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A geosynchronous orbit is one where a satellite's orbital period (more or less) matches a planet's rotation period. However, this is distinct from a geostationary orbit, where a satellite must remain over the same spot on the Earth. In an inclined geosynchronous orbit, a satellite would trace out an analemma over the course of a sidereal day. Such an orbit would also slowly decay, requiring adjustment over time.

Since you can have a geosynchronous orbit that is inclined, what is the maximum inclination you can have before it is no longer considered "geosynchronous"?

Asking because I found an interesting paper which considers the possibility of detecting aliens through geosynchronous satellites in orbit around their planets.

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    $\begingroup$ 180°, same as any other orbit. $\endgroup$
    – notovny
    Feb 28, 2018 at 21:24
  • $\begingroup$ @notovny Orbital inclinations go from -90° to 90°, not to 180°. $\endgroup$
    – Phiteros
    Feb 28, 2018 at 21:25
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    $\begingroup$ @Phileros Orbital inclinations of >90° are how retrograde orbits are traditionally expressed. I've never seen a negative orbital inclination in a list of orbital parameters, myself. $\endgroup$
    – notovny
    Feb 28, 2018 at 21:47
  • $\begingroup$ It would be important for a geosynchronous orbit, since it would correspond to the latitude of the spot it is geosynchronous over. $\endgroup$
    – Phiteros
    Feb 28, 2018 at 21:48
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    $\begingroup$ @Phiteros as the inclination is the angle between the orbit angular momentum vector and the Z axis of the reference frame, the domain is in fact 0-180, not -90-90. $\endgroup$
    – Chris
    Mar 1, 2018 at 1:50

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There is no maximum inclination for that. As you point out, and as "Fundamentals of Astrodynamics" by Bale et al. confirms in section 3.2.1, an inclined geosynchronous orbit traces out an analemma over a sidereal day. An equatorial geosynchronous orbit then simply traces out an analemma of height and width zero, i.e. a point-shaped one. At e.g. inclination 90° the orbit is over the poles, and traces out its analemma over the entire planet.

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  • $\begingroup$ @uhoh Check! Edited my answer correspondingly. $\endgroup$
    – Jan van Oort
    Aug 26, 2019 at 9:18

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