How to do a reverse orbit around the sun? (Inclination 180)

[tkraftson]

Do you have to do a giant reversal like with a mun or planet? How much DV would that take? I sometimes get offered contracts that require this and I never made the effort to try it.

[RoofCat]

You would need more delta v than the orbital speed of Kerbin just to get into other direction. And then some (a lot of some) to first rise Pe above Sun and then to your contract orbit. Due to the fact you start that burn from Kerbin orbit - slightly less for the node. Opposite to Kerbin at Kerbin altitude ~16ooom/s from LKO?

Easiest way is smart way. Go high for ~2500, switch direction for another 2500(?) at Ap, get back to circular with 2500 again at Pe. Use Jool+Tylo gravity underway for even more savings. Rather less than 10ooo m/s from LKO. It really depends on contract orbit as well of course. Retrograde at Moho level or retrograde at Eeloo level?

Those are very approximate numbers. The higher you go, the cheaper it will be and the more time take.

[tkraftson]

RoofCat, thanks for your reply, but I can't really understand it and it confirms my thinking that it is not worth the effort to attempt it.

[One Eye Jack]

You will need the dawn electric propulsion system. And even then it will be a chore. I had a mission that was on a polar orbit of the Kerbin Star and used the dawn engine. But even at that it was pretty difficult.

[ELK]

I'm confused, can't you just leave Kerbin to go around the sun either clockwise or counter-clockwise or is kerbin not on identical plane as the sun? If it's not couldn't you just leave the 2 times a year the planes intersect, start the orbit, then turn right back around? If you chose the better of the 2 times of the year you could go ahead of kerbins orbits so it would be better to go back. Unless the orbit you're trying to do is perpendicular then I guess it wouldn't matter to much when you started the mission.

I'm confused XD

[RoofCat]

brute force is reversing Kerbin orbital speed. You can see it in map mode - 9284.5m/s. Times 2. Because you have to reach 0 first and then the same again to create new orbit in the opposite direction. Minus LKO velocity launching in the right direction. Roughly. There may be Oberth somewhere.

At higher altitudes above Sun, orbital speed is much lower. That's why reversing there will cost much less! Getting higher is rather cheap (compared to getting lower).
Google also Bi-elliptical transfers.

You sure do know Mun and Minmus by now.
To reverse Kerbin orbit on low orbit, you will need 2x2290m/s
To reverse Kerbin orbit at Mun level, you will need 2x542.5m/s
To reverse Kerbin orbit at Minmus level, you will need 2x274.1m/s

To get to Mun level, you need 860m/s
To get to Minmus level you need 930m/s
Even if you don't do aerobraking and have to do the same transfer burn=slow down burn returning back to LKO, simple math can show you
2x2290 > (2*542.5+2*860)
4580 > 2805

those are simplified numbers, as you won't reach full orbital velocity (542.5) in the first place "jumping" up. You would have to circulize for that first! So reverse at Mun and Minmus level jumping from LKO is actually even cheaper - 200+m/s more saved at Mun and a bit less at Minmus

See the difference? That's how orbital mechanics work. Orbital velocity isn't linear. The closer to the gravity well, the quicker it grows. The further you go, the less energy you need to go even further. See the difference between Mun and Minmus transfer burns (860 vs. 930)? Almost nothing. While their altitude above Kerbin differs 4 TIMES!

Now the same (again, simplified) math with Minmus
2x2290 > (2*274.1+2*930)
4580 > 2408.2

So 4580 > 2805 > 2408.2
Or 180° reverse is more expensive on LKO than it is at Mun and that one is more expensive than it would be at Minmus.

In case of Kerbin you can save ~800m/s more using aerobraking on return. In solar orbit that is not an option. Encountering Jool or Laythe while doing so - at the right spot, it may be again. In addition to the huge gravity assist you get there for free as well.



KSP is about hard space turned easy challenges. Knowledge and creativity. But it is also sandbox. Don't like, don't do.

[ELK]

So it would take less fuel to retrograde than it would to normal / anti-normal to 180 your orbit on any celestial body?

[RoofCat]

using retrograde or normal should have the same effect in theoretical universe. No difference.

You need the same energy to reach the same changes! Orbit is accumulated energy (during launch). To balance 10t you need another 10t. No way around it. Straight or square
There is no leverage on the same orbit. You go for it higher where speeds are slower like described above.

In reality (also KSP) there may be slight differences between retrograde and normal approach. Due to burn times needed. Both will introduce bad drifts as you can't have perfect impulse burn. Drifts in different planes. One correction may be easier than the other one, not sure.

I prefer retrograde, it is cleaner for staying equatorial and KSP engines are strong enough in most cases to not fall down or touch atmosphere/some surface (or you should be smart enough to avoid that risk). Also I would never do full retrograde on LKO. Just launch West.
Solar is different. You have to be even smarter there due to higher orbital speeds and you not launching from the Sun.

[One Eye Jack]

Originally posted by ELK:

I'm confused, can't you just leave Kerbin to go around the sun either clockwise or counter-clockwise or is kerbin not on identical plane as the sun? If it's not couldn't you just leave the 2 times a year the planes intersect, start the orbit, then turn right back around? If you chose the better of the 2 times of the year you could go ahead of kerbins orbits so it would be better to go back. Unless the orbit you're trying to do is perpendicular then I guess it wouldn't matter to much when you started the mission.

I'm confused XD

It’s like launching a satellite in the opposite direction of Kerbins rotation. You have to have the normal delta V for an orbit plus you have to add the speed of Kerbins rotation.

Now imagine the speed of the planet going around the Star. You want to go the opposite way that the planet is already moving then it is going to take a huge amount of energy. You have to add that speed to the delta V needed. A huge difference.

I hope I am making sense. If not just say so.

Edit: It’s like you are already traveling around the Star very fast in one direction just because you are on a planet. Now you want to turn around and go the exact opposite direction.

[tkraftson]

RoofCat, thanks for the more understandable (to me) discussion above, and Astroknott58 for a basic analogy. I think I get it, but I still think it is much more effort and less fun than many other options in KSP. For example, I just finished putting a ship on Ike that had space for 12 Kerbals, mining equipment and a fuel maker and was on motorized wheels, then dug up 5000 units of ore and won a 1.3million contract. That was hard work and a very satisfying accomplishment, compared with reversing around the sun.