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The Clohessy-Wiltshire equations are a simplified (linearized) model that is applicable under few assumptions:

  • close range
  • circular target orbit
  • no disturbances

My question: Up to which boundaries are these assumptions valid?

In detail:

  • Up to which relative distance can CW equations be used?
  • Up to which eccentricity?
  • For which maximum time span is a prediction based on CW reasonable?

I guess the answers might not be absolute numbers, but maybe one can say something like: at a relative distance of 10 km the position error is about 10%. Or maybe there is an approximation or formula.

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    $\begingroup$ This is a bit tough to answer as is because, as is the case with any approximation, errors grow over time. Small errors becomes big ones given enough time. A calibration point: At the start of the ISS visiting vehicles project, the ISS office insisted that the ISS could have an eccentricity as high as 0.015 and that any visiting vehicle must be able to accommodate that. Back of the envelope calculations and detailed simulations said no visiting vehicle could use CW for rendezvous were that the case. (continued) $\endgroup$ – David Hammen Jun 22 '15 at 18:35
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    $\begingroup$ In actuality, except for ISS reboosts, the ISS eccentricity is about 0.003, well within the capabilities of a halfway decent spacecraft control system that uses CW. The ISS Program Office based that 0.015 eccentricity on the fact that the ISS altitude can range from 250 km to 450 km. That's the eccentricity of an orbit with a 250 km altitude perigee and a 450 km altitude apogee. That the Program Office would never let a vehicle rendezvous during an ISS orbit raising maneuver was irrelevant. It was ludicrous. $\endgroup$ – David Hammen Jun 22 '15 at 18:39
  • $\begingroup$ I found some information in the book "Automated Rendezvous and Docking of Spacecraft" from Wigbert Fehse (2008, p. 41): "Because of the linearisation, the accuracy of the Clohessy-Wiltshire (CW) equation decreases with the distance from the origin of the coordinate frame. In a LEO rendezvous mission, position errors will become significant at a distance of a few tens of kilometres from the origin. ... $\endgroup$ – Artas Aug 21 '15 at 13:48
  • $\begingroup$ ... For example, the error in the z-direction due to the curvature of the orbit will be dz = r (1 - cos(x/r)), which for an orbit with h = 400 km, r = 6766 km at a distance of x = 10 km, becomes dz = 7.4 m, and at a distance of x = 30 km becomes dz = 66.5 m." $\endgroup$ – Artas Aug 21 '15 at 13:49

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