# Optimal turning from equatorial to polar orbit

Yeah, I know, "launch into target orbit". But let's assume we need to perform the maneuver.

A satellite in LEO equatorial orbit would need excessive amounts of delta-V to enter polar orbit. The amount needed in a very high orbit, on the other hand, is very low. Thing is getting from low to high orbit is costly, so the answer probably lies somewhere in the middle. So - what would be the optimal maneuver of turning LEO 90 degrees?

• I think you may want to fly a weak stability boundary maneuver. – Rikki-Tikki-Tavi Jun 8 '15 at 20:49
• The short answer for circular orbits is that up to roughy 40° of inclination change it is best to perform it directly, but above that it is better to raise your apoapsis and change inclination there and above a required inclination change of 60° it would theoretical be most efficient to raise your apoapsis up to infinity. I will have to look up my derivation in order to write a full answer. – fibonatic Jun 9 '15 at 0:40
• Off topic but I couldn't resist mentioning it - the Dyna-soar program ( en.wikipedia.org/wiki/Boeing_X-20_Dyna-Soar ) claimed the capability to change inclination by going down into the upper atmosphere. This was for minor changes needed to get repeat photography of a region, but could in theory be used for more drastic orbit changes with a robust enough vehicle... – Andy Jun 9 '15 at 7:56
• @Andy: That's a very interesting twist on the concept. Instead of aerobraking, temporarily turn from a satellite into an airplane. – SF. Jun 9 '15 at 8:00

$\Delta{v_i}= {2\sin(\frac{\Delta{i}}{2})\sqrt{1-e^2}\cos(w+f)na \over {(1+e\cos(f))}}$
Assuming a 0 e orbit to begin with, this basically becomes $sqrt(2)*v$, where v is the orbital velocity. Thus, for a LEO velocity of around 7.8 km/s, that adds up to be quite a lot of fuel!
I believe the best strategy would be to do a nearly escape elliptical orbit, set so the $A_n$ or $D_n$ is at the peak of that elliptical orbit. When you reach the appoasis, your velocity will be very small, you can easily do the inclination switch there for minimal delta v. Then when you come back you give yourself the correct eccentricity. This will cost a significant amount of fuel still, but still reasonable. The escape velocity is about $sqrt(2)-1$ , so doubling that will give you an effective fuel budged of something less than your orbital velocity. The trick is to make the inclination change when your orbit is almost not moving, and that works best at extreme orbital heights on highly elliptical orbits.