Hi, I'm a bit of a noob to this site but I've been reading various things on here for a while and it seems like a good place to ask questions about Astronomy and Physics, etc.

Anyway, what I want to know about is Tidal Locking, specifically the kind that occurs with Moons around Planets and Planets around stars. You see I'm interested in creating Sci-Fi universes (realistic ones), not for publishing or anything, just for my own amusement.

So I've been doing a lot of leading about the physics of planetary orbits and habitable zones and stellar classification and all that, for a while now. I usually find handy calculators on the Internet for calculating interesting variables like planetary temperatures (for instance this nice one http://www.astro.indiana.edu/~gsimon...perature1.html), but recently I've come to the issue of Tidal locking and I can't seem to find much good info on it and certainly no online calculator programmes.

So what I need to know is; how do I calculate whether a planet or moon will be tidally locked or not. Is there a formula I can use, relating to the masses of the bodies involved?

In the Sci-Fi planets I've been trying to create I've had one that orbits around a red dwarf star, but as I'm sure you know the planet has to be very close to the star to be warm enough for life. For instance my planet is a large rocky world of 3 Earth masses, that orbits a M0 star with a mass of 0.45 Solar masses at a distance of 0.36 AU, will this planet be tidally locked or not? How do I calculate that?

Also for a moon, are all moons automatically tidally locked, or is it variable?
For instance I've come up with the idea of a Huge Super-Earth sized moon (11 Earth masses) orbiting a really big gas giant planet (4.3 Jupiter masses) at about 11 million km. Will this moon be tidally locked or not?

If anybody wants the exact numbers for these I can provide them, but I thought it would be easier for now to use common astronomical terminology.
So, any help on this would be much appreciated. Thanks.

Others will give you much better information than I can, but my very basic understanding is that ALL orbiting bodies are affected by tidal forces, though not all will become locked in the lifetime of the star. As I understand it, Mercury is "locked" in a 3:2 resonance with the sun.

There is a book that I have called World Building. It's one of a series of books aimed at SF writers for the purpose of getting the science part right enough to not distract from the fiction. I won't be home until late in the evening your time, but I'll try to remember to check and see if it has anything useful.

Edit to add: Sorry. Welcome to the board.

There a lot of very knowledgeable people around here, but the rules are bit off the norm for an internet forum. Make you you check them out.

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To answer the calc part... sorry cant help. To answer the part about do all do it then I refer you to Astronomy Cast (podcast) which has a nice description of these things... google it and look through their episode descriptions.

As I understand it (and I am not a scientist), you can think of the thing like this:

The moon is not going to be absolutely perfectly spherical with all of its matter perfectly evenly distributed. Now as it revolves the side that has fractionally more mass will come around to face the planet and at that particular moment it is experiencing fractionally more gravitational pull than when the lighter side is facing... more mass closer together will attract more... and then it rotates away again. But during that heavy side swing-by the moon lost a tiny fraction of its rotational energy because of the uneven pull.

Or think of a steel planet ball with a blow-up beach ball as the moon with a magnet pasted on one spot on the moon balls equator. Each rotating pass by the magnet will pull a slight bit off the rotational energy until eventually the beach ball stops rotating with the magnet facing the steel ball.

So eventually the moon ends up with its fractionally heavier side facing the planet. It is still rotating but much slower now... each full rotation takes exactly the same amount of time as the moon takes in its orbit around the planet... net effect is one side facing the planet.

Now how long that all takes depends on the initial conditions... but it is inevitable given enough time and no disturbances. So yes they all do it, but no not all will be like that 'currently', since they might not have had time to slow down yet.... depends on initial conditions.

Thats my two cents worth.... and it might not be worth even that much.

Oh... and by the way Murph.... your law sucks.

How long it takes a planet or moon to become tidally locked depends
in large part on the magnitude of the uneveness of the body's mass
distribution. The more lopsided it is, the more quickly it will become
tidally locked. A body with a perfectly spherical mass distribution will
never become tidally locked.

It also depends on the gravitational gradient of each body across
the other. The stronger the gradient across a body, the more rapidly
that body will become tidally locked. The gradient of course depends
on the mass and separation of the bodies.

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"

"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves

In the case of the Earth and moon, though, the "heavy side" of the moon is not pointed directly towards the Earth. It's a few tens of degrees off, if I recall correctly.
Tides make the distribution non-spherical, so it is possible to cause the body to slow down. Tidal slowing then operates on the tidal bulges, not the original mass distribution.

Another way to express this is that tidal locking (synchronization) happens faster for bodies with stronger internal energy dissipation - rotational energy is converted into heat (as in Io and Europa where perturbations of their orbits by the other big moons break the lock slightly and thus afford scope for energy transfer). This may happen by the friction and momentum transfer of water against rock flexure of rock in the interior, or whatever. For a giant planet it might be viscosity. A more rigid spherical body will take longer to synchronize.

I've wondered 'bout this. I'm going to guess that both bodies would have to be perfectly spherical and inelastic, or am I eating too many bananas?

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exactly what I was thinking a few minutes ago.
What would our Earth-Moon-system look like if the Moon was perfectly sperical (and inelastic)? (but not Earth)

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1 + e^i*pi = 0

Thanks for the response guys.

So if I follow what you seem to be saying, tidal locking is inevitable for any orbiting body which is not perfectly spherical (which no real planet or moon will be). So does this include the Earth? Will we eventually become tidally locked to the Sun?

So it’s all to do with asymmetry in the orbiting body's shape and mass? But does the mass/gravity of the primary and the distance to the satellite not have an influence also?

i.e. I assume that if Mercury were further from the Sun, then it would not be locked (or in 3:2 resonance), would this be correct? If the Earth was where Mercury is, would we be locked?

So if I understand correctly, the formula I should be looking for is how long it takes a hypothetical planet or moon to become tidally locked and if the answer is more than the age of the system then it won't be locked, but it is less then the current age, it will be locked, is this right?

The Wikipedia article on Tidal locking (http://en.wikipedia.org/wiki/Tidal_locking) appears to have such a formula but it looks very convoluted and confusing, does anybody know how to use this formula (or is there a simpler one)?

It sounds as if I don't like formulas any more than you do. For your needs- which I interpret as maintaining believability of your WIP by not breaking the laws of Physics TOO badly-try checking out this chart.

http://www.daviddarling.info/encyclo...H/habzone.html

It shows HZ of different stars. It includes an area where tidal locking would occur.
The tidal lock radius line is well past the HZ of the M star and zigs into Mercury. (OK Mercury has some type of 3:2 resonance, but for the reader of your book, saying Mercury is tidallly locked, imo, would be acceptable)

I interpret this graph to mean that all exosolar planets orbiting M stars within 0.1 AU would be tidally locked. (Perhaps that is why there is so much information/so many papers published about whether such a planet could maintain an atmosphere.)

Sanbeko ...who accepts payment in the form of dark chocolate unless you all think I am simplifying things too much and I have come to an erroneous conclusion, in which case I'd be interested to know. And I'd still eat the chocolate!

Well, the book I linked to sort of has a formula.

F=M/D³ where M is the mass of the large body and D is the distance.

The Earth Moon system is shown as having a tidal force of 177.6, while the example star (0.224 solar masses and the planet orbiting 0.058 AU) would generate 1130 times the tidal force of the Sun-earth system. I've no idea how that force would be applied to a timetable.

He does suggest that a 3:2 resonance could be used to allow for a day/night cycle that might be short enough to keep things alive.

One thing about his formulas. He uses a red dwarf for this example and his orbital distance is a lot less.

In the example in the book, he started with the luminosity and used that to figure the distance for Earthlike heating, then computed the orbit based on the mass. His example star was an M4.5, with a mass of 0.224 and there seems to be a huge difference in the comfort zone of the planets.

Maybe someone that knows this stuff a lot better then I do could check that site linked in the OP and make sure that it's correcting for everything to give a decent heating value. It seems to place a lot of significance on the greenhouse effect.

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so is the rotational energy being turned into heat by the deforming of the shape of the moon. Like warming up plasticine by molding it back and forth? Due to the shape of the moon always having to be slightly elongated along the axis pointing towards Earth(or what ever body)?

__________________

If a satellite was perfectly spherical and its density was uniform radially from the centre then the satellite would never loose rotational energy. It would be like a rock climber trying to climb a smooth wall. There would be nothing to try to grip. Now if the larger body was not also perfect then the moon would still become tidally locked with the planet but the face of the moon would change, it would just sit over the same point of the planet.

If I'm not mistaken at this point the satellite would orbit would be stable.

So you have 2 different locked aspect here.
1) the satellites rotation about its axis becomes synchronized with respect to the orbit around its parent body (what is happening with our moon)
2) the satellite becomes locked with the rotation of the parent body. This would eventually happen to our moon. The closer it gets to being tidally locked the further out its orbit will become.

So right now our moon is in synchronous rotation with its orbit. The tidal lock comes later? (do I have this straight?)

I forget when the moon is supposed to be tidally locked, but I know I'll be long gone before it ever happens.

Kind of like the sun going red giant on us. Not a big concern at this point. We have to get through the galactic merger with Andromeda 1-2 billion years before that happens anyway

What variables or constants in the equation would you like assistance?

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@ Sanbeko, thanks for that link that's the kind of thing I think I could use, but the graph is sort of hard to read as it's logarithmic. If I could find out how they derive that graph that would be useful.

Well where do I start? There's several variables in there that I don't quite understand, for instance I'm not sure about the "initial spin rate (revolutions per second)", does that just mean the rotational speed of the object (like the day length)? Or is it talking about the original rotational speed when the planet formed?

I'm not quite sure how to work out "moment of inertia" either, and I have no idea what the "dissipation function of the satellite" or a "Love number" is. The rest seems relatively easy.

The Wiki article seems to suggest that several of the variables are poorly known anyway even for planets in our own system, so I guess I could just use figures similar to those for the Earth-Moon system and try to scale them up, or something.

Yes, but it is its initial speed before tidal locking. Of course, it will be a fractional value of a second, but they do this to keep all your units consistent. Large equations require a little extra effort to make sure your units are the same. So, if you chose a rotation time of, say, 10 hours, then the revolutions would simply by the inverse of this in seconds. [Multiply hours by 3600 to get total no. of seconds, then take its reciprocal. 1/36,000 or 0.000028 rev/second.]

They show an approximation alternative: (0.4) x (moon's mass in kg) x (moon's radius)².

They suggest using a value of 100, though this will probably give you a longer time to achieve tidal locking than normal.

They offer an equation for that one, too, but it has two variables, density and rigidity. A moon's density will vary, but the big four around Jupiter range from about 1800 kg/m^3 to 3500 kg/m^3. They give you a range for the rigidity variable.

Bueno suerte!

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Ok ok, I think I'm starting to get this...

I've tried putting all the variables in, but I haven’t quite got the hang of this equation yet. At the moment I seem to be getting very small numbers or ridiculously big ones, by the way, what units is the end result supposed to be in, seconds or years, or something else?

Anyway, I'll try it a few more times and see if I can get this straightened out.

BTW, should I follow the Wiki advice and go for a relatively quick rotation for the initial spin? (like 12 hours?), or can I just try anything? i.e. is ~10 hours the sort of speed we would expect for a newly formed planet?

Murphy,

As George indicated, the units you put in determine the units that you
get out. If you put in seconds, the result will be in seconds. You should
use a consistent set of units such as SI metric unless you are sure the
formula can tolerate mixed units. (An equation that doesn't square any
values or find the square root of any values may allow you to put in hours
instead of seconds, for example, and get a correct result in hours. When
exponents are involved, that's less likely to work.)

The initial spin rate can be anything you want. If you're looking at the
Earth and Moon, for example, it could be your best guesses as to the
Earth's spin rate and the Moon's spin rate six months after the collision
which formed the Moon, or a million years later, or a billion years later,
or actual measurements from this morning.

The rotation rate of a newly-formed planet or moon is likely to be very
dependant on unpredictable events. However, as you can see by a look
at actual rotation rates of planets and asteroids in the Solar System,
they tend to end up in a relatively narrow range from a few hours to a
few days. Compression of a planet (by its own gravity) increases the
spin rate, while interaction with other bodies generally tends to reduce
the spin rate.

The Sun is thought to have lost most of its angular momentum to the
material in the disk of dust and gas that would later become the planets,
via strong magnetic fields that were generated while it was collapsing.
So the Sun rotates relatively slowly, while Jupiter and Saturn rotate
quite rapidly.

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"

"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves

Murphy my two cents worth is this:

Pick whatever you like. The universe is such a big place with so many things in it, that whatever you come up with is likely to be found somewhere or other. So long as you don't speculate on a 4 km diam world with a 100 km atmosphere or some such absurdity you are unlikely to be proved wrong.

The planet from your description is likely to be tidally locked according to wikipedia: http://en.wikipedia.org/wiki/Habitab..._dwarf_systems

Even things that shouldn't occur according to current theory, keep popping up unexpectedly in astronomy. Then everyone scrambles to work out how it happened and the theories and our understanding are advanced a tad further along the track towards a total description of everything.

Wont see it in our lifetimes....

That's just the proportion for the tide-raising force, directly proportional to the M of the large body, inversely proportional to the cube of the distance between the two bodies. It's also proportional to the radius of the smaller body.
There would have to be zero tides produced as well. If there are tides produced, there is something to "grip".

In other words, real objects always have something to grip. But sometimes things are just too strong to deform much, or too small to be that effected by tides (like a one centimeter steel ball bearing).Right now, the moon is said to be tidally locked to the earth, the earth is not yet tidally locked to the moon (and, the length of time that that would take is right up there with the life expectancy of the solar system).

Q, the specific tidal dissipation factor, is interesting. It's an inverse measure of the fraction of tidal energy dissipated per tidal cycle. Small values of Q indicate a large proportional dissipation; large values indicate a small energy dissipation. Hence the spin-down time varies directly with Q. It's also (usually) a measure of the lag angle between tide-raising body and tidal bulge: the smaller the value of Q, the larger the lag angle, and so the greater the tidal torque for a given size of bulge. Again, it's clear why smaller values of Q give shorter spin-down times.
Earth has a very low value of Q, around 12, presumably because of its oceans. Icy and rocky bodies are generally estimated around a 100, whereas gas giants are around 10000.

Grant Hutchison

In case anyone is wondering why I haven't seemed to reply yet, it's because I have but the moderators have not cleared my post yet.

I posted a long and detailed post last night about the calculations I did, but apparently because I included some links in it, it has to be viewed and passed by a moderator before it is put up, and they don't seem to have done that yet. (I just wish they'd speed it up, it's been like 24 hours now).

The wide range for Q is surprising.

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Yeah ok, I just wanted to confirm it was seconds and I only use all SI units anyway.

Ok so this is my calculations from the tidal locking formula on Wiki, it took me a while to iron out several mistakes I made in it, but I think I've got it now.

So I wanted to test the formula out on the fictional planet I mentioned in my first post, that is; a 3 Earth-mass planet orbiting a M0 red dwarf star with 0.45 Solar masses at a distance of 0.36 AU. And I wanted to know whether this planet would be tidally locked or freely rotating.

Here is all the data I've compiled for the fictional planet I've created...

* Planet:
- Diameter: 19,500 km (1.53 Earths)
- Mass: 1.816x10^22 Tonnes (3 Earths)
- Density: 4,678 kg/m^3 (0.85 x Earth's)
- Surface Area: 1,195,000,000 Sqr. km (2 Earths)
- Gravity: 1.3 gs (12.74 m/s)
- Escape Velocity: 15.8 km/s (1.4 x Earth's)
- Bond Albedo: 18%
- Average Surface Temperature = 19 Degrees C
- Solar Insolation: 594 W/m^2 (0.44 x Earth's)
- Atmospheric Greenhouse Effect: 2 x Earth's
- Distance to Star: 53,855,233 km (0.36 A.U)
- Length of Year (Orbital period): 117.6 days or 0.322 Years or ~ 4 months

And here is the data for the Red Dwarf...

* Star:
- Diameter: 844,944 km (0.607 Sol's)
- Mass 8.95x10^26 Tonnes (0.45 Solar Masses)
- Density: 2,694 kg/m^3 (1.91 Sol's)
- Surface Area: 2,243,000,000,000 Sqr. km (0.368 Sol's)
- Gravity: 32.44 gs (317.91 m/s) (1.16 Sol's)
- Escape Velocity: 518.4 km/s (0.84 Sol's)
- Spectral Type: M0 - Red Dwarf Star
- Luminosity: 0.056 times Solar Luminosity
- Surface Temperature: 3,439 K (3,166 C) (0.595 Sol's)
- Star's Life Cycle: 47.3 Billion Years

(Please note I created this scenario about a year ago from various Online calculator programs, such as these:http://www.geocities.com/albmont/mseqstar.htm and http://www.astro.indiana.edu/~gsimon...perature1.html, it is still entirely possible that I made mistakes in the calculations of the data, so if you can see any please point them out).

I don't know how to place the formula I'm working from on this page, but just follow this link to it on Wiki (http://en.wikipedia.org/wiki/Tidal_locking#Timescale).

The Variables of the Equation are as follows...

w is the initial spin rate (revolutions per second)
a is the semi-major axis of the motion of the satellite around the planet
I is the moment of inertia of the satellite => (0.4 x (Ms x R^2))
Q is the dissipation function of the satellite (~100)
G is the gravitational constant
Mp is the mass of the planet
Ms is the mass of the satellite
K2 is the tidal "Love number" of the satellite
R is the radius of the satellite

There is then another equation for working out K2, its variables are...

p is the density of the satellite
g is the surface gravity of the satellite
u is rigidity of the satellite

Note that I haven't used the correct Greek letters Rho and Mu, but instead p and u, to simplify things. Also note that the terms "Satellite" and "Planet" are mixed around here as the "Satellite" is in-fact a planet and the "Planet" (the primary) is a star (but I assume the equation works equally well for any two bodies regardless of what they are).

So anyway, here is what I was able to work all the variables out as...

w = 2.3148148148148148148148148148148x10^-5 Revs per second (12hrs)
a = 53,855,233,000 m
I = 6.90534x10^38 kg m^2
Q = 100
G = 0.0000000000667428 (6.67428x10^-11)
Mp = 895,000,000,000,000,000,000,000,000,000 kg (8.95x10^29 kg)
Ms = 18,160,000,000,000,000,000,000,000 kg (1.816x10^25 kg)
K2 = 1.0063953667809762626744247228517
R = 9,750,000 m

p = 4,678 kg/m^3
g = 12.74 m/s
u = 3x10^10 Nm-2

After much calculation the equation boils down to
3.9000374298489037646310090872017x10^100 Divided by
1.4222091054785234340300536876049x10^85

Which equals 2,742,239,108,739,694.1460203749960023 seconds, or 86,896,314 years.

So I think the conclusion I've had to come to is that the planet in question would take just 87 million years to become tidally locked to its star and since I've envisioned the planet as being several billion years old, it is indeed tidally locked.

This is rather bad news for my Sci-Fi planet as I'd hoped it could have a normal day night cycle, but being constantly facing the star will change everything. I haven't fiddled with the variables that much, but I found that speeding up the initial spin rate extends the time before tidal locking, for instance a 2 hour spin allows it to last for 521 million years, still not quite the 2 or 3 billion I was hoping for, but perhaps it will be possible to adjust it to make it like that, though that might stretch credibility.

So, did I make any glaring mistakes here? (it's quite possible since this formula is new to me and it's pretty complicated with so many numbers and opportunities to miss-type things).

Yes, it implies that the tidal bulge raised on a gas giant tracks the tide-raising body with an angular error of less than a thousandth of a degree. Gas giants seem to be very elastic.

Grant Hutchison

IIRC Uranus and Neptune seem to have Q of the order 1e5. Jupiter and Saturn on the other hand are more like 1e6. This may be due to the different internal structure - icy giants have thinner atmospheres and thicker water/ice and rock layers, whereas the hydrogen giants have much deeper atmospheres and different phases of hydrogen and helium in them.

I just plugged your numbers into a slightly different model I happen to have set up in MathCAD, and got 23 million years. This stuff is always a ball-park estimate, so we have effectively the same result.

Grant Hutchison

I haven't crunched your numbers, but a planet that large and that close to the star will definitely be tidelocked very rapidly.

The equation I'm using needs the initial rotation period of the planet when it was forming (I take the initial rotation period and then the solar tides act to reduce the rotational angular velocity of the planet - I based my equations on those given by Goldreich and Soter in their 1966 "Q in the solar system" paper. That said, I just noticed that the one you're using does too). Assuming a 10 hour initial rotation period, I get a value of 1.34 billion years for your planet to be tidelocked. Which has me worried, since that's rather different to yours.

I think Q may be too high - your planet is going to be a panthalassic ocean world with that density (it's pretty low for such a large world, so it needs a lot of volatiles to lower the density), which means that Q may have to be more like 10-12 (like Earth's) - I've seen values of 100 for icy satellites. Lowering it to 12 (earth's Q) in my equation brings the despinning time down to 161 million years. To match yours I need a Q of about 6-7.

I expect that Earth's very low value for Q of 12 is due in a big way to
the fact that the oceans are interrupted by land. If the oceans were
continuous and uniform depth, Q would certainly be much greater.

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"

"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves

Hang on... after further tweaks (just noticed your 12 hour initial rotation period, and the wiki page says that rigidity for rocky planets is 3e10 where I was using 5e10) and using those values knocks my result down to 914 million years for Q=100. If I use Q=10 then I get 91.4 million years.

Still not sure why I'm getting a despinning time that's 10 times longer than yours with Q=100 though.