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Here is my own translation of the Polish version of the Drake equation article in Wikipedia:

(...) They must provide the right amount of heat. The smaller the star, the less heat it gives off and the closer the planet must be until it reaches a limit. After this limit, it becomes so bound to the star that only one part of it will be illuminated and the other part will remain dark and cold, which prevents the emergence of life.

If I am getting the above correctly, if only half of a planet's surface is constantly illuminated and the other half remains in darkness forever, life cannot emerge and support itself on the entire planet, not even on the illuminated half.

What factors or reasons cause the above? Until reading the quoted passage, I was more than sure that in the presented scenario life is impossible on the dark side only, whereas it can emerge and support itself on the illuminated part.

Does this mean that (even ruling out all other factors) life could never exist on the Moon, if only because of the fact that half of it is always dark?

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    $\begingroup$ "The dark side of the moon" is the name for the half that always points away from earth. It gets sunlight, but we never saw that part of the moon until we started launching space cameras. $\endgroup$ Oct 5, 2023 at 18:50
  • $\begingroup$ Even @Arjuns link acknowledges that ribbon worlds can have a goldilocks zone (2nd to last paragraph). The bigger issue is emergence. Without tides what stirs the primordial soup? $\endgroup$ Oct 5, 2023 at 19:17
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    $\begingroup$ @MooingDuck: Isn't it the far side of the moon? $\endgroup$ Oct 5, 2023 at 20:14
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    $\begingroup$ "The far side of the moon" is the best name. "Dark side of the moon" is a nickname en.wikipedia.org/wiki/Dark_Side_of_the_Moon_(disambiguation) $\endgroup$ Oct 5, 2023 at 20:20
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    $\begingroup$ @MooingDuck That uses ‘dark’ in the sense of ‘unknown, hidden, mysterious’ — not ‘lacking light’.  That sense probably isn't used much these days (hence the confusion); think of e.g. a ‘dark horse’ (one unknown to gamblers), ‘darkest Africa’ (the parts that Europeans knew least about), or keeping someone ‘in the dark’.  — Of course, with all the probes, missions, and satellites we've sent, that side of the moon is now almost as well-known and -understood as the near side, so as you say, ‘far side’ is a much better term now. $\endgroup$
    – gidds
    Oct 6, 2023 at 11:23

5 Answers 5

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The phenomenon you are referring to is called "tidal locking", it occurs when the orbital period or one revolution of the planet is roughly equal to the time it takes to rotate in on its axis. Abiding to the sci-fi term coined by Isaac Asimov "Ribbon Planets", these worlds are common in orbits around type M dwarfs or red dwarf stars on the HR diagram.

The extreme conditions on these Ribbon worlds, make life impossible in the day and night hemispheres (except the terminator line where the temperatures are moderate for life to emerge). So, ribbon worlds, are tidally locked due to their close proximity to their host stars, because the tidal interactions, the action-reaction pairs of tidal friction forming between the star and the planet, it aligns itself into an equilibrium state like the Moon and becomes tidally locked to the major body.

Let us consider a moment where it is in fact, capable of hosting biosignatures like water, considering temperature for water, they would be close to the red dwarf because the habitable zone is close to the star as it has low stellar irradiance.

Continuing, one side of the planet would continue receiving extreme sunlight, and at least if a planet is rotating it would be capable of cooling down when it faces away from the Sun, so depending on its albedo, it would be absorbing the heat from the sun, and this would make its surface temperature extremely high and it would make all the water evaporate.

Now, since the only source of albedo has evaporated it will get hotter and also because water vapour is a greenhouse gas it will cause a runaway greenhouse effect and make the planet extremely hot, and no water means no life as we know it. This makes a temperature gradient, one is extremely hot one is extremely cold, making life uncommon on such a planet with uneven extreme temperatures. One related example would be Mercury, even though it has a non-synchronized rotation in 3:2 resonance.

Another major concern is the high amount of solar flares originating from the host stars, if the planet's magnetosphere can not protect it, it is fried

So, it is likely with our current understanding of life that life would be uncommon on ribbon planets. But there is a temperate region in the planets, this is the terminator zone, the boundary between day and night, where odds are life are better but that is an active topic of research whether it would be stable enough for life on it.

Regarding this, you may find this post interesting

Thank you, hope it helps you!

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  • $\begingroup$ Great answer. But it assumes that the part of the planet always facing the star receives extreme amount of energy and ends up with the very high surface temperatures. Why such assumption? Isn't it possible to have an Earth-like scenario only without the rotation? A planet that would have everlasting day on one side and everlasting night on the other side, but with medium temperatures similar to Earth's medium day and night temperatures? $\endgroup$
    – trejder
    Oct 5, 2023 at 13:34
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    $\begingroup$ But surely that just moves the goldilocks zone. Surely there is a radius from the star at which a day side tidally locked planet would be a comfortable temperature. Its just further from the star than for a normal planet $\endgroup$ Oct 5, 2023 at 22:22
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    $\begingroup$ You need to demonstrate, or argue, that in order to be tidally locked, the surface temperature on the illuminated side must be too hot, for any kind of central object. I don't think that is the case. $\endgroup$
    – ProfRob
    Oct 6, 2023 at 6:07
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    $\begingroup$ @RichardTingle if only the day side were a comfortable temperature, then all water that evaporates there would eventually diffuse over to the night side where it is so cold that the vapour resublimates to form a giant ice sheet. The process ultimately dries out the day side, even if the temperatures are perfectly harmless there. This can only be avoided if the atmosphere is so thick that it can equilibrate the temperature difference between day- and night side. $\endgroup$ Oct 6, 2023 at 15:40
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    $\begingroup$ @trejder "Earth's medium day and night temperatures" are only what they are because of the cycle. Night would lose its stored heat if the heat weren't replenished each day. Day would gain tons of heat if it didn't have a chance to lose the accumulated heat overnight. To give an idea of this effect, consider how much hotter noon is than dawn. Today where I live there was a difference of about 5°F over 5 hours, and I'm sure we've all seen more dramatic rises than that. The temperature began to fall mid-afternoon as the sun declined. But what if the sun never set and only kept pouring energy in? $\endgroup$ Oct 6, 2023 at 20:48
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Pierrehumbert (2010) modelled the possible climates of Gliese 581g. Although that planet has since been shown to probably not exist, the model of a planet with similar characteristics should still apply. According to his calculations, life may be possible on a tidally locked (=half-illuminated) planet.

The paper considers various models, including dry, rocky surfaces with no air or a pure N₂ atmosphere. You need only a moderate thick atmosphere to reduce the day-night temperature contrast:

With U = 5 m s⁻¹ , which is typical of Earth’s surface winds and somewhat below the typical winds in Joshi et al. (1997), the substellar point temperature, nightside temperature, and surface atmospheric temperature are 316 K, 173 K, 217 K for ps = 104 Pa; 269 K, 221 K, 233 K for ps = 105 Pa; and 240 K, 233 K, 235 K for p s = 106 Pa. Thus, as little as 1 bar of N2 is enough to substantially reduce the dayside/nightside temperature contrast.

That's for a dry planet. He then adds water to the equation, which introduces a greenhouse effect. He calculates that if the atmosphere is thick enough to equalize surface temperature, a water world would freeze over at a temperature of 192 K.

The paper goes on (emphasis mine):

However, if the atmosphere is so thin that the substellar point does not lose heat through dynamical heat transport, the substellar point temperature at this albedo becomes 270 K, and a slight reduction in albedo from dust or other contaminants might even be able to cause melting. Even if the ice crust fails to melt through, surface temperatures approaching freezing lead to very thin ice, supposing a sufficient supply of heat flux from the interior of the planet (Warren et al. 2002; Pollard & Kasting 2005). This should be considered a habitable state, insofar as similar solutions have been proposed for the survival of eukaryotic life during a Snowball event on Earth.

By adding CO₂ to the equation, we can raise the temperature at the sub-stellar point enough to get liquid surface water. Therefore, a tidally locked planet at a suitable distance from the star might be covered in ice everywhere except for its sub-satellite point:

Eyeball Earth
Source: Pierrehumbert (2010)

That's habitable.

The critical question then becomes: can we obtain and retain sufficient CO₂? The paper briefly refers to sea-floor weathering as a factor that is (or was in 2010) not well-enough understood, but that could potentially remove enough CO₂ to limit habitability (except in a sub-ice ocean such as speculated for Europa). But generally, the conclusion is that a tidally locked planet, "half illuminated", does not rule out a climate that allows for life.


Pierrehumbert, R.T., 2010. A palette of climates for Gliese 581g. The Astrophysical Journal Letters, 726(1), p.L8., doi:10.1088/2041-8205/726/1/L8. Available online (accessed 2023-10-06).

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  • $\begingroup$ How does it avoid excess heavy element loss (that is, elements heavier than Oxygen)? One of the known requirements is active vulcanism to recycle elements that sink to the bottom of the oceans. $\endgroup$
    – Joshua
    Oct 7, 2023 at 20:30
  • $\begingroup$ @Joshua That doesn't seem to be covered by the paper. $\endgroup$
    – gerrit
    Oct 9, 2023 at 6:47
  • $\begingroup$ @Joshua To answer your question: I have no clue. I'm aware of the opposite problem (solar wind stripping away light elements from the top of the atmosphere), but know essentially nothing about problems with heavy elements becoming out of reach without vulcanism. Can heavy elements be sustained without vulcanism? Can life exist without easy access to heavy elements? Is either related to be half-illuminated, or are those orthogonal problems? Many questions, which I don't think are all conclusively answered. $\endgroup$
    – gerrit
    Oct 9, 2023 at 7:02
  • $\begingroup$ Your particular rig-up here looks unstable in the presence of active vulcanism. It is known that sulphur and prosperous are mandatory (yes there's that paper involving arsenic; it's less available than phosphorous here) I'm more familiar with the eyeball planet which loses too much hydrogen. $\endgroup$
    – Joshua
    Oct 9, 2023 at 13:53
  • $\begingroup$ @Joshua Why would active vulcanism be a problem for the rig-up sketched here? Some volcanic islands might pop up under the ice layer or through the ice layer (nunataks) and have occasional eruptions. Large eruptions can temporarily disrupt the climate, but I don't know why it would make the situation illustrated in the image unstable. $\endgroup$
    – gerrit
    Oct 9, 2023 at 14:01
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I wish to add a couple of comments to the good answer provided by Arjun.

A popular fallacy has it that close-in planets are always tidally locked in a 1:1 spin-orbit resonance (SOR), i.e., are always showing the same face to their host stars -- like the Moon is showing the same face to the Earth.

In reality, the situation is more complex. If an inner planet is perturbed by sufficiently massive outer planets, its eccentricity may be pumped up considerably. In that case, the planet is likely to get entrapped into a higher SOR (like our own Mercury). What happens after that will depend on the proximity to the star, ratio of masses, and the rheology of the planet. On some occasions, the higher SOR may become the end-state.

[ On other occasions, the capture may be temporary, because the tidal dissipation rate in higher SORs is higher by orders of magnitude than in the synchronous spin mode. In that case, a thermal runaway may change the rheology and, therefore, the tidal response. The planet will then leave the higher spin state, and will continue despinning towards synchronism. This happened to the TRAPPIST-1 planets $b$, $d$, and $e$, which were captured in the 3:2 or higher SORs during the initial spin-down, but got overheated and slipped further down to synchronism -- or possibly to pseudosynchronism, in the case of $b$. ]

Another mechanism capable of keeping even a close planet in spin state nonsynchronous with the star is synchronisation by a massive moon. Modeling demonstrates that such events are relatively rare -- but not vanishingly rare.

All in all, we should be prepared to encounter, from time to time, close-in planets outside synchronism.

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    $\begingroup$ Why would such a planet be half-illuminated (which is the question)? $\endgroup$
    – ProfRob
    Oct 6, 2023 at 6:05
  • $\begingroup$ If a planet is synchronised, then it of course will be half-illuminated. I just wanted to emphasise that not all close planets are synchronised. That's all what I wanted to add. $\endgroup$ Oct 6, 2023 at 15:06
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    $\begingroup$ The 3:2 case with high eccentricity is also quite interesting. $\endgroup$ Oct 6, 2023 at 15:51
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As several people have indirectly noted, the premise of this question is controversial. There are plenty of studies* suggesting that an Earth-like planet tidally locked to its star can be habitable as long as its atmospheric and/or ocean circulation are sufficient to transport heat from the day side to the night side and keep the temperatures reasonable on both sides.

If this circulation was not present or was insufficient, however, then the temperatures on the side of the planet permanently facing the star would quickly climb above the boiling point of water, while those on the side facing away from the star would drop far below freezing. This would quickly cause all water on the planet's surface to accumulate as ice on the cold side, making both sides inhospitable to water-based lifeforms.

Furthermore, if the temperature on the cold side got low enough (and without any heat circulation it would), the same thing could also happen to nitrogen and oxygen and any other gases necessary for an Earth-like atmosphere. They, too, would freeze on the cold side and fall to the surface, leaving the planet with no atmosphere at all to speak of.

(For comparison, there are craters on the poles of the Moon and even on Mercury that are permanently shielded from sunlight, just like the whole night side of a tidally locked planet would be. In the absence of an atmosphere to circulate heat, these craters are some of the coldest places in the whole solar system.)


To some extent these are both potentially runaway processes: the less liquid water there is on the surface, and the thinner the atmosphere gets, the weaker the heat transport via air and water circulation will become. Thus, while a planet with a sufficiently thick atmosphere and deep oceans may be able to stay habitable, if the night side ever gets cold enough the whole system may cross a critical point from one stable state to another and the whole hydro- and atmosphere may freeze out on (geologically) very short timescales.

You could compare this with the runaway greenhouse effect that is believed to have occurred on Venus at some point in its early history, transforming it from a vaguely (proto-)Earth-like state to the super-hot hellscape it is today. The end state of the "runaway freeze-out", while quite different, is no less inhospitable to life.


*) See for example:

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    $\begingroup$ If I remember correctly, most of the computer models using an earth-like planet showed that the necessary wind movement to keep heat transport going established itself quite quickly, forming large cyclonic systems but they weren't nearly as hurricane-like as predicted, it only took a constant moderate wind to keep plenty of heat moving around to warm the dark side and keep the sun-side cool (except for the subsolar point which was pretty horrible). $\endgroup$ Oct 6, 2023 at 13:42
  • $\begingroup$ Why would the star-facing side necessarily exceed the boiling point of water? I wouldn't expect a tidally locked planet orbiting the Sun at 100AU to overheat even on its sun-facing side, so by continuity it seems like some radius ought to give an equilibrium temperature that's conducive to life. $\endgroup$ Oct 6, 2023 at 18:40
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    $\begingroup$ @RavenclawPrefect indeed it wouldn't necessarily exceed the boiling point, but the argument holds true regardless. Water evaporates also way below the boiling point, and that's enough for all of it to eventually accumulate on the night side. $\endgroup$ Oct 6, 2023 at 19:28
  • $\begingroup$ @RavenclawPrefect It is impossible for a potentially habitable planet to 1) orbit a star at a distance off 100 AU, 2) be in the star's habitable zone, 3) be tidally locked, and 4) have even the least bit advanced lifeforms. 1) and 2) combined are incompatible with 3) and 4). if a planet orbiting a star at a distance of 100 AU is in the star's habitable zone, the star be 10,000 times as Luminous as the Sun. A star that luminous would would be about type B1V and should last on the main sequence for only tens of millions of years before becoming a red giant. Continued. $\endgroup$ Oct 6, 2023 at 21:49
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    $\begingroup$ @RavenclawPrefect Continued That is even more the case with stars many times as luminous as the Sun. Their habitable zones will be so far from those stars that their planets can not possibly be tidally locked. So there will never be a tidally locked planet orbiting a star at 100 AU. $\endgroup$ Oct 6, 2023 at 22:02
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The other answers have done a marvelous job of explaining a variety of reasons life would have a hard time on a tidally locked planet, but there's a more controversial reason as well... Heat engines (which photosynthesizing life are) require a temperature difference to operate. If the temperature doesn't change over the course of a day, they can't extract any useful work (energy for life) from the incoming solar energy which means that a tidally locked, or even very slow day planet would have a very difficult time getting life-like things started. While this seems fairly straightforward from a physics perspective, it's interrogating the origins of life itself, and there have been very few papers on the topic - thus controversial.

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  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
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    Oct 10, 2023 at 16:16

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