What I want to know is if it is possible for two planets to be close enough together to tidally lock each other (as in both planets are tidally locked), but in such a manner that they don't orbit each other. More specifically, could this happen in such a manner that the one further from their star would be perpetually in the shadow of the other?

I know, weirdly specific question. I'd be interested in the reasoning/science behind the answer, but a simple yes/no answer would actually suffice. (in case anyone's wondering, backstory for two a WH40k Wolf successor armies).

  • $\begingroup$ You might want to elaborate a little more. For example, you could say you want two planets in the habitable zone of your star, with both of them being tidally locked to each other and rotation at a speed where one always remains in the shadow of the other. Or something like that. $\endgroup$ Feb 26, 2016 at 5:12
  • $\begingroup$ The only logical but immensely unlikely way to achieve such a situation would be to build a super strong and 100% rigid tower which has zero chance of bending or swaying and connect the two planets. I am not sure how you would prevent the planets from floating through space with the same grace as a squash that was thrown like a football. $\endgroup$
    – MonkeyZeus
    Feb 26, 2016 at 14:03
  • $\begingroup$ Granted this is the 40k universe we're talking about here, so just make them (inactive) Necron tomb worlds that have some super-tech hand-wavy reason to stay locked to each other and no one will question it any more than the other bizarre things that happen in 40k. $\endgroup$
    – thanby
    Feb 26, 2016 at 15:20
  • $\begingroup$ That doesn't make sence. Of one orbits the star, and the other always casts a shadow on it, then they must also rotate around each other. Draw a sketch at 3 month intervals and see. $\endgroup$
    – JDługosz
    Feb 27, 2016 at 1:12
  • $\begingroup$ Although this is a duplicate, it does have some nice answers. It would be a shame to throw them away. $\endgroup$
    – k-l
    Feb 27, 2016 at 3:50

6 Answers 6


It's not possible.

Ye Be Warned
Really bad MSPaint skills were involved in the making of this answer. Many artists turned over in their graves. Also, the real physics are a bit more complex, but I'm not a real astronomer so this will have to do for a first-order approximation.

What is Tidal Locking?
Tidal locking is explicitly a function of orbiting another body.

Earth/Moon before tides.

The tides from one body (say, our moon), cause the near side of the other body (Earth, in this case) to be slightly closer than the rest of the planet. Because that mass is closer, it has more gravity than the rest of the planet.

Earth/Moon after tides.

But if the planet is rotating, the bulge moves forward a bit, so it's not directly between the center of the Earth and moon.

Earth/Moon after tides and rotation.

This means the force on the east side of the Earth (orange line) is slightly stronger than the force on the west side of the Earth (purple line), resulting in a small, but measurable, net torque. This torque means the Earth is gradually slowing down, and will continue to slow down until one side of the Earth is always facing our moon (which is already tidally locked because it's a lot less massive, so it didn't take as long to slow down).

Of note, this means a slowly-turning planet with a fast-orbiting moon will actually speed up until tidal locking occurs. Also, not shown, as Earth slows down, the distance between Earth and our moon increases. If the Earth were speeding up, the distance would decrease over time.

Tidal Locking by Sharing an Orbit
So for two planets to be tidally locked, but not orbiting each other, they have to be tidally locked to something else. The local star is a good candidate. You could have something like this:

Large planet orbiting star, with three smaller planets orbiting at its Langrangian points.

The blue, green, and purple planets are orbiting 60° from each other, and they could be tidally locked to the star (so one face of the planet is always looking at the star), which means they'd also be "tidally locked" to each other in the sense that each planet would always see the same face of the other planets.

Note that, to the best of my knowledge, you need a more massive planet orbiting halfway between two of the smaller planets for this to work; though you don't need all three planets -- the grey one plus any of the other three would still orbit. Addendum: it seems the L3 point (purple) is unstable so you don't want to use it anyways, but L4 and L5 (blue and green) are both stable and don't require a large primary (grey). So you could ostensibly have just the grey and blue or green planets of about the same size. Still, it doesn't solve the problem of shading the second planet.

To Your Idea
Ok, so we can sort of get tidal locking without the planets orbiting each other, but it's not what you wanted. You're looking for this:

Two Planets Orbiting Star at Different Distances

You can do that, but there's a problem. The orbital period is directly related to it's distance. In order for a planet to be in orbit, the centrifugal force from inertia has to exactly offset the centripetal force from gravity. There's less gravity farther from the star, so the more distant planet has to orbit more slowly or it will just fly off into space. Additionally, the length of the orbit is proportional to it's distance ($circumference=2\pi\times radius$). So it's going farther, and it's doing it more slowly, which means the time taken has to be longer.

Geometrically, there's just no way for the outer planet and the inner planet to orbit the star at different distances and orbit in the same amount of time.

So let's say you put the blue planet in a proper orbit, then magically slow the green planet down to the same speed. Now, gravity overcomes centrifugal force, causing the planet to fall, and you end up with a highly elliptical orbit (if it's slow enough, the "orbit" will involve crashing into the star). At the highest point in the orbit, it's traveling the same speed as the other planet, but as soon as it gets closer to the star, it speeds up. The extra speed, combined with the even shorter orbital distance, means it's still orbiting the star faster than the outer planet.

Two planets orbiting star. Outer circular, inner elliptical.

A similar effect happens if you speed up the outer planet. Either it ends up in a highly elliptical orbit, or, if it's fast enough, it flies straight out of the star system, nevermore to be seen. Regardless, it still orbits more slowly than the inner planet, if at all.

Of note, there are a variety of weird orbital characteristics, such as the outer planet orbiting exactly three times for every twice the inner planet orbits, or their "day" period being some simple ratio of each other's. But none of these involve tidal locking, or one planet always being in the other planet's shadow.

The L2 Point
From comments and other answers, it looks like I need to also address the L2 point. In addition to the above configuration (the grey primary and the blue/green (L4/L5) and purple (L3) secondaries), there are two more Lagrangian points, L1 and L2. L1 is between the sun and the primary, while L2 is behind the primary, opposite the sun.

Small planet sitting at primary's L2 point.

In this case, the gravity from the nearby planet adds to the gravity from the far away star, meaning the small planet has to go faster to balance out the extra gravity. Right at the L2 point, the extra speed is exactly enough so the outer planet stays right behind the inner planet.

Just what we wanted, right? Well, no. Sorry.

First, I assumed that "twin planets" (from the title) meant the two planets needed to be similar size. In this case, the L2 point is right out, because it requires the outer planet to be much smaller than the inner planet.

Second, for Earth, the outer planet is too far away, and isn't actually shaded by the inner planet, so it doesn't meet the requirements. However, this space.SE answer shows that it is possible, with the correct geometry, to solve this (in our solar system, everything from Mars to Pluto has a perpetually-shaded L2 point; Mercury to Earth do not). So you're ok here as long as you do the math.

Third, most problematically, is the L2 point is extremely unstable. A little nudge in any direction, the the outer planet falls out of the L2 point. From this NASA article, it takes about 30 days to get to Earth's L2 point by coasting. This means it takes about 30 days to fall out of the same point, back to the starting altitude (and much less time to fall out of the primary's shadow).

You get a bit more time if you're sitting in a halo or Lissajous orbit, but you still need to make corrections about once a month. With a different-sized primary, you might get a little more time, but we're still talking maybe a few years of stability. Also, both of those orbits move you (and possibly keep you) completely out of the primary's shadow, so they don't really meet the requirements.

There's just no way to keep a planet in the L2 point for any reasonable length of time without constant, planet-scale thrusters, at which point you might as well just call it magic. My understanding is the WH universe has plenty of techno magic that could accomplish this, but it's not really feasible under normal circumstances. And I'm really hard-pressed to come up with a reason to bother; it would be far cheaper and simpler to just build your colonies on a planet farther from the star (or just accelerate the planet farther from the star) than to keep one planet in another planet's L2 point for any significant amount of time.

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    $\begingroup$ Upvote for your MSPaint skills. I'd llike to add that with such changes in the distance separating the green planet from its star, it would be unsuitable for life. Temperature changes would get too extreme... $\endgroup$
    – Nico
    Feb 26, 2016 at 8:32
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    $\begingroup$ You could have a planet on the L2 lagrange points. Not sure if that is close enough that the smaller planet would always be in the shadow of the larger though. Also the L2 point is not stable so the orbit would devolve fairly quickly. $\endgroup$
    – Taemyr
    Feb 26, 2016 at 8:47
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    $\begingroup$ +1 For understanding that centrifugal force is not always wrong. So many people have been brainwashed into believing that "centrifugal" is equivalent to "wrong" by repeating it 6.4x10^4 times in high school. All they do is replace one misconception (youtube.com/watch?v=UTWk4pu9m8I) with another misconception ("centrifugal force is not real, get it off your free body diagram! I am not listening to you, IT IS NOT A FORCE, you are wrong!"). Also, there is the trend of blindly crossing out "centrifugal" and instaid writing "centripetal" ("Centripetal force flings mud off tires.") $\endgroup$
    – k-l
    Feb 26, 2016 at 11:57
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    $\begingroup$ Your last image makes me strangely hungry for breakfast... can you throw some bacon in there too? $\endgroup$
    – sǝɯɐſ
    Feb 26, 2016 at 17:09
  • 2
    $\begingroup$ @KiranLinsuain: Obligatory XKCD $\endgroup$ Feb 26, 2016 at 18:38

Yes, but the orbit will drift over time.

This may be possible if and only if there are no other (major) planets in your solar system. They won't be tidally locked, but the smaller one might be mostly in the shade. Consider a Halo orbit or Lissajous orbit around the L₂ point of a sufficiently low-density planet. The smaller planet may be in the shade most or even all of the time, if the configuration is right.

Lagrange points
(Source: Wikimedia commons)

It is possible to orbit L₂, in an orbit known as a Halo orbit. In our solar system, those orbits are unstable, because of disturbances by other bodies such as the Moon and Jupiter. But if you envision a solar system where no other bodies of significance exists, such a configuration could temporarily exist.

Imagine such a world. At the larger planet, gravitation is too large and climate is too hot for life to exist, but by being much smaller and shady most of the time, the smaller planet has a reasonable climate, such that in the period of less than 50 million years, intelligent life emerges. Life goes on, until at some point scientists on the smaller planet discover that their orbit is very slowly diverging. It's not a problem for the next several thousand years, but the periods in the scorching Sun are getting slightly longer every century. Models predict this will only get worse, perhaps due to a passing star less than a light year away. Within the next 20,000 years, life will become unbearably hot. Can science come up with a solution? Can we generate enough thrust to keep our planet close enough to L₂ to manage our climate? Can science save the world? Stay tuned for this exciting new novel Life at L₂!

See also those questions on Space Exploration SE:

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    $\begingroup$ Don't forget that even if the smaller planet is more livable, being perpetually in the shade would make it impossible for anything resembling plants to grow, and would likely cool it down to the point of freezing. We have the benefit of rotating fast enough to keep both sides of our planet reasonably warm, that wouldn't be the case here. The same goes for the dark side of the larger planet. $\endgroup$
    – thanby
    Feb 26, 2016 at 13:59
  • $\begingroup$ @thanby Depends. If it's in the full shade 80% of the time, partial shade 10% of the time, and full sunlight 10% of the time, that may be enough to heat the atmosphere and make it liveable. And/or perhaps the large sister planet is so hot that it alone radiates enough heat to keep the smaller planet liveable. Life would be quite different from how it is on Earth, but life adapts. $\endgroup$
    – gerrit
    Feb 26, 2016 at 14:14
  • $\begingroup$ Hmm... a T dwarf as the larger "planet"? $\endgroup$
    – DevSolar
    Feb 26, 2016 at 14:55
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    $\begingroup$ For the system to possess "stable" Lagrange points, the masses of the body must be something like this: $M_{sun} > 25 \cdot M_{planet1}$ and $M_{planet1} > 11 \cdot M_{planet2}$. Minimum mass for a planet to retain water for geologic ages would make $M_{planet 2} = 0.4 \cdot M_{Earth}$. This makes $M_{Planet1} = 4.4 \cdot M_{Earth}$. I guess that could make life interesting. $\endgroup$
    – Jim2B
    Feb 26, 2016 at 14:58
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    $\begingroup$ @Jim2B Interesting. I didn't make any quantitative calculation. How did you reach those figures? Assume where you write $=$ you rather mean $\ge$? $\endgroup$
    – gerrit
    Feb 26, 2016 at 15:33

Yes, but only for a few moments or hours before they crash into each other.

In the absence of a mutual orbit, there is no force to prevent their mutual gravitational attraction from pulling them into each other, and this will happen very quickly.

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    $\begingroup$ as long as you don't install planet level anti tractor beams at each of them ;) $\endgroup$ Feb 26, 2016 at 6:00
  • $\begingroup$ That is not accurate. We are talking about a three-body problem, not a two-body one. In the L₂ point, gravitation between the two smaller problems is offset by the gravitation to the larger body. $\endgroup$
    – gerrit
    Feb 26, 2016 at 10:53
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    $\begingroup$ @gerrit tidal locking requires distance small enough for L points to be irrelevant. Or, if masses are big enough to tidal-lock against such distances, we are talking about crashing with the star itself. $\endgroup$
    – Mołot
    Feb 26, 2016 at 11:39
  • $\begingroup$ There's no such thing as tidal locking without orbiting. You can get the orbiting part through L₂ and forget about the tidal locking part. $\endgroup$
    – gerrit
    Feb 26, 2016 at 11:42
  • $\begingroup$ @gerrit I just want to clarify that you can definitely have tidal locking without orbiting. Two bodies that exert gravitational pull on each other will cause tides, and therefore tidal friction, and therefore will end up in tidal lock. It doesn't require that they are orbiting each other. It just so happens that the only way for them to stay at a stable distance from each other for a long time is to orbit each other or for them to both orbit a nearby bigger object. See MichaelS's answer for a more in depth explanation $\endgroup$
    – Kevin
    Feb 26, 2016 at 22:06

This sort of configuration may be possible if the smaller planet is located at the L1 or L2 Lagrange point of the larger. I doubt this would work out in real life (and haven't done the math to determine how close or far apart they'd need to be if it did), but, fortunately, Warhammer 40k is not a scientifically rigorous setting, so you can probably get away with simply invoking Lagrange points along with a bit of handwavium to stabilize that configuration.

Also, technically, the planets would be tide-locked to their shared sun, not to each other, but the practical effect would be the same as what you're looking for.

  • $\begingroup$ This is the only accurate answer so far. What remains to be calculated is whether we can create a stable pair of orbits in which the smaller planet is in the perpetual shade of the larger planet. $\endgroup$
    – gerrit
    Feb 26, 2016 at 10:55

If the two planets are close to the same size, it would be possible to give them each a relative velocity such that neither is technically orbiting the other, but rather both are orbiting a point in space between the two. Such an orbit could easily be stable as long as there aren't too many other large bodies passing by, and the planets would tend to become tidally locked to each other over time. To my knowledge, we haven't seen any planets arranged like this yet, but there are binary star systems with such a configuration.

With a bit of juggling of the math, and the mass of the star and the planets you could probably find a point where the orbital period of the two planets around their common point matched the orbital period around the sun, leading to one being in perpetual (or near perpetual) shade. Mental approximations of the math, however, make me think that you'd likely end up with some combination of the planets being too light or the star too heavy for such planets to support terrestrial conditions. (I think you'd end up being 1/8th the mass of Earth per planet to do this with a sun-type star at 1AU and the second planet roughly as far away as our moon.) So the previously-suggested halo orbit is probably the more believable (albeit less stable) option.

If terrestrial conditions aren't required, then you can just look up the basic orbital mechanics equations on Wikipedia. First calculate the orbital period of your desired planetary set, and then see how far from your desired stellar mass you'd have to put it to make the periods match. (You can adjust the surface gravity of the planets more-or-less independently of their mass by adjusting their density, just keep in mind that density is based on what they're made of, and a solid ball of uranium wouldn't be conducive to supporting life.)

Do keep in mind that, if the outer planet really is always completely in the shade, it's going to get really cold. Like, nitrogen snow kind of cold. Calculating how much of the outer planet would actually be shaded can be done from the relative diameters of the sun and the inner planet and their distance for the shape of the inner planet's shadow, and the distance between the two planets for what diameter that shadow will be when it gets to the other planet. You probably want at least some sunshine, or else your people will have to live around the terminator of the inner planet instead.


There is a non-orbital solution as well. We've not got the technology to build it but 40k is considerably more advanced in many ways.

Have the larger planet orbiting at it's natural orbit.

Have the smaller planet orbiting in the shadow of the larger one further away from the larger planet.

As already discussed there is a problem with this, they will drift apart.

Embed a tether made of unobtanium through the cores of both planets and through space between them. That tether is strong enough to stop them from drifting apart. They now orbit as a joint unit. You would need to arrange the dynamics so that they hold the same position relative to each other (good thing you wanted tidally locked) and keep a modest tension on the tether.

  • $\begingroup$ Wouldn't the outer planet lag behind unless it also had some sort of planet-sized jet engine to propel it? $\endgroup$
    – thanby
    Feb 26, 2016 at 14:01
  • $\begingroup$ You would need to position the tether appropriately to compensate for each effects. I'm having trouble visualizing just how the forces would work in my head (orbital dynamics is tricky) but basically you need to get the two planets into a position where naturally they would drift apart and then use the tether to stop them. $\endgroup$
    – Tim B
    Feb 26, 2016 at 14:43

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