3
$\begingroup$

So I am building a tidally locked world, but I want the shadow zone to vary a bit. One way to achieve this is through axial tilt. A planet with orbital period = day period, so tidally locked and 0 degree axial tilt will have the same area illuminated at all times. But a planet with the same parameters and say, 22.5 degrees axial tilt will have the shadow zone move by 22.5 degrees from the poles in both directions every year.

If we want to introduce another type of movement we can play with the eccentricity of the orbit. Look up libration of the moon to learn more. My question is this: given the orbital parameters of any planet, how can I calculate the libration in the east/west direction?

EDIT: Sketch of my understaning

Starting at the bottom of the elliptical orbit, east and west are illuminated equally. A point on the globe just behind the terminator, so in the dark will experience a sunrise at some point between the bottom position and the right hand position. Then a sunset between right and top. Then this point will stay in the shade until the planet is at the bottom position again, after which a new orbit beings and it will experience another sunrise. So one sunrise and one sunset per orbit. What am I missing?

$\endgroup$
  • 2
    $\begingroup$ "Tidally locked" means that the planet has no own rotation, or (more precisely) rotational axis is at 0 degrees and rotational speed equals its orbital speed. $\endgroup$ – Alexander May 14 '18 at 20:14
  • $\begingroup$ Hi and welcome to Stack Overflow! Questions like this can often be answered by playing around in Universe Sandbox until you get something you like. $\endgroup$ – Schwern May 14 '18 at 20:14
  • $\begingroup$ Welcome to worldbuilding.SE! When you have a moment, please take our tour and visit our help center to learn more about us. I'll leave the details to our participants who know much more about orbital mechanics than I do, but I'm pretty sure that "tidally locked" means little to no axial tilt. Earth's tidally locked moon, for example, only has an axial tilt of 1.5°. $\endgroup$ – JBH May 14 '18 at 20:22
  • $\begingroup$ Maybe I didn't explain my planet correctly. I meant a planet with a period of 1 year and a rotational period (day length) of 1 day also, but on a slight axial tilt. Such a world would still be classified as a tidally locked one I think. $\endgroup$ – nAUTILUS May 14 '18 at 20:25
  • $\begingroup$ While I don't have any idea, I usually don't care about such things when building worlds, but I've noticed that you wrote "some precision". This is a somewhat strange statement to make since it's very vaguely saying that you want something to be precise. Can you perhaps quantify that one? Anyhow, good job on your first question, I think it's the best one I've seen this month (which could be due to me not having a clue about the topic though) $\endgroup$ – Raditz_35 May 14 '18 at 20:27
1
$\begingroup$

What you're looking for is called the "Equation of Time". That's the term of art for how much a primary, i.e. the Sun for Earth, is east or west of overhead at local average noon. That in turn is how much (in degrees longitude) your shadow zone would have moved east or west.

To separately calculate the eccentricity component (there's a nice animation of that component here), there are several approaches. The Wikipedia derivation in the prior link seems particularly opaque, there's a more pedagogical one here, but the bottom line is that small values of eccentricity $\epsilon$ goes straight to the motion: The peak of the eccentricity component will be $\pm 2 \epsilon \times 360^{\circ} / 2\pi $. For Earth, with an eccentricity of 0.0167, that's $\pm 1.9^{\circ} $. (Larger values of eccentricity, over say about 0.1, require more complicated analysis)

Please note that, if you have both eccentricity and tilt, the effects can either add coherently or partially (but not totally) cancel, depending on how the axis of tilt is related to the major axis of the orbit. They can't completely cancel because one is once per orbit and one is twice, so they have different frequencies:

enter image description here

(from Wikipedia, but note that the labels on the top two plots are swapped)

The primary moves north and south once per orbit (per "year"), so that's one "day": It rises and sets once.

The analemma (curve in bottom-right of figure) shows where the Sun appears in the sky at nominal local noon on the equator. For your tide-locked moon, that's where it'll always be, so the "at noon" part isn't needed as it is for the rotating Earth. You can see it move north and south due to the tilt, and east and west due to the eccentricity. When the sun is to the east, the east side will be a little more lit; when it's to the west, ditto for the west.

Note that the primary moves east-and-west twice per orbit (per "year"), so that's two "days" for the east and west pole-dwellers: They'll see it rise, reach a peak, set, and just start to rise again twice per orbit, i.e. per year. The east pole sees primary-rise just after both aphelion and perihelion; for the west pole, those are when two primary-sets happen, but both places will see two rises and two sets per orbit.

There's a (Java-based) calculator and plotter for all this here.

$\endgroup$
  • $\begingroup$ Thank you so much for your answer! Is it really as simple as that equation at the bottom? Say with an eccentricity of 0.25 my terminator (line where the shadow starts) will vary by +- 29 degrees? If so then that's fantastic!!! Many thanks :) $\endgroup$ – nAUTILUS May 15 '18 at 8:42
  • $\begingroup$ For very high eccentricity, like your 0.25, you need to consider higher-order terms to get the exact value. See Eqn 44 in the Muller link above. The motion gets complicated, with various speed-ups and slow-downs over a year, but the next terms are -1.25/sqrt(2) e^2-5/6e^3 But there's a larger issue: Above an eccentricity of 0.14 (IIRC, but it's something like that) tide locking will work to a harmonic of the orbital period, not the period itself. Mercury is an example of that. Bottom line: For very large eccentricity, it's complicated and needs to be calculated out in detail. $\endgroup$ – Bob Jacobsen May 15 '18 at 14:42
  • $\begingroup$ I settled on an eccentricity of 0.0823 giving me 10 degrees of "wobble". with it being exactly out of phase with my axial tilt of about 22 degrees I can a nice, wide, illuminated area. I also choose an orbital period of just 140 days, so the dark areas only spend 70 days in the dark on average. Again, thank you so much for your help, it really helped me with my world. Do I need to edit my original post in any way now that my question has been answered so expertly? $\endgroup$ – nAUTILUS May 15 '18 at 15:57
  • 1
    $\begingroup$ @nAUTILUS I don’t really see a need to edit your question, but simplifying it for later readers can sometimes be good. Also, note that the tilt gives 1 day per year to the north/south zones, while the eccentricity term gives two days/year to the east/west zones (and clock makers in the NE, SE, SW and NW zones tend to have nervous ticks). That might be useful for your story lines... $\endgroup$ – Bob Jacobsen May 15 '18 at 16:22
  • $\begingroup$ I don't understand what you mean by "gives 1 day per year"? $\endgroup$ – nAUTILUS May 16 '18 at 9:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.