Would it be possible for a planet with an elliptical orbit to be tidally locked to a star, only for a certain part of it's orbit?

If this is possible, then:

  • could the tidally locked side vary from orbit to orbit?
  • would it be possible to have the same side facing the star each orbit?
  • would it necessarily have to be the same side facing the star each orbit?

(I assume that if the answer to this question about a planet orbiting a star, would be equally applicable to a moon orbiting a planet?)

  • $\begingroup$ Please take a look at this planet: worlddreambank.org/L/LIB.HTM $\endgroup$
    – Galaxy
    Jul 21, 2023 at 16:03
  • $\begingroup$ The hard-science tag is not meant to be used to "force" a science-based answer. The science-based tag is more than suitable for that. The hard-science tag is meant to be used when you unequivocally want proof that the answer is correct. It's ruthless. Answers that don't meet the tag's requirements are subject to deletion. Is that really what you were looking for? $\endgroup$
    – JBH
    Jul 22, 2023 at 23:58
  • $\begingroup$ Well, I'm looking for a hard science answer. I don't know where you get the 'force' part from? $\endgroup$
    – Jacco
    Jul 23, 2023 at 7:57

1 Answer 1


There are a lot of misconceptions going on with tidal locking. When really it is just a configuration of orbiting bodies in the lowest energy state possible. True tidal locking only occurs when you have a perfectly spherical moon orbiting a perfectly spherical planet in a perfectly circular orbit in an otherwise completely empty universe. In any other situation what you are looking at is a spin-orbital resonance.

enter image description here

In this little Newtonian simulation i just cooked up you can see the idea. I hope at least. The orbiting object is in a resonant spin. For ever orbit it completes around the central object, it spins arounds its own axis twice.

Now, in reality these resonance frequencies will probably be a bit more complex than 2:1. Also, obviously this orbit in the sim is a tiny bit dramatic but it illustrates the idea very visually.

Such an orbit is tidally locked in the sense that each time the moon is at its lowest point, you see the same face. Really, at any time in the orbit you always see the same face for that point. For instance, at the highest point in this setup you always see the backface. Mercury is a pretty good real life example, as it is in a 3:2 resonance. And what is Mercury, if not a quaismoon of the sun ? Indeed, even our lovely moon is in a resonance, its just 1:1. But you can see the same effects, the Moons face does wobble around a bit. But there is nothing saying you couldn't have a more interesting setup.

Now, lets tackle the other questions one by one.

"could the tidally locked side vary from orbit to orbit?"

Depends on what you mean. The pattern cant change. You might have some weird resonance where it takes 10 orbits for the moon to return to its original orientation. But there is always a repeating pattern involved.

For instance, you might have a setup where it takes 7 orbits of the Moon for the exact same face to show up in the sky. Which might be an continent way for a Week as a unit of time ?

"would it be possible to have the same side facing the star each orbit?"

Not really, orbits drift relative to the star. They do so in their own resonance. We can see that with Mercury again, its called precession;

enter image description here

The exact same thing happens to the Moons orbit, see Lunar Precession, and it would happen in your scenario. I would show a sim but accurate precession can only be modeled in General Relativity, and while i do have a program for that, i dont have the nerve xD

enter image description here

The main take away is really that over time the side which sees the sun in any resonance other than 1:1 will shift a bit.

"would it necessarily have to be the same side facing the star each orbit?"

Isnt that the same question as the one before ?

  • 1
    $\begingroup$ Wow, stunning visuals : ) Does your answer take the mass of the orbiting body (and thus its inertia) into account? (I guess so, but just making sure) $\endgroup$
    – Jacco
    Jul 23, 2023 at 8:13
  • $\begingroup$ @Jacco , naturally :D Though it is still Newtonian in nature, so there is some inherent inaccuracy. $\endgroup$
    – ErikHall
    Jul 23, 2023 at 10:34
  • $\begingroup$ What does the last image visualise? $\endgroup$
    – Jacco
    Jul 24, 2023 at 8:58

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