# Can a planet's axial tilt turn to always be facing its star?

In my particular scenario, There is an habitable exoplanet in which the axial tilt turns to always face the system's star, and the planet is not in tidal lock, this allows for a certain range of scenarios which i would like to explore, such as the perpetual days and nights in the poles, and the much shorter (or longer) daytime periods in any point above or below the equator, and how any native lifeforms would have adapted to survive such a peculiar system.

In short, i'd like to know if it is possible for the tilt to always be the same relative to the stars center of mass (instead of the position of background stars which are relatively unmoving) and if the planet would need or can absolutely not have any other orbiting bodies? would the planet have to orbit the star, or would the star have to orbit the planet? what would be the planet's required mass? what would be the stars required mass? would for that particular system the star and the planet have to be orbiting each other in a pseudo-binary system?

For any other clarification, this particular scenario is a rough mash of light sci-fi and magic fantasy, so if theoretically this planet could exist with some changes to the basic rules of the universe without completely tearing the universe apart, please do tell me! even if this planet is completely impossible, any information that i could obtain will come useful in any future worldbuilding scenarios.

• “axial tilt turns to always face” <- if it were not for the pictures, I would have no clue what that means. Perhaps you could say some variation of (A) periodicity of precession is same as its orbit with star; (B) axis of rotation maintains consistent angle with the axis of orbit. – can-ned_food Sep 11 '17 at 13:27
• "Not actual scale" I think that needs to be on any model of the solar system that is meaningful to the human brain. Space is big. Really big. You think it's a long way down the road to the chemists, but that's just peanuts to space. Listen.... – MikeTheLiar Sep 11 '17 at 14:56

Probably not, because the planet would either have to be too close to its star, or its moon would have to be too close to the planet.

Let's take the Earth as an example.

The simple, first-order calculation of precession -- ignoring for the moment the effect of the Moon -- shows that the precession rate depends on the mass of the star and on the orbital semi-major axis $a$ to the minus third power ($a^{-3}$). The precession period the Earth would have if only the Sun mattered would be about 40,000 years. The only feasible way to increase the precession rate is to reduce $a$ by moving the Earth closer to the Sun. ("But won't this make the Earth hotter?" I hear you ask. Yes, it will.) Reducing the Earth's orbit by a factor of 34 should do it ($34^{3} \approx 39,000$) -- except that now the orbit is so small that the Earth's year is only 2.2 days long! So we have to move the Earth closer still....

There is, in principle, a magic point where precession rate = orbital period, because precession rate goes up with $a^{-3}$, while orbital period goes down with $a^{-1.5}$. Using the simple formula, they equalize if you reduce the Earth's orbit to about 1/1200 of its original size. However, at this point, the Earth's "year" is only about 12 and a half minutes. Also, the Earth is deep inside the Sun -- only about 20% of the way out from the center to the Sun's surface -- so the simple approximations we've been using are no longer valid. (And it's very, very hot.)

In reality, the Moon also contributes to precession, and its contribution can be amplified by moving the Moon closer to the Earth, which doesn't have the problem of simultaneously reducing the length of the Earth's year. So if you move the Moon about 34 times closer, you could in principle boost the precession rate due to the Moon to about once/year. This would put the Moon just barely outside its Roche lobe, so it probably wouldn't break up (unless its interior was molten, and then maybe it would). Of course, having something like the Moon that close would mean enormous tides, which would brake the Earth's rotation (and cause the Moon to move outward); a more stable situation would have the Earth's rotation tidally locked to the Moon's orbit, so the Moon would never move in the sky as seen from the Earth, and the day would be about 3 and half hours long. But having the Earth rotate faster would reduce the precession rate (which is proportional to $1/\omega$, where $\omega$ is the Earth's rotation rate), so you'd have to move the Moon even closer.... In the end, I don't think you can avoid having the Moon so close that it be torn apart by tidal forces. Which would be a shame, unless you like rings.

• Wouldn't the total tidal forces at opposition exceed the moon's Roche limit if it were that close to the Earth? – Ash Sep 10 '17 at 18:11
• @Ash -- do you mean with the additional tidal force from the Sun? Possibly, yeah, although it would be pretty small compared to the Earth's. – Peter Erwin Sep 10 '17 at 18:14
• Yeah sorry opposition with the Earth on one side and the Sun on the other, if your at the lope limit you wouldn't need much extra force to tip over. – Ash Sep 10 '17 at 18:27
• Could the exoplanet for some reason have an unusually heavy equatorial bulge? Maybe it would have to be very small for that (more asteroid-sized) – Hagen von Eitzen Sep 10 '17 at 21:29
• Hagen -- I looked at that; the problem is that the shape of the planet has a very weak effect; moving it closer to the Sun (or moving the moon closer to the Earth) has a much stronger effect. – Peter Erwin Sep 11 '17 at 15:58

Sure it can but you'd quickly (geologically speaking) evaporate the planet, you'd need about 1.4*10^27Nm of torque annually to get the axis of rotation to precess 360 degrees per year which is what you're asking for, that's going to produce 1.4*10^27j of waste heat. To put that in perspective one megaton is 4.18*10^15j so you get the equivalent of 291,666,666,666 million tons of TNT worth of waste heat annually (it's almost exactly 0.1% of the sun's total constant output). Earth weighs 5.97*10^24kg so that's 234.5j per kilogram per annum for the entire planet from core to exosphere, that will heat the whole planet by roughly 0.25K a year every year, you'll melt the whole planet inside of 6800 years and evaporate it en masse after only 11800 years. That assumes you start from a completely solid lump of pure silica which the Earth is not, it will take far less time to vapourise an Earthlike world.

• Hah, good point... – Peter Erwin Sep 10 '17 at 18:15
• Although I could see earth begin to radiate heat away faster as it warmed up, reaching a point where the torque being applied + incoming radiation (mostly solar) would be equal to the radiation rate. However, I have no idea where this "magic point" would be, although my guess would put the earth far into the evaporated stage. – Gryphon Sep 10 '17 at 18:20
• @Gryphon You'd have to look at the "blackbody' behavior of Silica to get a good idea of the possible heat radiation to see if you could cool the planet fast enough to do the job. Geochemically Earth is for all intents and purposes Silica with "stuff" in it, the stuff is mostly metals and lowers the specific heat and melting/boiling point of the whole. – Ash Sep 10 '17 at 18:24
• Yeah, that's a whole lot of math for me to do when I've got some time (if I remember...) – Gryphon Sep 10 '17 at 18:25
• @Gryphon Yeah sorry I'm not touching it, I had these numbers lying around from a similar question or they wouldn't be here. – Ash Sep 10 '17 at 18:29

It seems to be clear from the other answers that if the planet's axial tilt were non-zero, then it would be pretty difficult to achieve what you're looking for.

I'd like to know if it is possible for the tilt to always be the same relative to the stars center of mass

Well you could make the planet have an axial tilt of zero degrees. In that case, the "tilt" would always be the same with respect to the star. Of course, this wouldn't create some of the possibilities that you wanted to explore:

such as the perpetual days and nights in the poles

However, the absence of axial tilt (and consequently, seasons) would result in other peculiar phenomena that you might be interested in. What if an Earth-like planet had no axial tilt?

• not if you want to live there, +1 – Mazura Sep 10 '17 at 23:30

No, it cannot.

It would mean to change in the course of a single year the rotational momentum of the whole planet having it spin around a whole turn.

To say it in another way: it would be to have a complete Axial precession in one year (instead of the standard 25,765 years).

Axial precession is due to "flattened" Earth shape due to rotation and would be faster for planets spinning more quickly; it wouldn't be possible to have it as short as a single year because planet would disintegrate first.

• So, I believe your claim that the planet would disintegrate first but if you have numbers that you can use to justify your claim, I'd be very curious to see them. :) – Green Sep 10 '17 at 15:12
• Standard axial precession for the Earth is about 26,000 years, not 385,000. – Peter Erwin Sep 10 '17 at 15:12
• @PeterErwin: you are absolutely right. I pasted a number, but apparently I had copied the wrong one. I edited to correct. – ZioByte Sep 10 '17 at 15:19
• Except that "a year" can be a rather long period of time, depending on distance from the sun. – Peter Sep 10 '17 at 19:55
• @Peter: unfortunately precession is caused by interaction with star gravitational field (trying to force "flattened" planet to rotate on the same plane of the orbit), so if year is longer then also force causing precession would be less. You wouldn't gain anything. – ZioByte Sep 10 '17 at 21:50

Yes you can. While precession of earth's axial tilt is 26,000 years, the sun's SOI is nearly a light year in radius. Even using only the moon's component of precession it would be easily possible to place an earth-sized planet and moon far enough out to pull this off.

Of course you end up with a dark ice-ball of a world this way.

Are you willing to have a circumbinary system where the orbital period of the planet is similar to that of Saturn (30 years)? Then, observably, yes.

Kepler-413b is in a binary star system, in circumbinary orbit. The tilt of its axis changes by up to 30 degrees per 11 years. It receives between 1.7 and 3.8-times the stellar flux as we do on Earth, so it's a bit hot. If you could move it 1.4-times further out, the flux would be 0.85 - 1.9-times here, so maybe survivable substantially away from the equator. The axial tilt, orbital precession, and orbital tilt all seem to be changing in a currently unpredicted way. This may be due to additional gravitational actors in the stellar neighbourhood (planets or stars). But, if useful, allows the rapid axial precession to be

• fixed if you do not have additional actors and have synchronized star-star and planet-star orbits,
• disrupted by small random amounts due to additional actors : planets or small/far stars, and/or
• disrupted by large amounts by one big actor that is on a highly eccentric orbit around the CoM.

Edit (details from references):

• There isn't actually any observational evidence on axial precession of Kepler-413b though, is there? The 2014 paper does (very clearly) constrain the orbital precession. It then goes on to make some arguments about Cassini states, secular resonance etc., but all this seems quite speculative to me. There has also been an erratum, which again doesn't give a clear statement but does say that it's also plausible for the axial precession to have a period of 300 years. – leftaroundabout Sep 11 '17 at 15:49
• @leftaroundabout : I find that Kostov et al. are very consistent in claiming an 11 year spin axis precession, even in the errata. – Eric Towers Sep 12 '17 at 4:11
• Still this spin precession is completely derived from theoretical calculations based on the observed orbit alone, if I haven't misunderstood the paper. Nothing wrong with that; my point is mainly that you can't argue “observationally, yes” – these calculations could as well have been carried out before Kepler-413b was ever discovered. All observation it provides is that there exist real planets in an orbit whose properties can, theoretically, support such funny effects. That's a great finding, but whilst we haven't actually measured the spin precession I would be wary. – leftaroundabout Sep 12 '17 at 8:02
• @leftaroundabout : I have very little concern about estimating the effect of binary tides on precession and libration of a planet. Their range of planet masses and temperatures are consistent with a homogeneous planet, so the usual simplifying assumptions work. The rest is straightforward. It would now be vastly more surprising if the planet somehow avoids having resonant obliquity and resonant libration. – Eric Towers Sep 12 '17 at 15:14
• It would be vastly more interesting if the planet behaved differently. I would not find it very surprising, intuitively, if the tidal forces would cause so violent internal dynamics that any rigid-body derived model would stop working. That planet isn't just a lump of rock after all. But we don't have a very clear understanding of the internal dynamics of gas planets, certainly not under such extreme conditions, so I wouldn't make any conjecture as to what would happen. – leftaroundabout Sep 12 '17 at 15:52