Fixed Rotation?
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Re: Fixed Rotation?
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).
Lazarus dF star
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Re: Fixed Rotation?
Sirius_Alpha wrote: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.NuclearVacuum wrote:But what is the defining point between an eccentric orbit and a circular one?
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 pseudosynchronous rotation?
Or I should ask, what is the "eccentric borderline" for a planet to have a pseudosynchronous rotation?
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NuclearVacuum Terrestrial Planet
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Re: Fixed Rotation?
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.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.
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 pseudosynchronous 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 cutoff 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 subsolar point on the surface free to drift about the planet's longitudes.
http://www.oklo.org/wpcontent/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.
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Sirius_Alpha Admin
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Re: Fixed Rotation?
Sirius_Alpha wrote:Tidal locking has no cutoff distance, it just takes longer and longer to happen over larger semimajor 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 ?
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Sedna
Sedna Planetary Embryo
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Re: Fixed Rotation?
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
I obtain 0.47 AU for Earth.
My whole wandering off about Gliese 581 d can be ignored now
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Sirius_Alpha Admin
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Re: Fixed Rotation?
Yes that's good now, Earth is spinning free. Mercury is still tidally locked, but its orbit is pseudosynchronous (if I remember this term right). It shows that unities are very important: if one of them is not good, the whole formula is wrong.
Actually the formula with g/cm^3 gave a resonable result for young planetary system, such as 2M1207.
Bye
Sedna
Actually the formula with g/cm^3 gave a resonable result for young planetary system, such as 2M1207.
Bye
Sedna
Sedna Planetary Embryo
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no pseudosynchronism
Lazarus wrote:Of course Gliese 581 d is on an eccentric orbit. Leads to pseudosynchronisation not synchronisation. For e=0.38, pseudosynchronous period is just over half the orbital period, at 34.6 days.
Dear Lazarus,
Browsing this Forum for topics related to tides, I saw your aforequoted old post, and I cannot help from commenting on it. As you will see from http://iopscience.iop.org/0004637X/764/1/27 , pseudosynchronous spin of terrestrial planets and moons is unstable and therefore cannot exist.
I am not sure how to extend this statement to objects with surface or internal oceans. However, I have strong arguments in favour of extension of this statement to purely viscous objects (stars, jupiters).
Best regards,
Michael
efroimsk Micrometeorite
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Re: Fixed Rotation?
Indeed, as I stated later in the thread...
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.
Lazarus dF star
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Very likely, fluid planets never get pseudosynchronous either
Lazarus wrote:Indeed, as I stated later in the thread...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.
Very likely, the fluid planets will not get into pseudosynchronism either.
If you are interested, I can explain my point in more detail, but this will require some maths
efroimsk Micrometeorite
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Re: Fixed Rotation?
If you feel you can contribute to a discussion then go ahead and contribute. You don't need to start asking people for permission!
Lazarus dF star
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Re: Fixed Rotation?
No need to ask Besides, this is a mathfriendly board.efroimsk wrote:If you are interested, I can explain my point in more detail, but this will require some maths
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Sirius_Alpha Admin
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Why pseudosynchronous rotation is an unlikely fate for many (perhaps, all) fluid bodies.
All right guys. You have opened the lid.
1. To warm up, recall the good old formula for the static Love number:
k_2 = (3/2)/[ 1 + (19/2) \mu/( g R \rho ) ] ,
where g, R, and \rho are the surface gravity, radius, and mean density of a nearspherical homogeneous body, while \mu is the relaxed (static) rigidity modulus. Similar (up to numerical factors) are the formulae for all other k_{\el} with \el > 2.
I prefer to rewrite the formula in terms of the relaxed (static) compliance J = 1/{\mu} :
k_2 = (3/2)/[ 1 + (19/2) 1/( J g R \rho ) ]
As we remember, J interconnects the strain tensor u with the tidal stress \sigma:
u = J \sigma ,
while k_2 interconnects the degree2 additional tidal (static) potential U_2(r) with the tideraising potential W_2(R) of a (stationary) perturber:
U_2(r) = (R/r)^3 k_2 W_2(R) .
Here r is some exterior point, while R is a surface point right beneath it.
For details, see https://i.servimg.com/u/f14/18/07/03/56/figure14.png
As Peter Golrdeich pointed out back in mid60s, in the denominator of the expression for k_{\el} we observe competition between selfgravitation (the `1' term) and rheology (the term with J).
2. With a lot of math, it can be proven that in many situations similar machinery works for timedependent tides.
First, we have a similar formula for the Fourier components of the potentials:
\bar{U}_2(\omega) = (R/r)^3 \bar{k}_2(\omega) \bar{W}_2(\omega) .
Here I omitted the selfevident arguments r and R, but inserted the argument \omega, which is the Fourier tidal mode. I also put overbars, to remind that we are now talking about the Fourier components of U and W appropriate to the tidal mode \omega, and about the complex Love number \bar{k}_2(\omega) corresponding to this mode.
Second, and most important, the complex Love number \bar{k}_2(\omega) will be expressed through the complex compliance \bar{J}(\omega) in the same algebraic way as the static Love number was expressed via the static compliance:
\bar{k}_2(\omega) = (3/2)/[ 1 + (19/2) 1/( \bar{J}(\omega) g R \rho ) ] ,
where the complex compliance interconnects the Fourier components of the strain and stress tensors:
\bar{u}(\omega) = \bar{J}(\omega) \bar{\sigma}(\omega)
The old comment by Peter Goldreich remains in force. Were it not for the '1' in the denominator, the complex Love numbers would behave exactly as the complex compliances  and there would be no difference between the frequency dependencies of materials and the frequency dependencies of celestial bodies made of those materials: \bar{k}_2 ~ \bar{J} = 1/{\bar{\mu}}. In reality, though, we have this term `1', wherefore selfgravitation comes into play, and the tidal response differs from the seismic response.
3. Suppose you tell me that a celestial body is composed of a material with rheology \bar{J}(\omega), and you ask me to tell you what its k_2/Q gonna be.
Well, this a turnofthecrank situation. I take your \bar{J}(\omega), plug it into the above formula for \bar{k}_2(\omega)  and into all the similar formulae for \bar{k}_{\el}(\omega) with \el > 2  and I obtain the complex Love numbers.
Then I calculate the negative imaginary parts of of my complex Love numbers, i.e., k_2 \sin\epsilon_2 , where k_2 = k_2(\omega) is the absolute value of the complex Love number, while \epsilon = \epsilon(\omega) is the negative phase angle (negative argument) of the complex Love number.
The quantity k_2(\omega) =  \bar{k}_2 (\omega)  is what they often call the dynamical Love number, while \sin\epsilon_2(\omega) is often misdenoted with 1/Q. In fact, one should use 1/Q_2 (and, likewise, 1/Q_{\el} for \el > 2). Recall that the tidal Q is not the same as the seismic Q ! See the image https://i.servimg.com/u/f14/18/07/03/56/figure16.png
4. What remains is to grab the expansion of the tidal potential (or force, or torque) into the Fourier series over the tidal modes \omega_{lmpq} , and to insert the right k_l(\omega_{lmpq}) \sin\epsilon_l(\omega_{lmpq}) into each {lmpq} term of the series. This is now rheology will enter the tidal theory in a consistent manner. See the image
https://i.servimg.com/u/f14/18/07/03/56/figure15.png
Now let us limit ourselves, for simplicity, to the degree2 Love number and to the principal (semidiurnal) tidal mode {lmpq} = {2200}. Instead of \omega{2200}, I shall simply write \omega.
Whatever realistic rheology \bar{J}(\omega) you take, its insertion into the expression for the complex Love number will render you the k_2(\omega) \sin\epsilon_2(\omega)  or, as many will say, k_2/Q  shaped as a kink https://i.servimg.com/u/f14/18/07/03/56/figure13.png
The kink goes continuously through nil when the tidal mode goes through zero  so we cross spinorbit resonances smoothly.
5. Now, the final point. Let us take a viscous body lacking rigidity completely. Its complex compliance will read as
\bar{J}(\omega) =  i/(\chi \eta) ,
where \eta is the viscosity, while \chi = \omega is the physical forcing frequency. (It is positive definite, while the tidal mode \omega can be of either sign.)
Plug this into the formula (**) given at https://i.servimg.com/u/f14/18/07/03/56/figure16.png , then take its negative imaginary part (which is k_2/Q)  and get a kink!
6. The bottomline is that a purely viscous rheology is incompatible with the linear tidal law k_2/Q ~ \omega. The latter rotten law stems from the assumption that \Delta t is the same for all frequencies. We now see that this law and therefore its underlying assumption are mighty unphysical.
At the same time, it is this CTL (`constant time lag') law which underlines the calculation leading to pseudosynchronism. (Mind a mistake in the book by Murray & Dermott who derived pseudosynchronism from a `constant geometric lag' assumption http://iopscience.iop.org/0004637X/764/1/26 )
I would say that all depends upon the viscosity of the body. The peak frequency is about G \rho^2 R^2/\eta. If it so happens that the peak frequency much exceeds n e^2 , then perhaps pseudosynchronism still has a chance. But not otherwise.
Important disclaimer:
In realistic stars and jupiters, complex hydrodynamical processes are going on. Among other things, these entail deviation from the aforedescribed wonderful similarity between the static and dynamical formalisms. Therefore the above backoftheenvelope estimate cannot be taken as a serious calculation of the k_2/Q. Nonetheless, even this, simplified treatment is sufficient to demonstrate how far the resulting k_2/Q is from being linear in frequency. In reality, its frequency dependence will be even more complex.
7. Although you did not ask me about that, I cannot help from mentioning that employment of the correct kinkshaped k_2/Q (or, better to say, k_2/Q_2) renders dramatic consequencies for the theory of entrapment into spinorbit resonances. For example, Mercury had absolutely no chance in the world to avoid the 3:2 trap, as was demonstarted by Valeri Makarov http://iopscience.iop.org/0004637X/752/1/73
Thank you for urging me to write all this discourse. I now have to go and shape it into a short Letter to the Editor, and probably send it to the ApJ Letters
1. To warm up, recall the good old formula for the static Love number:
k_2 = (3/2)/[ 1 + (19/2) \mu/( g R \rho ) ] ,
where g, R, and \rho are the surface gravity, radius, and mean density of a nearspherical homogeneous body, while \mu is the relaxed (static) rigidity modulus. Similar (up to numerical factors) are the formulae for all other k_{\el} with \el > 2.
I prefer to rewrite the formula in terms of the relaxed (static) compliance J = 1/{\mu} :
k_2 = (3/2)/[ 1 + (19/2) 1/( J g R \rho ) ]
As we remember, J interconnects the strain tensor u with the tidal stress \sigma:
u = J \sigma ,
while k_2 interconnects the degree2 additional tidal (static) potential U_2(r) with the tideraising potential W_2(R) of a (stationary) perturber:
U_2(r) = (R/r)^3 k_2 W_2(R) .
Here r is some exterior point, while R is a surface point right beneath it.
For details, see https://i.servimg.com/u/f14/18/07/03/56/figure14.png
As Peter Golrdeich pointed out back in mid60s, in the denominator of the expression for k_{\el} we observe competition between selfgravitation (the `1' term) and rheology (the term with J).
2. With a lot of math, it can be proven that in many situations similar machinery works for timedependent tides.
First, we have a similar formula for the Fourier components of the potentials:
\bar{U}_2(\omega) = (R/r)^3 \bar{k}_2(\omega) \bar{W}_2(\omega) .
Here I omitted the selfevident arguments r and R, but inserted the argument \omega, which is the Fourier tidal mode. I also put overbars, to remind that we are now talking about the Fourier components of U and W appropriate to the tidal mode \omega, and about the complex Love number \bar{k}_2(\omega) corresponding to this mode.
Second, and most important, the complex Love number \bar{k}_2(\omega) will be expressed through the complex compliance \bar{J}(\omega) in the same algebraic way as the static Love number was expressed via the static compliance:
\bar{k}_2(\omega) = (3/2)/[ 1 + (19/2) 1/( \bar{J}(\omega) g R \rho ) ] ,
where the complex compliance interconnects the Fourier components of the strain and stress tensors:
\bar{u}(\omega) = \bar{J}(\omega) \bar{\sigma}(\omega)
The old comment by Peter Goldreich remains in force. Were it not for the '1' in the denominator, the complex Love numbers would behave exactly as the complex compliances  and there would be no difference between the frequency dependencies of materials and the frequency dependencies of celestial bodies made of those materials: \bar{k}_2 ~ \bar{J} = 1/{\bar{\mu}}. In reality, though, we have this term `1', wherefore selfgravitation comes into play, and the tidal response differs from the seismic response.
3. Suppose you tell me that a celestial body is composed of a material with rheology \bar{J}(\omega), and you ask me to tell you what its k_2/Q gonna be.
Well, this a turnofthecrank situation. I take your \bar{J}(\omega), plug it into the above formula for \bar{k}_2(\omega)  and into all the similar formulae for \bar{k}_{\el}(\omega) with \el > 2  and I obtain the complex Love numbers.
Then I calculate the negative imaginary parts of of my complex Love numbers, i.e., k_2 \sin\epsilon_2 , where k_2 = k_2(\omega) is the absolute value of the complex Love number, while \epsilon = \epsilon(\omega) is the negative phase angle (negative argument) of the complex Love number.
The quantity k_2(\omega) =  \bar{k}_2 (\omega)  is what they often call the dynamical Love number, while \sin\epsilon_2(\omega) is often misdenoted with 1/Q. In fact, one should use 1/Q_2 (and, likewise, 1/Q_{\el} for \el > 2). Recall that the tidal Q is not the same as the seismic Q ! See the image https://i.servimg.com/u/f14/18/07/03/56/figure16.png
4. What remains is to grab the expansion of the tidal potential (or force, or torque) into the Fourier series over the tidal modes \omega_{lmpq} , and to insert the right k_l(\omega_{lmpq}) \sin\epsilon_l(\omega_{lmpq}) into each {lmpq} term of the series. This is now rheology will enter the tidal theory in a consistent manner. See the image
https://i.servimg.com/u/f14/18/07/03/56/figure15.png
Now let us limit ourselves, for simplicity, to the degree2 Love number and to the principal (semidiurnal) tidal mode {lmpq} = {2200}. Instead of \omega{2200}, I shall simply write \omega.
Whatever realistic rheology \bar{J}(\omega) you take, its insertion into the expression for the complex Love number will render you the k_2(\omega) \sin\epsilon_2(\omega)  or, as many will say, k_2/Q  shaped as a kink https://i.servimg.com/u/f14/18/07/03/56/figure13.png
The kink goes continuously through nil when the tidal mode goes through zero  so we cross spinorbit resonances smoothly.
5. Now, the final point. Let us take a viscous body lacking rigidity completely. Its complex compliance will read as
\bar{J}(\omega) =  i/(\chi \eta) ,
where \eta is the viscosity, while \chi = \omega is the physical forcing frequency. (It is positive definite, while the tidal mode \omega can be of either sign.)
Plug this into the formula (**) given at https://i.servimg.com/u/f14/18/07/03/56/figure16.png , then take its negative imaginary part (which is k_2/Q)  and get a kink!
6. The bottomline is that a purely viscous rheology is incompatible with the linear tidal law k_2/Q ~ \omega. The latter rotten law stems from the assumption that \Delta t is the same for all frequencies. We now see that this law and therefore its underlying assumption are mighty unphysical.
At the same time, it is this CTL (`constant time lag') law which underlines the calculation leading to pseudosynchronism. (Mind a mistake in the book by Murray & Dermott who derived pseudosynchronism from a `constant geometric lag' assumption http://iopscience.iop.org/0004637X/764/1/26 )
I would say that all depends upon the viscosity of the body. The peak frequency is about G \rho^2 R^2/\eta. If it so happens that the peak frequency much exceeds n e^2 , then perhaps pseudosynchronism still has a chance. But not otherwise.
Important disclaimer:
In realistic stars and jupiters, complex hydrodynamical processes are going on. Among other things, these entail deviation from the aforedescribed wonderful similarity between the static and dynamical formalisms. Therefore the above backoftheenvelope estimate cannot be taken as a serious calculation of the k_2/Q. Nonetheless, even this, simplified treatment is sufficient to demonstrate how far the resulting k_2/Q is from being linear in frequency. In reality, its frequency dependence will be even more complex.
7. Although you did not ask me about that, I cannot help from mentioning that employment of the correct kinkshaped k_2/Q (or, better to say, k_2/Q_2) renders dramatic consequencies for the theory of entrapment into spinorbit resonances. For example, Mercury had absolutely no chance in the world to avoid the 3:2 trap, as was demonstarted by Valeri Makarov http://iopscience.iop.org/0004637X/752/1/73
Thank you for urging me to write all this discourse. I now have to go and shape it into a short Letter to the Editor, and probably send it to the ApJ Letters
Last edited by efroimsk on 24th January 2013, 3:27 pm; edited 1 time in total
efroimsk Micrometeorite
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Re: Fixed Rotation?
I would like to know if these formulae and the situation they describe could also explain the slow rotation of Venus... or Venus is realy a diferent situation envolving colisions with other solar system bodies? A Venuslike planet in a Venuslike orbit always will have that slow rotation or could it have a rotation period more earthlike?
Stardust Micrometeorite
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Re: Fixed Rotation?
Stardust wrote:I would like to know if these formulae and the situation they describe could also explain the slow rotation of Venus... or Venus is realy a different situation envolving colisions with other solar system bodies? A Venuslike planet in a Venuslike orbit always will have that slow rotation or could it have a rotation period more earthlike?
Not sure. Have to cogitate on this.
I am now looking at the paper by Correia and Laskar in Icarus 163, 24 (2003). There they did a lot of good work, but unfortunately their tidal model was not very advanced: k_2/Q linear in the tidal frequency at low frequencies, and a constant k_2/Q at higher frequencies.
It would be surprising if their results stand the test of the consistent tidal model.
(Compare their results on Mercury with the computation by Makarov http://iopscience.iop.org/0004637X/752/1/73 )
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