Fixed Rotation?

You're quite welcome =)

A speculative guess. If we assume a radius of 1.638 R_e (guess, as I don't have the equation to approximate the radius of a terrestrial planet handy), we're given a density of 9.07 g cm^-2. Plugging it into the equation, I get a semi-major axis of 1.03238 AU, which again, GJ 581 d falls within.

At 1.346 R_e, the tidal lock distance is 0.923432 AU.
At 1.122 R_e, the tidal lock distance is 0.842988 AU.
At 0.897 R_e, the tidal lock distance is 0.754003 AU.

This continues in an asymptotic manor, so I tried to figure out what radius GJ 581 d would have to be to be at the limit of the tidal lock radius. It would require a density of 95,000 g cm^-2, which is somewhere around 0.077 R_e = 462 km.

I think it's fair to say that for any reasonable radius for Gliese 581 d, it's probably tidally locked.

Of course Gliese 581 d is on an eccentric orbit. Leads to pseudo-synchronisation not synchronisation. For e=0.38, pseudo-synchronous period is just over half the orbital period, at 34.6 days.

Sirius_Alpha wrote:Ah, it's quite alright. And I think the answer is "most certainly". I can't think of a reason why stars would behave differently. I would furthermore assume tidal locking to be more efficient for tight binary stars due to their higher mass and increased tidal effects.

This is my last question for this topic (I swears). I am interested in the Delta Trianguli system (a close orbiting pair of Solar stars). With their close proximity, they should be locked onto each other. But since they are stars, what does that necessarily mean? Will they (themselves) be locked, while their plastic surfaces will rotate?

Also, I wonder what this means for the sunspots? If they are all locked onto each other, I think the sunspots will always face outward, while it stays calm in the center. Does that sound right?

I would guess that they would be in synchronous rotation, but their upper surface would be free to move about. That's a good question though. I think some hot Jupiters are in a similar situation. Though they rotate synchronously, their upper cloud layers are wind driven, and blow around the planet.

The movement of Jupiter's cloud layers is not 100% directly determined by the rotation of the planet, as we see Jupiter's clouds rotating in opposite directions and such. Stars also have this differential rotation, for the sun, the poles rotate slower, around 30-ish days? And the equator rotates once every 25-ish days?

Sunspots would be affected by each star's magnetic fields as well as, perhaps, gravitational effects if the stars are really close. I would expect sunspot activity to be increased on the side of the stars that face each other.

I'm not really sure what you mean by your description about the sunspot locations.

And as always, don't be afraid to ask questions.

Sirius_Alpha wrote:I'm not really sure what you mean by your description about the sunspot locations.

I already know that any planets around Delta Trianguli will be hit with superflares, so i guessed outward. I just wanted to know will there be magnetic activity (i.e., sunspots) closer to the barycenter (facing each other), the opposite side of each star (facing outward), or will there be an equality of activity all over the stars?

Sirius_Alpha wrote:And as always, don't be afraid to ask questions.

I asked questions almost all the time years ago, and everybody just criticized me. So you can understand why I am being cautious. But since you insist:

Lazarus wrote:Of course Gliese 581 d is on an eccentric orbit. Leads to pseudo-synchronisation not synchronisation. For e=0.38, pseudo-synchronous period is just over half the orbital period, at 34.6 days.

First off, thank you for the page. I have been wondering what the Mercury rotation was all about. Secondly, I original thought of Gliese 581 d to have a "Mercurian rotation," but I never understood that it had to do with eccentricity. I always assumed it had to do with its proximity. Now that I think about it, I think planet d might have a pseudo-synchronisation with its star. But what is the defining point between an eccentric orbit and a circular one? Because I now think planet c might also have a rotation like this.

NuclearVacuum wrote:I already know that any planets around Delta Trianguli will be hit with superflares, so i guessed outward. I just wanted to know will there be magnetic activity (i.e., sunspots) closer to the barycenter (facing each other), the opposite side of each star (facing outward), or will there be an equality of activity all over the stars?

I would assume that if the stars are interacting, the sunspot activity would be closer to their barycenter. I don't know quite what would happen if they aren't interacting, as far as sunspot activity goes.

NuclearVacuum wrote:I asked questions almost all the time years ago, and everybody just criticized me. So you can understand why I am being cautious.

I'm afraid I don't remember that. If you've felt we've treated you poorly, I do apologize.

NuclearVacuum wrote:But what is the defining point between an eccentric orbit and a circular one?

The only thing I can think of is that an eccentric orbit would have an eccentricity greater than zero. An orbit with e=0.000001 would thus be eccentric. But such orbits are often called circular because they might as well be.

I'm not sure Mercury is in a pseudosynchronous rotation (yet?). Mercury rotates three times every two orbits. I'm unsure if this describes a pseudosynchronous state.
Here's about the time were I go do research .

Okay, I plugged Mercury's eccentricity into the equation on the Oklo post. I got 0.849511 * P_orbit. Mercury's rotation period is ~0.66 * P_orbit.

Is Mercury protected in this orbit / rotation ratio? Will tidal effects continue to slow its rotation period? And if so, will it slow it down to ~0.849511 * P_orbit? Or will it slow it down exactly to P_orbit?

NuclearVacuum, this paper is interesting, and pretains to the topic of close binaries and tidal evolution.
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1981A%26A....99..126H&data_type=PDF_HIGH&type=PRINTER&filetype=.pdf

In a detached close binary, tidal evolution will continually change the orbit and rotational system parameters. Ultimately either an equilibrium state will be reached asymptotically, or the two stars will spiral in towards each other at an increasing rate, leading to a collision. An equilibrium state is characterized by coplanarity (the equatorial planets of the two stars coincide with the orbital planes), circularity (of the orbit) and corotation (the rotation periods of the stars equal the revolution period). Such an equilibrium state is stable (unstable) if more (less) than 3/4 of the total angular momentum are in the form of orbital angular momentum.

Sirius_Alpha wrote:

NuclearVacuum wrote:I asked questions almost all the time years ago, and everybody just criticized me. So you can understand why I am being cautious.

I'm afraid I don't remember that. If you've felt we've treated you poorly, I do apologize.

More times, I asked questions that others would respond something like "That's elementary!" But I was criticized on my ideas (primarily my "planetary naming" ideals). But than again, I am a renegade.

Thank you for the apology, but you don't have to. You did not criticized me at all. In fact, you are the first one to actually encourage me to keep asking questions. Sorry to put in this sappy speech, but I had to get that off my chest.

Sirius_Alpha wrote:NuclearVacuum, this paper is interesting, and pretains to the topic of close binaries and tidal evolution.
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1981A%26A....99..126H&data_type=PDF_HIGH&type=PRINTER&filetype=.pdf

In a detached close binary, tidal evolution will continually change the orbit and rotational system parameters. Ultimately either an equilibrium state will be reached asymptotically, or the two stars will spiral in towards each other at an increasing rate, leading to a collision. An equilibrium state is characterized by coplanarity (the equatorial planets of the two stars coincide with the orbital planes), circularity (of the orbit) and corotation (the rotation periods of the stars equal the revolution period). Such an equilibrium state is stable (unstable) if more (less) than 3/4 of the total angular momentum are in the form of orbital angular momentum.

That is really funny! I was just thinking about this, now I can get the answer. Thank you so much!

Haha, aww =).

I'm glad I could help ^_^.

Mercury is complex because the orbital eccentricity varies significantly over long timescales: the current rotation is not so favourable in its current orbit but it is more likely the planet was captured into the resonance at a higher eccentricity. If I understand it correctly (and the tidal stuff is not my cup of tea!) a planet with a permanent asymmetry - which implies a solid planet can get captured into resonance, but fluid planets end up pseudosynchronised.

And oceans/atmospheres can make a difference too: apparently the massive Venusian atmosphere (which could also be regarded as a supercritical carbon dioxide/nitrogen ocean if you want) makes the 1:1 spin/orbit resonance unstable for Venus, favouring the current rotation conditions.

Fascinating.

How does the Venusian atmosphere affect Venus' rotation? I can sort of imagine it perhaps being a buffer to prevent some tidal circularization, but why does the atmosphere make the 1:1 spin/orbit resonance unstable?

This might be worth reading for answers to that (pdf), but I'm not particularly sure I understand it myself. Something to do wtih thermal tides I think (the atmosphere expanding and contracting as it heats up and cools down).

Sirius_Alpha wrote:

NuclearVacuum wrote:But what is the defining point between an eccentric orbit and a circular one?

The only thing I can think of is that an eccentric orbit would have an eccentricity greater than zero. An orbit with e=0.000001 would thus be eccentric. But such orbits are often called circular because they might as well be.

lol, mind if I rephrase that? What is the defining line between a circular orbit and an oval orbit? I am more into imagery when studying, so this GIF helped me out.



As far as I can tell, anything below an eccentricity of 0.4 has a (fairly) circular orbit, while anything above that has an oval orbit.

Gliese 581 c has low eccentricity (e=0.16). But in comparison to the two inner planets, c's orbit looks oval and eccentric. Would planet c pseudo-synchronous rotation?

Or I should ask, what is the "eccentric borderline" for a planet to have a pseudo-synchronous rotation?

NuclearVacuum wrote:What is the defining line between a circular orbit and an oval orbit? I am more into imagery when studying, so this GIF helped me out.

I don't believe there is one. As the eccentricity approaches zero, it gets less oval. As the eccentricity approaches 1, it gets more oval. An orbit with e=0.01 is still an oval, because it isn't a perfect circle.

Your gif shows that the orbit looks circular at e=2. If you actually measure the width of the orbit, and then its height, you'll see that it, too, is not circular.

NuclearVacuum wrote:Or I should ask, what is the "eccentric borderline" for a planet to have a pseudo-synchronous rotation?

My admitedly poor understanding of pseudosynchronous rotation leads me to believe that if the orbit has any eccentricity whatsoever, it will be in a pseudosynchronous rotation. I don't see any need for there to be a cut-off limit to the eccentricity.

I graphed the pseudosynchronous rotation equation in the Oklo blog and here's what I got.


My interpretation of this is that for a rotation whose length has been tidally determined, any eccentricity will result in pseudosynchrous rotaton.

Edit:
And the more I think about it, the more pseudorotation makes sense. Tidal effects are going to try and force the planet to rotate such that it is synchronous. If the orbit is eccentric this is impossible. When the object is nearest the sun, it will have to spin the fastest to keep the sun at a constant longitude. So tides will cause the rotation of the planet to be such that, for a brief time when the planet is nearest the sun, it actually is nearly synchronously rotating. But at this point in its orbit, it's moving quickly around the star, so as it gets farther away, since the rotation period doesn't change, it's free to have several rotations before it swings back around the star.

You can imagine that right at the moment of periastron, HD 80606 b's rotation is such that it keeps the same longitude toward the star. But since the orbit is eccentric and the planet rotates at a constant speed, when it's no longer at periastron, it rotates in its pseudosynchronized state, with the sub-solar point on the surface free to drift about the planet's longitudes.

http://www.oklo.org/wp-content/images/hd80606bgeometry.jpg

If you were to somehow "catch" HD 80606 b right at periastron, and drop the eccentricity to zero, it would probably be spinning in perfect synchronous rotation.

Sirius_Alpha wrote:Tidal locking has no cut-off distance, it just takes longer and longer to happen over larger semi-major axes.

An (oversimplified) approximation can be given by:
0.0483 (T M^2 / p)^(1/6)

Where T is the age of the system,
M is the mass of the star in solar masses
p is the density of the planet.

This equation gives you the distance from the star in AU within which a planet with the parameters you provide would be tidally locked. Note that this distance increases over time, as it takes a while for planets to tidally lock when they're farther out from the star. If you allow the star to survive long enough, every planet would become tidally locked.

I did the math, but there's a problem somewhere. The formula tells that Mercury is in the tidal lock area. Even the Earth, our pretty blue planet, is tidally locked. This is the same thing with the extrasolar planets. I suppose there are missing parts in the formula or some unities are wrong. And one question: what is the real formula, if you have it ?

Bye

Sedna

Toying with the equation, I think I figured out what went wrong. For the density of the planet, instead of using density in terms of g/cm^3 (as has been done earlier in this thread), kg/m^3 should be used instead.

I obtain 0.47 AU for Earth.

My whole wandering off about Gliese 581 d can be ignored now