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Due to the Oberth Effect, it is most efficient to accelerate a spacecraft along its orbit at or as close-as-possible-to periapsis, and decelerate within close proximity to apoapsis. But all discussions of this seem to be centered around pure tangential orbital acceleration, for example performing transfers within the same plane or reducing speed to allow orbital re-entry.

Does the same logic apply to non-tangential maneuvers, say, adjusting orbital inclination?

Assume we have two spacecraft in stable orbits of a planet that wish to rendezvous, and that these two orbits that are not co-planar. One must perform a transverse engine burn to adjust its own inclination to match the other.

At two points on each orbit, the orbital planes intersect. It is at one of these two points that acceleration must be applied to align the orbits. Regardless of whether this maneuver is performed at the ascending or descending node (defined by comparing their individual inclinations relative to the equator of the planet), at least some of the acceleration will be against the spacecraft's actual velocity at that time, because the goal is to reduce the relative inclination. Because of this, does the Oberth Effect make it most efficient to perform this maneuver at the node closer to apoapsis since the vessel's speed will be lower than at the opposing node?

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  • $\begingroup$ Great question! The answer is that it depends on which orbital element you need to change. I'll write a full blown answer in a few hours unless someone else does so earlier than me. $\endgroup$ – ChrisR Jan 16 '17 at 17:42
  • $\begingroup$ Actually, now that I think about it, the acceleration vector to change the inclination would optimally be perpendicular to the velocity of the craft, so is proximity to an apside actually meaningless in this example? $\endgroup$ – Omaha Jan 16 '17 at 20:37
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While delta-V - or change in velocity - remains the same, the effect on the trajectory depends in large part on the absolute velocity.

While a small change of speed (a small modification of the velocity vector) at periapsis can convert to large apoapsis change, maneuvers like plane change (normal/antinormal burn) or moving argument of periapsis (rotating the orbit major axis around the central body) require changing direction of the velocity vector by a significant angle - and the larger the velocity, the more delta-V is needed to change that.

An easy way to picture it is: you want to change your orbital inclination by 45 degrees by a burn perpendicular to the current trajectory - so you need to turn the velocity vector by 45 degrees. If your current speed is 20m/s, it's a 20m/s burn. If your current velocity is 2000m/s, you need a 2000m/s burn.

Thus inclination changes are best performed at speeds as low as possible - e.g. near a distant apoapsis of a highly eccentric orbit - or through gravity assist of an extra body, e.g. the Moon. Oberth effect is actively detrimental to these maneuvers.

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  • $\begingroup$ Your original velocity and delta V are legs of an isosceles right triangle? In which case your new velocity is greater than your starting velocity by a factor of $\sqrt{2}$. Increasing apogee speed raises your perigee. Raising your perigee reduces the Oberth benefit. $\endgroup$ – HopDavid Jan 26 '17 at 16:26
  • $\begingroup$ @HopDavid: Yes, that maneuver was chosen for picturing the problem most easily. The burn for turning 45 degrees and retaining absolute velocity value would contain a retrograde component and a value of 2-sqrt(2) of current velocity. The point was a clear argument though, not description of optimal maneuver. $\endgroup$ – SF. Jan 26 '17 at 16:44
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I think I've answered my own question. As I mentioned, acceleration to alter inclination would be perpendicular to the orbital velocity, so any effect that is a function of that velocity should not apply.

...right?

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    $\begingroup$ More so. The same delta-V creates the same velocity change but buys you far less inclination change. Imagine you want to change your inclination by 45 degrees by burning perpendicular to current trajectory. If you travel 20m/s, your burn needs to be exactly 20m/s perpendicular. If you travel 2000m/s... ? $\endgroup$ – SF. Jan 17 '17 at 19:56
  • $\begingroup$ I see, I see. Write it up as an answer and I can accept it. $\endgroup$ – Omaha Jan 25 '17 at 20:00

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