What's the orbital period of an asteroid which has synodic period of 429 days?

I know that the formula of orbital period for inner planets is $$P_{orb} = \frac{P_{syn}}{P_{syn} + 1}$$ and for outer planets is $$P_{orb} = \frac{P_{syn}}{P_{syn} - 1}$$ where $P_{syn}$ is the synodic period of the planet measured in years. However, I don't know whether the asteroid should be counted as an inner planet or outer planet.

  • $\begingroup$ Are you assuming the asteroid has a not overly eccentric orbit? If so, it would be outer and you could even estimate its semimajor axis using R^3/T^2 being a constant. If it crosses the Earth's orbit, I'm not sure synodic period is well-defined: it might conjuct the Sun twice in 3 days once and then almost a year without conjucting again. $\endgroup$
    – user21
    May 19, 2017 at 0:35

1 Answer 1


Just to clarify, inner planets usually refers to Mercury, Venus, Earth, Mars. But in relation to your question, the "inner" planets are just Mercury and Venus and the outer planets are Mars and out, because the +1 / -1 formula is in relation to whether the Earth is passing the planet as it orbits the sun or being passed by the planet.

Someone else can do the math if they like, the gist is explained here. Because it's the motion relative to the stars in the background that you're tracking, anything that orbits closer to the sun and faster than Earth would move in a different direction and require a different formula than orbits further out and slower.

If your asteroid remains outside the Earth's orbit, then it would obey the outer planet formula.

If your asteroid is an Earth crosser, which isn't uncommon, then it would have a different type of orbital period that would follow neither formula, and I'm not sure what formula would apply for an earth-crossing asteroid.

  • $\begingroup$ Yeah, I can just see trying to calculate synodic for Halley's Comet ! I bet it can complete quite a few of its sideral orbits before showing up in exactly the same place in the Terran sky. $\endgroup$ May 17, 2017 at 15:05
  • 1
    $\begingroup$ @CarlWitthoft When Halley's remains well outside the orbit of the Earth it should be mostly similar to an outer planet with a sidereal orbit similar to Uranus or Neptune. It just gets weird when it gets close and crosses the Earth's orbit. The apparent motion from our point of view, might be like part of a horse-shoe orbit (I would guess). The sidereal formulas mentioned above, don't account for eccentricity, so they're just approximations. (and sorry for the "I thinks", but I prefer to go that route when I'm not certain.) $\endgroup$
    – userLTK
    May 17, 2017 at 18:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .