# Navigation on a tidally locked planet

I have a tidally locked planet on a circular orbit, with an axial tilt of 22.5° That means one side is always in the sun, the other one in the shade. The subsolar point moves along a north-south line during each rotation around the sun, 22.5° each way. The night side can see the stars.

How would someone on this planet determine their position? Both on the night and the day side? What tools would be necessary? Sextant, some sort of table of celestial positions, a time-keeping device? How precise would it have to be?

• are you sure that tidal locking and sensible axial tilt can occur together?
– L.Dutch
Commented Jun 13, 2018 at 19:05
• Axial tilt means the planet is not fully tidally locked. On geological timescales, it is "nearly" tidally locked, and will achieve tidal lock in the near future. On human time scales, it's far enough in the future that no one will care for the lifetimes of a few civilizations. Commented Jun 13, 2018 at 19:20
• Obviously I missed this day of astronomy class. How does the axial-tilt have to do with one side of the planet being in constant sunlight? Does it not rotate? is there something special about 22.5 that makes the spin & rotation of the planet equal-out?? Commented Jun 14, 2018 at 6:03
• Tidal locking. Got it. Commented Jun 14, 2018 at 6:10
• You can't have a meaningful axial tilt and tidal lock (as opposed to orbital resonance). It's impossible. Commented Jun 14, 2018 at 6:34

If the person is wandering across the broiling wastes of the sunward side, then a gyrocompass (since there's probably no magnetic pole anymore) is needed to determine a north-south line, and a simple stick-and-shadow sundial can be used with that north-south line to gauge both latitude and longitude. No need for a reliable clock, since the sun is always in a constant, predictable position.

If the person is wandering across the frozen wastes of the night side, then the traveler will need the gyrocompass and a small ephemeris with key equatorial star positions to use in place of the sun, and a sextant to use in place of the stick. A reliable calendar (not clock) is needed here, since the star positions do not change significantly each hour.

If the person is wandering along the terminator, then they already know their longitude, and can use the sun-glow or a sextant-with-equatorial-star to calculate latitude.

A simple square-and-protractor to measure the angles would be handy, as most folks aren't very good at estimating angles by eye.

Of course, since the traveler is likely to have space travel (how else did they get there?), they could simply have a GPS satellite constellation in orbit, which makes the answer much easier.

• Thank you so much for your great answer! Could you elaborate a bit more on how the gyrocompass works and how you would go about calculating lat and long using a sundial? I'm not sure I understood that part completely. Commented Jun 13, 2018 at 20:57
• "since the traveler is likely to have space travel (how else did they get there?), they could simply have a GPS satellite constellation in orbit, which makes the answer much easier." I wish I could give you an extra +1 for that. Commented Jun 13, 2018 at 21:37
• This is a setting where the planet was colonized by humans long ago from an interstellar generation ship. shortly after colonization begins some calamity happens that destroy pretty much all human tech/irradiates the rest. Only very few small communities survive and in the following apocalyptic conditions lose all their knowledge. Then after many thousands of years they don't know where they're from and basically invented everything from the beginning. This is all only so that I can have humans in a realistic setting without having to reuse earth. Commented Jun 14, 2018 at 0:08
• @nAUTILUS Well, since this kind of planet is unlikely to have a magnetosphere, it won't have a naturally occurring atmosphere anymore either. The humans need to retain some knowledge (space suit maintenance, how to get oxygen from the rocks, hydroponics, etc) since your Eden will be quite airless. Commented Jun 14, 2018 at 2:01

If it's tidally locked to the sun, by definition there's no such thing as axial tilt. That can be disregarded.

The first thing is to not think of latitude and longitude in the way it's oriented on Earth. For simplicity, we'll assume the same numbering system as on Earth (360 degrees = full circle), and for timekeeping, we'll assume one orbit around the sun is 100 days (Earth standard).

The center of your navigational grid on the lit side is going to be the sun. Directly overhead marks the zero position, the hot pole. Instead of 90 degrees, it's going to be labelled 0. The terminator--ignoring atmospheric diffraction of light--is going to be 90. That's going to be your latitude. On the daylight side, easy to find; you simply use a sextant to measure the height of the sun over the horizon, and better than the Earth, you don't have to wait until midday to do it; you can do it at any time.

On the cold side, you rotate the grid. Now, your reference is a recognizable star or constellation closest to your planet's orbital axis, so it's the closest thing to unmoving you can find in the sky. Essentially, instead of finding a pole star for your planet's rotation as on Earth, you find it for your planet's revolution around the sun. Find one above and below the orbital plane, and you've got a way of determining stellar latitude by measuring the distance of one (or other features, as the sky is mapped) above the horizon.

Picture it this way: Imagine that on the Eurasian side of the Earth, latitude and longitude are exactly like they are now, but on the side with the Americas, it's turned sideways so that the "pole" is centered on the Galapagos islands. Now 90 degrees latitude, as centered on the Galapagos, would be the same great circle around the planet as 0/180 degrees longitude based on the Eurasian Grid. The Eurasian side is the dark side of the planet, the Galapagos side the light side.

Okay, so on the Eurasian night side, further measurement is relatively easy, once you've invented reasonably accurate timekeeping mechanism (ie, clocks). You've got astro-north, and astro-south. You know that stars near the astro-equator (which would be the plane of the planet's orbit) take a half-year (50 days) to go from horizon to horizon. At day 1, it rises above the horizon, is at its apex at day 25, and set on day 50.

If you have accurate timekeeping, you know that at a specified location at a given date a given star should be a certain number of degrees above the horizon. If you measure the height, you can measure the difference between where it is and where it should be on a given date as seen from your latitude, and that can be used to calculate the longitude.

Now, this only works if you have accurate timekeeping devices so you know your date/time compared to the reference point, but really, not a lot of difference from the problem of measuring longitude on Earth before the advent of accurate portable clocks.

On the day side, it's trickier. Solar latitude is trivial to determine, as mentioned. Solar longitude becomes harder without the stars. You're initial assumption (with the sun "bobbing" up and down, as seen from the planet) would make it easier, but that can't happen if the planet is tidally locked.

• Thank you very much! I guess my planet is nearly tidally locked then, as it still has an axial tilt. I think on the day side I would need sun azimuth and altitude as well as the current date and time to get my lat and long, right? Commented Jun 13, 2018 at 23:23
• No, it doesn't have axial tilt worth mentioning. If one face of the planet is constantly facing the sun, the only way for the sun to appear to "bob" in the sky is if the planet itself is rotating in a northerly direction, then stopping, reversing its rotation, stopping, reversing its rotation, and doing it over and over again. How is this happening? Commented Jun 14, 2018 at 22:49
• I guess it's not technically tidally locked. It has a day length of exactly one year and an axial tilt of 22.5 degrees. That means that the subsolar point moves along a North south line from the equator, back and forth. The analemma is a straight line. Commented Jun 15, 2018 at 12:57
• Having a rotational period of a year is the definition of tidally locked. Commented Jun 15, 2018 at 14:41
• Some commenters further up said that you have to have an axial tilt of 0 as well. Commented Jun 16, 2018 at 16:43

With a south pointing chariot

https://en.wikipedia.org/wiki/South-pointing_chariot

This Chinese invention is a chariot with a pointer linked to the wheels with a clever mechanism. It would point in the same direction no matter which way you turned it.

The Wikipedia article says that it was probably not used for navigation and was really just a curiosity. The Chinese also invented the magnetic compass for one thing. The main issue with it is the impossible amount of precision that would be needed to make it work over long distances. If you can manufacture it to a level of precision that is almost unheard of even today, run it on perfectly level ground, suffer no wear on the wheels whatsoever and if you happen to live on a world that is flat, it works quite well. Otherwise, not so good.

That said, it also mentions that if you are traveling on long, straight roads, you could disconnect the pointer, and when you get to the end of the road you can carefully line up the chariot with the road, connect the pointer, turn onto the next really straight road, line up your chariot again and disconnect the pointer. This greatly reduces mechanical error, although at some cost of handling error, also at the cost of making every road the Roman way.