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How would one go about determining the apsides of an orbit, knowing only the current distance between the 2 bodies, the orbital velocity, and the period? Every formula I've seen requires eccentricity, but conveniently, every formula for eccentricity requires the apsides. Surely there must be some way to find one without knowing the other?

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  • $\begingroup$ Do you have measures of the distance as a function of time perhaps? Or are the bodies co-orbital? In latter case, there simply isn't enough information to infer their orbit, unless you also have ratios (in which case it might be possible, but tricky). $\endgroup$ – TildalWave Dec 12 '13 at 3:45
  • $\begingroup$ @TildalWave I do not have a formula like that, else I would be able to find the apsides just by minimizing and maximizing the function. I have the relative velocity, relative position, and semi-major axis, as well as the gravitational parameters of the planet and the vessel in orbit. $\endgroup$ – Trey Stone Dec 13 '13 at 0:19
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You mention this is a system in which a vessel is orbiting a planet. In that case it's fair to assume the mass of the vessel is negligible compared with the planet mass and the planet will sit in one of the focal points of the ellipse.

You can then use Kepler's second law to find the semi-minor axis as $$ \frac{1}{2} r v = \frac{\pi a b}{P} $$ where $r$ the current distance and $v$ the current velocity, $P$ is the period and $a$ and $b$ give the respective semi-major and semi-minor axes.

From $b$ you find the eccentricity as $$ b=a \sqrt{1-e^2} $$

And the apsides follow as $$ \mathrm{periapsis} = a ( 1 - e ) $$ $$ \mathrm{apoapsis} = a ( 1 + e ) $$

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  • $\begingroup$ This formula is incorrect. By the vis-viva equation, for all orbits with the same semimajor axis around the same body, the scalar velocity is the same for all radial distances they share. The substitution done to get v assumes that the velocity vector is perpendicular to the distance vector r, in which case the magnitude of r is already the periapsis or the apoapsis. $\endgroup$ – notovny Oct 10 '19 at 20:47

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