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I am world-building an adventure for the Alien RPG.

As a world approaches being tidally locked, and by this, I mean in the process of slowing down, but not stable.

In such a world that no longer completes a rotation on it's own axis (apart from that induced by being tidally locked rotation), is it possible to have a partial turn, oscillating back and forth on it's own axis?

I picture the characters view from their location as being able to watch the sun come up, rise to about midday, then appear to rotate back towards the direction of "sunrise", to become sunset. I hope I am describing this clearly enough.

Is there a way for this to take days instead of months in the process, but not sure what to call this type of action. I am not sure how fast I could have this oscillation occur, either. I picture an Earth-size rocky world as the planet in question.

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    $\begingroup$ Just make your planet fully tidally locked, but in a significantly elliptical orbit. The sun will hop up and down by many degrees each orbit/day $\endgroup$
    – PcMan
    Sep 28 at 19:48
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TL;DR: probably not.

In such a world that no longer completes a rotation on it's own axis (apart from that induced by being tidally locked rotation), is it possible to have a partial turn, oscillating back and forth on it's own axis?

Nope. This would involve slowing the planet's rotation to a stop, and speeding it back up again. There's a lot of energy stored up in a rotating planet, and trying to stop it promptly will involve some Really Interesting things happening, and those are the sorts of things that you want to be viewing from high orbit.

I picture the characters view from their location as being able to watch the sun come up, rise to about midday, then appear to rotate back towards the direction of "sunrise", to become sunset.

I believe it is possible to get this sort of effect over the length of a day on a tidally locked (or near tidally locked) world, due to the phenomenon of libration

There's a nice gif of this effect on wikipedia, showing libration of the moon as seen from Earth:

Lunar libration

You can see an apparent rocking back-and forth and "breathing" motion... the motion isn't the moon really rocking, but slight tilts and eccentricities in the orbit mean that during rotation slightly different parts of the surface are visible from Earth. The reverse is also true.

On the moon this would manifest on certain portions of the surface (around the terminator) as the Earth rising above the horizon, describing part of an analemma which might indeed involve an apparent backwards movement at some periods, from some positions.

Is there a way for this to take days instead of months in the process, but not sure what to call this type of action. I am not sure how fast I could have this oscillation occur, either. I picture an Earth-size rocky world as the planet in question.

Unfortunately the libration effects occur over an entire orbit, ie. over the whole year. On the moon they take about a month. On a gas giant moon the might occur over a few days... there are obviously other issues with gas giant moons, and by the time your peeps visit them they'll be long tidally locked. There are plenty of other questions and answers on here about exomoons so I won't duplicate them.

If you do go thus route, consider that it isn't necessarily stable for the long term as tidal effects will act to circularise the orbit which will reduce libration effects. This is important for gas giant moons... in order to induce a reasonable eccentricity for interesting libration effects you'll need to arrange an orbital resonance which has its own problems as Io's exciting geology will show.

If you had a smaller, cooler primary you might be able to find just the right combination of habitable zone and tidal-locking timescales for your needs. Suitable large outer planets might be able to give the orbit of your world the eccentricity it needs, but there's a risk of ending up with a spin-orbit resonance of the sort that Mercury has... the day length is 2/3rds of the year length, and the situation is reasonably stable. I'm not sure what libration would look like there.

There's a good chance you're out of luck, but the situation isn't so implausible that you couldn't handwave it in.

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  • $\begingroup$ Thank you very much for the information! That is a great explanation. $\endgroup$
    – RixRam
    Sep 27 at 18:44
  • $\begingroup$ So, for this story I'm dealing with, if a colony were built on the day/night terminator, the terminator would cycle a little back and forth, based on what appears to be fluctuations induced by the axial tilt and orbital plane? Is that a ballpark correct statement? I wonder is a small solar system, such as Trappist-1, with orbits measured in scant days, might work. Or am I still out to lunch? Oh, and no TLDR needed. I do the R. ;-) $\endgroup$
    – RixRam
    Sep 27 at 19:00
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    $\begingroup$ @RixRam I've had complaints before about length answers, so I do try to summarise with a TLDR ;-) I think you've got it, yes. I'm not sure if anyone has made a simulator suitable for showing the effects of libration so you can double check, but such a think might exist. $\endgroup$ Sep 27 at 20:32
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You either have two possibilities:

  1. the planet is approaching tidal lock from revolving around its axis in a shorter time than it takes to complete an orbit around its star. The tidal forces are slowing down the rotation of the planet around its axis.
  2. the planet is approaching tidal lock from revolving around its axis in a longer time than it takes to complete an orbit around its star. The tidal forces are accelerating the rotation of the planet around its axis.

You can't switch between 1 and 2, since the tidally induced momentum will become smaller and smaller the more you approach tidal locking, so you won't have an "overshoot".

A case similar to 2 happens with Mercury

Mercury rotates in a way that is unique in the Solar System. It is tidally locked with the Sun in a 3:2 spin–orbit resonance, meaning that relative to the fixed stars, it rotates on its axis exactly three times for every two revolutions it makes around the Sun. As seen from the Sun, in a frame of reference that rotates with the orbital motion, it appears to rotate only once every two Mercurian years. An observer on Mercury would therefore see only one day every two Mercurian years.

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    $\begingroup$ Worth noting that Mercury does this, in part, because of a frozen-in tidal bulge combined with the most eccentric planetary orbit in the Solar system. Each perihelion, one of the bulge poles points through the sun (and next time, the other), but they swap off so often (every 88 days) that there's little or no impetus to smooth out the bulges. $\endgroup$
    – Zeiss Ikon
    Sep 27 at 17:54
  • $\begingroup$ Thank you both. It seemed like a long-shot, but trying to look up the proper celestial mechanics is difficult when you cannot determine the right terms to use. I really appreciate your responses. $\endgroup$
    – RixRam
    Sep 27 at 18:49
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    $\begingroup$ Also the rotation of Mercury is sufficiently slow (55-day-long sidereal rotation, 176-day-long solar day), and its orbit sufficiently eccentric, that the Sun does appear to reverse direction for the 8 earth-days surrounding perihelion, Not quite long enough for the Sun to completely rise and then completely set and rise again, though. $\endgroup$
    – notovny
    Sep 28 at 20:46
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Doubtful. You say Earth-sized planet and the Earth weighs ~6 x 10^21 tons.

  • To get that sort of pendulum motion, the planet would need have a mass distribution sufficiently asymmetrical to act as a pendulum and the masses involved are so stupendous that the gravity of the pendulum-like planet would cause it to crush itself back into a rough sphere.
  • Even if the planet held its shape, the momentum involved for that much mass is equally stupendous and the force of gravity by the central star is only 0.0006 Gs at 1 AU. Even without doing the math, the oscillation cycle would be something on the order of hundreds of thousands of years (at least!) rather than months, let alone days. The sort of gravity necessary to cause oscillation even on a scale of centuries would probably simply tear the planet into an asteroid belt.
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  • $\begingroup$ Okay, that is a very certain answer, and greatly appreciated. On a slightly different note, this does make me think of using rings instead. Thank you very much! $\endgroup$
    – RixRam
    Sep 27 at 18:46
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Something similar to this happens on earth near the poles.

solstice in alaska

https://news.uaf.edu/time-stands-still-on-winter-solstice/

In temperate regions we are used to seeing the sun traverse the sky east to west. In the polar winter, the sun peeks up out of the east, rises, and then "turns around" and goes back down, still in the east.

If you want your characters to see a sun doing as you describe you need a planet with an axial tilt as ours has. Day length will of course depend on how fast your planet is rotating.

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