Fixed Rotation?

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Fixed Rotation?

Post by NuclearVacuum on 30th April 2009, 1:58 pm

How close would a planet need to be from its parent star to loose its rotation and be tidally locked to the star? Would close orbiting stars (like DEL Tri and IOT Peg) be tidally locked onto each other?

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Re: Fixed Rotation?

Post by Sirius_Alpha on 30th April 2009, 5:47 pm

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.

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Re: Fixed Rotation?

Post by NuclearVacuum on 1st May 2009, 6:58 pm

This helps me out, thank you very much. But you didn't answer my other question.

Would close orbiting stars (like DEL Tri and IOT Peg) be tidally locked onto each other?

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Re: Fixed Rotation?

Post by Sirius_Alpha on 1st May 2009, 10:14 pm

Embarassed Sad I'm afraid I don't know much about those stars.

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Re: Fixed Rotation?

Post by NuclearVacuum on 2nd May 2009, 10:16 am

Sirius_Alpha wrote:Embarassed Sad I'm afraid I don't know much about those stars.

I only used them as an example (sorry for that). I was asking would circumbinary stars be locked onto each other? Two stars in close orbits.

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Re: Fixed Rotation?

Post by Sirius_Alpha on 2nd May 2009, 12:13 pm

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.

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Re: Fixed Rotation?

Post by NuclearVacuum on 2nd May 2009, 4:06 pm

Thank you so much for your answer. That question has been bugging me for a long time. Also, sorry for my stupid questions.

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Re: Fixed Rotation?

Post by Sirius_Alpha on 2nd May 2009, 10:26 pm

I'm glad I could help. And your questions weren't stupid Smile.

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Re: Fixed Rotation?

Post by Edasich on 3rd May 2009, 7:06 am

If this can help, cataclysmic variables have the low-mass secondary quite tidally locked. There was also a peculiar one (I don't remember at the moment, maybe LY Aqr) where the (probably substellar) seconday would show strong temperature differences between "dayside" and "nightside", just alike Upsilon Andromedae b.

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Re: Fixed Rotation?

Post by Sirius_Alpha on 3rd May 2009, 11:45 am

Edasich wrote:There was also a peculiar one (I don't remember at the moment, maybe LY Aqr) where the (probably substellar) seconday would show strong temperature differences between "dayside" and "nightside", just alike Upsilon Andromedae b.
Probably not the one you're thinking of, but HW Vir is like that.

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Re: Fixed Rotation?

Post by Edasich on 3rd May 2009, 5:23 pm

Yeah, also.
But I have to find that paper again, because there I've read this interesting coincidence Laughing

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Re: Fixed Rotation?

Post by marasama on 25th May 2009, 1:14 am

Can be any distance if you factor in collisions.
Some impact that slow (instead of speed) the rotation could cause the planet to lock at even further distances.

But on the gravity itself, I don't know.

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Re: Fixed Rotation?

Post by NuclearVacuum on 26th May 2009, 8:56 pm

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 SUCK AT MATH!!!!!! But let me see if I got it right.

I wanted to calculate whether Gliese 581 d rotated or was locked to its star. But since there is no confirmed density of the planets, I made another calculation. But in order to calculate the density, you need the volume. So I had to calculate the volume. No

Just to see if I got the idea, I tested out Gliese 581 c (which is widely thought to be locked). Here is what I got

Code:

T = 9000000000 (9*10^9) (9 billion years old)
M = 0.31 (solar masses)
p = 1.4 (g/cm^2)
x = 1.7 (AU)

This means that Gliese 581 c would have to be orbiting its star within 1.7 AU (which it does) to be tidally locked. This calculation seems to be accurate, not to try Gliese 581 d.

Code:

p = 0.84 (g/cm^2)
x = 1.53 (AU)

According to this, Gliese 581 d would be locked to its star. Does this sound right?

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Re: Fixed Rotation?

Post by Sirius_Alpha on 27th May 2009, 8:06 am

I confirm your result for Gliese 581 d, using your value for the density.

However, the density seems a bit low for a 7 M_e planet.


Last edited by Sirius_Alpha on 27th May 2009, 11:22 am; edited 3 times in total

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Re: Fixed Rotation?

Post by NuclearVacuum on 27th May 2009, 11:05 am

Thank you so much for the help.

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Re: Fixed Rotation?

Post by Sirius_Alpha on 27th May 2009, 11:22 am

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.

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Re: Fixed Rotation?

Post by Lazarus on 27th May 2009, 6:12 pm

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.

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Re: Fixed Rotation?

Post by NuclearVacuum on 29th May 2009, 4:25 pm

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?

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Re: Fixed Rotation?

Post by Sirius_Alpha on 29th May 2009, 4:31 pm

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.

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Re: Fixed Rotation?

Post by NuclearVacuum on 29th May 2009, 4:47 pm

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.


Last edited by NuclearVacuum on 29th May 2009, 4:57 pm; edited 2 times in total (Reason for editing : wanted to answer my "sunspot question")

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Re: Fixed Rotation?

Post by Sirius_Alpha on 29th May 2009, 5:04 pm

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 Razz.

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.


Last edited by Sirius_Alpha on 29th May 2009, 5:19 pm; edited 3 times in total (Reason for editing : Adding more stuff, lol.)

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Re: Fixed Rotation?

Post by NuclearVacuum on 29th May 2009, 5:23 pm

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!


Last edited by NuclearVacuum on 29th May 2009, 5:28 pm; edited 1 time in total (Reason for editing : commenting new stuff)

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Re: Fixed Rotation?

Post by Sirius_Alpha on 29th May 2009, 5:30 pm

Haha, aww =).

I'm glad I could help ^_^.

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Re: Fixed Rotation?

Post by Lazarus on 29th May 2009, 5:31 pm

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.

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Re: Fixed Rotation?

Post by Sirius_Alpha on 29th May 2009, 5:37 pm

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?

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