See Topic. I know it is true for Sol, but in another solar System, is it possible that the mid or outer planets are completing one full orbit in a shorter amount of time than what a (smaller?) planet, closer to the star, would require ?

  • $\begingroup$ Not if there is just single central star - see Kepler's laws. Not sure if there exists any arrangement of multiple stars or something to allow it but I guess no. $\endgroup$ – jkavalik Jul 22 '16 at 7:03
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    $\begingroup$ There is one interesting exception I thought I might mention - a second, small "planet" orbiting at the L2 Lagrangian point of a much heavier planet. It will be further out than the larger planet but will orbit in the same period. This is not a stable, natural orbit though, so I'm just mentioning it out of academic interest. $\endgroup$ – Andy Jul 22 '16 at 7:43
  • $\begingroup$ @Andy An object in L4 or L5 does also have a larger distance to the barry centre if the mass of the planet is some significant fraction of the star's mass. $\endgroup$ – SE - stop firing the good guys Jul 22 '16 at 9:31

Yes, it is possible.

While the other answers show the correct observation that the orbital period is directly derived from the semi-major axis (half the distance of the longest axis of an ellipse), that is not exactly the same as the average distance. Consider the following example:

eccentric orbit

Here, both the circular and the eccentric orbit have the same semi-major axis, and thus the same orbital period. The eccentric orbit is divided roughly in half, with one part closer to the Sun than the circular orbit, and one farther away (red part). However, an object moves faster in its orbit when close to the Sun, so the object is going to spend a lot less time in the inner part than in the outer part. The average distance is then larger than that of the circular orbit. Trading a little of that average distance to get a slightly lower orbital period, your scenario is achieved.

While what you describe is is technically possible in this way, what you are really after is what the others have explained. This is just a remainder of how orbital mechanics can surprise you once in a while.

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    $\begingroup$ Such orbits would likely be horrenduously unstable (and technically not "planets" as the orbit region is not cleared) $\endgroup$ – Chieron Jul 22 '16 at 9:59
  • $\begingroup$ @Chieron Eccentric orbits are perfectly stable. Why are you assuming there are other objects in the region? $\endgroup$ – SE - stop firing the good guys Jul 22 '16 at 10:01
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    $\begingroup$ you are passing at least the second planet. $\endgroup$ – Chieron Jul 22 '16 at 10:04
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    $\begingroup$ Your answer compares an excentric orbit to a circular one with a slightly larger semimajor axis. The excentric body needs to cross the other orbital region when it approaches the star and again on the way back. This is not impossible, given orbital resonances. But any third planetary body will greatly complicate matters. $\endgroup$ – Chieron Jul 22 '16 at 10:51
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    $\begingroup$ @Chieron It does not really have to cross the other planet's orbits, consider for example a large inclination. $\endgroup$ – SE - stop firing the good guys Jul 22 '16 at 11:03

No, the laws of physics are the same wherever you are. Kepler's Laws govern what kinds of orbits planets have. In this case, we are interested in Kepler's Third Law. It states that the square of the period is proportional to the cube of the semimajor axis. Mathematically, this is $P^2\varpropto a^3$.

Basically, this means that the further away a planet is from its star, the longer it takes to complete an orbit. This is a universal law, and it does not matter whether the system in question is our own solar system or some other planet going around a faraway star.

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    $\begingroup$ Shouldn't that be just $P^2\varpropto a^3$? (No equal sign.) $\endgroup$ – a CVn Jul 22 '16 at 11:34
  • $\begingroup$ Oh yeah, my mistake. I put that in but forgot to take it out when I looked up what the proportionality symbol in MathJax was. $\endgroup$ – Phiteros Jul 22 '16 at 16:34
  • $\begingroup$ Such things happen; that's what comments are for. $\endgroup$ – a CVn Jul 22 '16 at 20:31
  • $\begingroup$ "The laws of physics are the same wherever you are" - says the guy that's only been in one solar system... pff... $\endgroup$ – corsiKa Jul 23 '16 at 5:59
  • $\begingroup$ @corsiKa It's dangerous to make assumptions like that. $\endgroup$ – Phiteros Jul 23 '16 at 6:27

No it's impossible.

The size of an orbit is relative to the speed of the orbiting object.

If a planet moved faster, it will be further, and completing a revolution will take longer because of the increased distance.


No, inside a single solar system this is not possible. The larger a circular orbit the longer it will take to complete.

Two catches:

If we have two different solar systems, one with a massive sun (and thus higher speeds needed for the same orbit) and you compare that to a system with a lighter sun then you will see different speeds. But then we are comparing two orbits around two different suns.

Secondly, I have no idea how this will work for non-cicular orbits. Especially if you take one circular orbit and one highly elliptical one. Also, how do you declare which is the 'outer planet' if the ellipse takes it both inside and outside the other planets orbit?

  • $\begingroup$ I think that your two caveats are somewhat beyond the scope of the question, and may confuse a reader. I understand what you're getting at, but they are a bit extraneous. $\endgroup$ – Phiteros Jul 22 '16 at 7:25
  • $\begingroup$ Also, your second consideration doesn't even hold. What determines a planet's orbital period is the length of its semimajor axis. It is independent of the eccentricity of the orbit. So for a planet on a highly elliptical orbit, though it may come closer to the star at some point in its orbit, its orbital period will still be longer. The planet with the longer semimajor axis would be considered the 'outer' planet in most cases. $\endgroup$ – Phiteros Jul 22 '16 at 7:36
  • $\begingroup$ I was wondering about that. Hence the ' I have no idea how this will work for non-cicular orbits`. I guess I should remove that part. That leaves the answer with almost no added value over the two others. (it only stresses the law within the same system). I might as well delete it. $\endgroup$ – Hennes Jul 22 '16 at 7:41

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