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In a recent question, the slow drift of a landing site on a rotating body relative to an orbiter came up.

This would sometimes require a plane change correction when one desires to recover the landing party.

What would the $\Delta v$ cost of such a maneuver be?

Some initial considerations:

  • The landing site does not uniquely define a plane. While the latitude sets a minimum inclination, the inclination of the orbiter's initial orbit may very well be larger.
  • Similarly, an entire family of planes are possible for the pickup orbit. Optimisation would require picking the closest one.
  • Landing from an equatorial orbit would never require a plane change, and neither will a polar landing.
  • When the initial inclination is low and/or the stay is short, the cost is low.
  • For all configurations of inclination and latitude, there would be no need for a plane change exactly twice during the body's rotation.
  • While the plane change can usually be done as a simple impulse, that's not always optimal. I would imagine calculation just the inclination change angle instead of the velocity change would be simpler

While not exactly a hairy problem, the 3-dimensional geometry is not quite trivial, so it would be nice to have an answer to this for future reference.

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  • $\begingroup$ While not in the spirit of the question, if launching from Earth, just wait 24 hours for the Earth to rotate back into the same position. That's what Soyuz 6, 7, and 8 did. Launched each 24 hours, 6 took movies of the attempted docking of 7 and 8. They then landed at about the same spot in 24 hour intervals. Zero delta-v required. $\endgroup$ – DrSheldon Sep 6 '19 at 12:04

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