Assume that we have a earth like planet,orbiting a sun like star. On which have all the right ingredient:atmosphere,liquid water,molten metal core.....etc for the existence of the life as we known it.

And assume orbiting said planet is a moon , with which the planet is reciprocally tidally locked (the same side of one planet always facing the other,for both of them ).

We can't have the planet rotating too slow or one side would be baked and the other frozen, so its rotation period must be similar to earth(plus couple of days max). And we can't have a moon traveling faster at the distance such as earth's now is, not without it escape the planet's gravity well , so our moon need to be much closer to the planet.

My question is : what would the atmospheric condition on such a planet be like? And especially at the sub-lunar point?

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    $\begingroup$ Isn't the moon already tidally locked to the earth? $\endgroup$
    – Oxy
    Mar 15, 2017 at 10:59
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    $\begingroup$ where is the difference between your question and the Moon-Earth system? Unless you make it clear, I think this question is "unclear what you are asking". $\endgroup$
    – L.Dutch
    Mar 15, 2017 at 11:32
  • $\begingroup$ Hello and welcome to WorldBuilding.SE! As others have suggested you should try to add a few details to differentiate your question from the moon-earth system. Otherwise this question might get put on hold for some time until you provide the necessary details. This is a standard precaution if questions don't quite fit the sites guidelines to ensure that they are edited and edits won't invalidate existing answers. So please edit your question. If you got questions about the site please take the tour and visit the help center or have a look at the Meta Site. Have fun! $\endgroup$
    – Secespitus
    Mar 15, 2017 at 11:42
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    $\begingroup$ Do you mean geostationary as well as tidally locked moon? As in, the moon is always in the same place over a particular landmass/waterbody no matter the day or night? If not, I don't see a difference to the earth-moon system $\endgroup$ Mar 15, 2017 at 11:44
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    $\begingroup$ @Azuaron, if you look at the edit history you can see that the question had ''geostationary'' added to it, and then was edited to say ''the planet and the moon are tidally locked to each other''. So it's still different to the Earth-Moon system and the question idea is still focussed but I think it's now a little unclear due to language difficulties. I recommend that w.n. edits again and puts geostationary back in. They can literally just add "i.e. geostationary moon" or something similar after their sentence, that way they won't mess up their sentence structure. $\endgroup$ Mar 15, 2017 at 20:21

3 Answers 3


The Moon and Earth being tidally locked won't do much (given the Moon is already tidally locked) - the only difference if they're both locked to each other is you'd basically almost completely lose the tides. I say "almost" because the Sun does have a small effect on the tides. But they'd definitely be much less strong than they are now.

From the point of climate, the Sun / Earth's tilt / Earth's rotation are the major drivers of climate, not the Moon, so the Earth and Moon being tidally locked again wouldn't have much effect on that. In fact, the Moon and Earth will eventually be tidally locked to each other. They're both very, very slowly changing each other's rotation, as well as the Moon changing its distance.

TL;DR: The beaches would be more boring. That's pretty much it.


The constraint that the day not be too long is fairly stringent one.

A 24 hour orbit -- geosynchronous for today's earth puts the moon at 40,000 kilometers. This is also very close to roche's limit -- The moon is under serious stress.

A 100 hour day would make for very cold nights, and very hot afternoons. This would put the moon far enough away that it was in no danger of splitting into chunks and developing rings.

I think your orbit/day has to be between 36 and 100 hours. Note that this requires an orbital height of 55,000 to 100,000 km. (Calculator here: http://www.calctool.org/CALC/phys/astronomy/earth_orbit )

The larger temperature swings would make violent weather events -- thunderstorms, hurricanes, tornadoes more common. On our present planet this would make some parts of the world uninhabitable, without technical aid. (

A slower rotation speed means smaller coriolis forces. Not sure what this would do the Hadley circulation. I thinkit would mean fewer, but larger Hadley cells. I think this would make for larger weather system that moved more slowly.

Note that ALL of these changes are due to the longer day. The moon being closer doesn't make any difference.

The tidal forces of the moon would distort the equipotential surface (what we call 'level'. But all that means is that your planet would be more egg shaped. It would be hard to detect until you got decent at surveying and figure out that the curvature was smaller going to/from the moon, than it was around the other way.

The side facing the moon would have a natural navigation system. The angle of the pole star gives you latitude, the angle of the moon gives you longitude.

If you want to stir things up, give in a geosynchronous period, but in an eccentric orbit. From the surface of the planet it would appear to drift backward slowly while doing the outer half of it's orbit, then drift forward rapidly during the inner half.

Don't get to carried away. Tides change with the 3rd power of distance. Flex the crust too much, and you get more volcanism. Possibly more tectonic plates, smaller continents.


Alright, so I get that there's been some confusion as to exactly what orbital scenario it is that you're describing, but I'm going to assume you're talking about a situation in which the moon in question is orbiting its home planet in a geosynchronous fashion, like an artificial satellite.

Now, SpaceMouse's answer in pretty good (i.e., tides would be a lot milder), but it's my understanding that one of the most notable effects of a geosynchronous orbit for an Earth-like moon would be that the tides on the particular side of the planet it's orbiting above would be drastically higher than those of the side it's never above. Basically, the increase in gravitational attraction would "pull" the waters of the oceans towards the moon, thus resulting in perpetually higher tides on one side of the planet. But since oceans can't actually be higher on one side of an Earth-like planet than the other (hence the global standard we call "sea-level"), I imagine this would actually result in increased ocean activity on the non-lunar side of the planet, so as to compensate for the higher tides on the side facing the moon. Atmospherically, you might have a real problem with the production of phytoplankton, which are the Earth's primary producer of oxygen (around 70%). This would stem from the possible lull in activity as described by SpaceMouse, and could render your world inhabitable (at least for humans).

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    $\begingroup$ Oceans can be perpetually higher than mean sea level in one area. That would just mean that they're perpetually lower in another area. Even here on Earth, sea level at the equator is perpetually higher than sea level at the equator due to the planet's rotation. This is why it's a mean sea level. See psmsl.org/train_and_info/faqs for more details. $\endgroup$ Apr 2, 2017 at 6:56
  • $\begingroup$ Ah, good point. I imagine the gravitational pull would still have the same basic effects, but I hadn't factored in the planetary rotation. $\endgroup$ Apr 2, 2017 at 19:58

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