Mars has much more trojans in L5. Does it reflect a common pattern?

  • $\begingroup$ I think, likely, the answer is no. Earth has an L4 asteroid, none in L5. Uranus also has an L4, no L5 and Neptune has more L4 than L5 objects. en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/Earth_trojan Jupiter is the king of trojans and it appears to have similar amounts in it's L4 (Greek camp) and it's L5 (Trojan camp). en.wikipedia.org/wiki/List_of_Trojan_asteroids_%28Greek_camp%29 and en.wikipedia.org/wiki/… $\endgroup$
    – userLTK
    Jul 10, 2015 at 23:40
  • $\begingroup$ @userLTK what is about theoretical proof based on Newtonian mechanics and/or GR? Is there at least a subtle difference in likelihood? $\endgroup$
    – Anixx
    Jul 11, 2015 at 20:42
  • $\begingroup$ That's why I made a comment and not an answer. I don't know of any proof, though I suspect mathematically L4 and L5 are the same, though Jupiter's orbital resonance could affect Mars' L4 and L5 I think. That's possible, but that would be a coincidental and non permanent effect. $\endgroup$
    – userLTK
    Jul 12, 2015 at 3:05

1 Answer 1


There is a symmetry in the rotating gravitational field, which means that capture of an asteroid to the L4 is just as likely as to the L5 Lagrange point.

In the case of Mars the split is 1:6, and a simple binomial model suggests a probability of 0.0625 (the probability of a 1:6 or 0:7 split given the hypothesis that they are distributed randomly is 0.0625) This doesn't give a reason to suppose there is anything unusual happening, and as noted in the comments, there is no favouring of L5 when other planets are considered.

As noted by UserLTK: "Earth has an L4 asteroid, none in L5. Uranus also has an L4, no L5 and Neptune has more L4 than L5 objects."

The conclusion is that No, the greater number of L5 martian trojans is just a random effect.

  • $\begingroup$ See the comment above. Earth has 1 L4 and no L5, Neptune also has more L4 than L5 $\endgroup$
    – James K
    Mar 10, 2016 at 18:21
  • $\begingroup$ edited with the information from UserLTK's comment $\endgroup$
    – James K
    Mar 11, 2016 at 21:05

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