400 m/s to reverse direction. Leaving SOI costs 950 m/s with almost zero to reverse, SAVING 360 m/s!

THAT'S METHOD FOUR!

How are you getting 400m/s to reverse? Your orbital speed will be much less than that at Minmus altitude if your Pe is anywhere close to Kerbin. Eg. an orbit with a periapsis of 100km and an AP at Minmus' altitude will have a speed of just 47m/s at apoapsis, so just 94m/s to reverse.

Edited August 17, 2015 by Red Iron Crown
Fixed silly multiplication error. :/

[Brainlord Mesomorph]

How are you getting 400m/s to reverse? Your orbital speed will be much less than that at Minmus altitude if your Pe is anywhere close to Kerbin. Eg. an orbit with a periapsis of 100km and an AP at Minmus' altitude will have a speed of just 47m/s at apoapsis, so just 92m/s to reverse.

KerikBalm said so.

I didn't check his math

[KerikBalm]

Well, I was just throwing numbers out there "we actually have a rough idea of the numbers involved without doing much math"

I would trust the number from sharpy of 52 m/s more... which is quite different than my numbers (its also at the very SOI edge, not minmus, butthe dV needed to go to minmus or the SOI edge is pretty much the same)

I just knew that the speed at apoapsis would be significantly less than the speed at which minmus orbits in a near circular orbit.

400 or 52 sounds like a big difference (maybe 400 is more like if you only went to Mun's distance from Kerbin... like say you want the plane change to happen sooner)... but even overestimating by ~350 m/s, the eliptical orbit-> plane change is the very clear winner.

My intent was to show beyond a doubt that method 2 was superior to method 1.

As to the newly proposed "method 4":

The problem is that as soon as you leave the SOI, you won't enter it again for a long time unless you make a burn... I think just the SOI edge and a 52 m/s reversal is best for practical purposes (practically I don't go out that far, as I often have a reason to value in game time in addition to RL time)

Edited August 17, 2015 by KerikBalm

[Brainlord Mesomorph]

Wait, we forgot something: the retrograde part.

Leaving the SOI about 90 degrees to prograde, you can almost match the planets orbit, and orbit parallel to it.

Leaving SOI retrograde though, you will be way slower than the planet. and you would have to burn fuel to catch up to it.

RIGHT??

EDIT: or am i confusing retrograde relative to the planet and retrograde relative to the sun. I could be orbiting the planet backwards, but still fling the ship ahead of the planet of in solar orbit, just on the sun-side -

Edited August 17, 2015 by Brainlord Mesomorph

[KerikBalm]

if you leave going slower, then your pe is lower than kerbin's, but your AP will intersect kerbin's orbit.

Your orbital period will be shorter, and it may be many years before an intercept with no burn.

The opposite happens if you leave going prograde.

Either way, ideally, you'd cross the SOI going just a few m/s relative to kerbin (assuming perfect execution... which won't happen in reality)

But I can't see how you'd cross the SOI with less velocity to cancel than if you had you Ap just barely inside the SOI.

What is happening is that you are taking your Kinetic energy (ie, your velocity), and converting it into Potential energy (height).

The reduced plane change costs is because of gravity... when you cross the SOI boundry, that gravity is gone, so I don't see how you'd get it to be any cheaper by crossing the SOI as opposed to just going right up to the edge.

[Sharpy]

Well, I was just throwing numbers out there "we actually have a rough idea of the numbers involved without doing much math"

I would trust the number from sharpy of 52 m/s more... which is quite different than my numbers (its also at the very SOI edge, not minmus, butthe dV needed to go to minmus or the SOI edge is pretty much the same)

I just knew that the speed at apoapsis would be significantly less than the speed at which minmus orbits in a near circular orbit.

400 or 52 sounds like a big difference (maybe 400 is more like if you only went to Mun's distance from Kerbin... like say you want the plane change to happen sooner)... but even overestimating by ~350 m/s, the eliptical orbit-> plane change is the very clear winner.

274.1 m/s is Minmus orbital speed according to the wiki, so twice that to reverse = 548,2. The difference in delta-V to reach various distant orbits is really tiny but the speed differences are quite big. Also, the times - the ships by the apoapsis are sloooooow. 52 is experimentally measured 26 doubled - the speed at the apoapsis nearly touching the SOI. Imagine going the distance from Minmus to SOI edge at back at a pace of a decent rover on the surface.

Leaving the SOI would reduce the burn even more but finding the right burn to reenter becomes quite tricky; sometimes it's better to go for an easier but less optimal maneuver (example: launching to other planets from Kerbin instead of Minmus orbit (after Minmus refueling) with Kerbin assist; the Oberth effect is not worth the headache of setting up the maneuver.)

Also: seriously reconsider Mun Oberth maneuver. I don't have the numbers with me but I believe they are pretty good.

[GoSlash27]

You're Right!

using KerikBalm's Kerbin/Minmus analogy, it would cost 930 m/s to reach Minmus orbit, and 400 m/s to reverse direction. Leaving SOI costs 950 m/s with almost zero to reverse, SAVING 360 m/s!

THAT'S METHOD FOUR!

That's funny. The best way to reverse orbital direction is to leave orbit!

Yeah... I definitely wouldn't leave orbit. You'll waste more DV trying to get a new encounter than you'd save. a highly- elliptical orbit to the edge of SoI is darn- near stationary at Ap. It'd cost tenths of a m/sec to reverse direction there.

Best,

-Slashy

[Nich]

I am going to have to disagree if you have an highly essentric Mun orbit your are still going 100+m/s at the edge of the SOI. To leave the SOI you only need .001 more dV and you will leave with +1 m/s prograde at year 1000 you do a 1 m/s correction burn and reenter the Muns SOI in the opposite direction.

Alternatively you can leave retro and slightly in towards kirbin to set up a slightly elliptical orbit. Since you are traveling slower across a faster route and you left behind you should intersect next orbit ahead and in front. Then all that is required is a .1-2 m/s burn at PE

[Brainlord Mesomorph]

Yeah... I definitely wouldn't leave orbit.

Your right, some overly-round math led us to some incorrect conclusions. The most fuel-efficient thing is to go to the edge of SOI. but that will also take the most time. That highest orbit could take days or weeks.

If life support or any other consumables were at issue, you wouldn't want that. But this is KSP, we have time.

- - - Updated - - -

. Since you are traveling slower across a faster route and you left behind you should intersect next orbit

A Kerbal YEAR later?!

Some people take fuel efficiency too far.

[ajburges]

274.1 m/s is Minmus orbital speed according to the wiki, so twice that to reverse = 548,2. The difference in delta-V to reach various distant orbits is really tiny but the speed differences are quite big. Also, the times - the ships by the apoapsis are sloooooow. 52 is experimentally measured 26 doubled - the speed at the apoapsis nearly touching the SOI. Imagine going the distance from Minmus to SOI edge at back at a pace of a decent rover on the surface.

Leaving the SOI would reduce the burn even more but finding the right burn to reenter becomes quite tricky; sometimes it's better to go for an easier but less optimal maneuver (example: launching to other planets from Kerbin instead of Minmus orbit (after Minmus refueling) with Kerbin assist; the Oberth effect is not worth the headache of setting up the maneuver.)

Also: seriously reconsider Mun Oberth maneuver. I don't have the numbers with me but I believe they are pretty good.

If you are in a circular orbit that far out, you are doing something wrong.

A 80 x 47000 km orbit around Kerbin gives you a speed of 45.72 m/s at Ap (period of 110 hours) (921.19 m/s to boost Ap). Near SOI boundary with a 80 x 80000 km orbit around Kerbin gives you a speed of 27.08 m/s at Ap (period of 241 hours) (930.45 m/s to boost Ap).

A gravity assist around a Moon is generally preferable because the dV is pretty good for the time it takes.

[KerikBalm]

274.1 m/s is Minmus orbital speed according to the wiki, so twice that to reverse = 548,2. The difference in delta-V to reach various distant orbits is really tiny but the speed differences are quite big. Also, the times - the ships by the apoapsis are sloooooow. 52 is experimentally measured 26 doubled - the speed at the apoapsis nearly touching the SOI. Imagine going the distance from Minmus to SOI edge at back at a pace of a decent rover on the surface.

Yea, I know, I was just pulling rough numbers where there would be no doubt that its not actually a higher number. I suspected it was actually much much less, but for the purposes of comparing #1 to #2, it was adequate.

I apologize for any misinterpretations this may have caused when comparing #2 to other methods.

ajburges.... yes, you shouldn't be in a circular orbit... oh how my rough numbers have detoured this...

The point of using the velocity of minmus (in a circular orbit), was to establish an upper bound for what the reversal maneuver would cost.

Yes, it is actually even lower than that, method #2 wins even more.

I am going to have to disagree if you have an highly essentric Mun orbit your are still going 100+m/s at the edge of the SOI. To leave the SOI you only need .001 more dV and you will leave with +1 m/s prograde

I don't see how you can leave the SOI with less relative velocity than you'd have right at the edge of the SOI.

Gravity is slowing down the relative velocity as you travel out of the gravity well, it can't do that any more once you leave the SOI

at year 1000 you do a 1 m/s correction burn and reenter the Muns SOI in the opposite direction.

Did you type that right? year 1000????

Edited August 17, 2015 by KerikBalm

[Laie]

A highly- elliptical orbit to the edge of SoI is darn- near stationary at Ap.

I think I got it down to like 3m/s on a high apoapsis near Kerbin. But not all SOIs are created equal... IIRC it is where the parent body's gravity becomes stronger than the local one. So the high-AP approach should work best around Eeloo, and not very well at Laythe or Moho.

[ajburges]

I don't see how you can leave the SOI with less relative velocity than you'd have right at the edge of the SOI.

Gravity is slowing down the relative velocity as you travel out of the gravity well, it can't do that any more once you leave the SOI

Did you type that right? year 1000????

I don't think he is. I think he is proposing a long transfer to Kerbin. Eject and wait until the minimal difference in period causes you to enter on the opposite side of Kerbin SoI. You can do a correction burn a quarter orbit to SoI change.

[Nich]

I had incorrectly assumed you would keep moving away from the Mun but the more I think about it no mater which way you leave you should intersect the Mun again in 1 orbit assuming the difference in relative velocities is small. I am going to agree that the SOI boundry depends on the size of the relative bodies in question.

If Minimus was orbiting Kerbin at 100km its SOI would be tiny, at this point it would make a lots more sence to leave the SOI to change orbit.

Fk = G*mk*ms/rk^2

Fm = G*mm*ms/rm^2

SOI is defined as when Fk=Fm

mm/rm^2 = mk/rk^2

mm = mass of minimus (2.6457897×10¹⁹ kg)

mk = mass of kerbin (5.2915793×10²² kg)

rm = SOI of minimus

rk = 100km

sqrt(mm*rk^2/mk) = rm = 2236m

thus as i suspected Minimus's SOI would be inside the surface that close to Kerbin.

[MarvinKitFox]

Method 1:

A simple 2000 m/s retrograde burn. Problem, unless you have a ridiculous amount of thrust, you’re probably going to deorbit and crash before you finish the burn.

Method 2:

A series of perigee kicks to raise the apogee into an elliptical orbit

Method 3:

A series of plane-change burns?

But my guess is that if you’re going to 1000 m/s in the wrong direction no matter what we do its going to cost 2000 m/s to turn around. Right?

Method 1: works fine, but burns 2 * 1000 + whatever you need to maintain altitude while u-turning... bad idea

Method 3: Good grief!! You need to be build out of Fuel of Awesomeness to consider this!

Method 2: Yes

A bit of a boost, increasing apogee altitude and reducing apogee velocity, is the way to go. It also increases your Apogee altitude, thus reducing the "hover in place" cost of a u-turn.

The higher you kick your apogee, the better.

For best results, kick your apogee to infinity-1 (or a good approximation thereof)

With a starting orbit speed of 1000, this will require a kick prograde to a velocity of sqrt(2) * Vstart = 1414.21m/s. = Burn of 414.21/s from a start of 1000m/s

With apogee at basically infinity, your velocity at apogee will be basically zero. Apply the same zero twice as a u-turn.

At perigee, apply another 414.21m/s of braking, to reenter your perfect circle orbit in the opposite direction.

Total cost: 828.42m/s.