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I've seen many questions about tides and tidal variations on a wide variety of interesting single star, multi star, single planet, binary planet, single moon, multiple moon, and even ring system combinations and variations.

I've also seen many questions about co-orbits including trojan, Lagrangian, horseshoe, and more exotic orbits.

But I was not able to find one about the tidal effects of horseshoe orbits, so here it is: What would tides be like on an Earthlike planet with a horseshoe orbital partner body?

For simplicity, let's just assume that it's actually Earth, and that the object in the horseshoe orbit with Earth is the Moon, with no other changes to them other than the new orbital configuration. Also, assume that the Moon's closest approach in this new orbit is the same as it's closest approach in it's current, actual, real life orbit.

EDIT An explanation of how often the Moon would approach Earth, to cause these tides to take effect, isn't necessary but would be interesting if you feel so inclined to include it. For this question, the minimum information is just: a description of how the tide coming in and going out for each approach of the Moon in this new configuration would be effectively different from how current tides come in and go out for an average tide cycle in real life. Things like how long does it take to come in, how long does it stay at high tide, how long to go back out to low tide, is high tide in this new configuration higher, lower or the same height as a real tide, etc.

EDIT 2 I'd guess that spring and neap tides would not be a thing, as I understand those to be when the moon is in line with, or opposite (respectively), the sun to either combine or counteract (respectively) each others' tidal effects, and this orbit would never cause them to line up. But I could be mistaken in my assumption.

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  • $\begingroup$ IIRC, I think a question is only allowed 5 tags. The 5 you have now look reasonable but I found some others that might be more specific. I'd keep [science-based] and [moons] for sure. But consider swapping out some for [tides] or [orbital-mechanics]. $\endgroup$ – Cyn Jan 11 at 21:00
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    $\begingroup$ @Cyn Noted, and thanks for the heads up $\endgroup$ – Dalila Jan 11 at 21:10
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Tidal effects are based on how big something is and how far away it is. The Sun is pretty far away, but really big and thus has an effect on the tides. The Moon is not very big (on a planetary scale) but is pretty close and has an effect. Venus is bigger than the Moon, but much further away, and consequently has very little effect on the tides. This change would put the Moon further away than Venus for the majority of it's orbit, thus we can infer that for most of it's orbit it will have very little effect on the Earth's tides. When it does come closer in it's horseshoe path it would have less effect on the Earth's tides than currently, unless it is coming closer than it currently orbits.

Tidal timing is dominated by the rotation of the Earth. You can see this in the ~12 hour timing of the tides (half rotation). This would not change, in fact without the Moon in it's current place the tides would become much more regular.

This leads into how often would it come closer. Something in a horseshoe orbit with Earth is going to be going slightly faster or slower than the Earth around the Sun and will take a long time to move in a full horseshoe with respect to Earth.

A good example is found on wikipedia, with nice labels for times, which demonstrates an actual horseshoe orbit of a small asteroid with respect to Earth.

https://en.wikipedia.org/wiki/File:Animation_of_(419624)_2010_SO16_orbit.gif

From the animation you can clearly see the horseshoe with respect to Earth, and you can see that it takes several hundred years to complete a full "horseshoe".

So the Earth's tides would be much more regular, except for approximately every hundred years when the Moon approaches the end of the horseshoe and you would have tidal effects similar, but smaller than current tides for a few years until it moves further away back along the horseshoe.

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  • $\begingroup$ You mentioned "When it does come close...it would have less effect on the Earth's tides than currently, unless it is coming closer than it currently orbits." To which I'd ask "Why" if it's equal distance, as specified in my question why would it still have less effect (other than spring/neap tides I already mentioned)? To which you seem to answer "Tidal timing is dominated by the rotation of the Earth", but the Earth still rotates in this scenario, so again, "Why" would Earth "have tidal effects similar, but smaller than current tides" instead of equal? $\endgroup$ – Dalila Jan 11 at 22:28
  • $\begingroup$ If at closest approach it is as near as the moon is currently it would have the same effect on the tides, but the moon's effect at these close approaches would never line up with the sun's effect as it currently does making the overall tide less. $\endgroup$ – Josh King Jan 11 at 22:38
  • $\begingroup$ Oh, I see. Yes, that's the neap and spring tide thing I was talking about. Thanks for the clarification. $\endgroup$ – Dalila 2 days ago

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