# Tidal Locked to a Red Dwarf? What Would the Calendar Look Like?

1. Jul 15, 2011

### Nick Writes

Okay, I'm nervous. I feel a bit like I'm standing in front of an auditorium and am about to make a speech...

I'm a fiction writer...

Wait! Don't go yet!

I'm one of these hillbilies who *needs* some help!

I'm working on a science fiction bit and I do believe in research. I'm always bothered when writers have access to the science, but opt to ignore it. This is worse for hard science fiction, which is what I'm working on...

Given that Epsilon Indi is considered a solid candidate to have a tidally-locked planet in orbit around it. I dubbed it Epsilon Indi A and located it about .28 AU from the star.

Due to the fact that I'm writing about colonizing some of these red-dwarf worlds, I need it to be quasi-Earth-like.

And yes, I realize that means the world has much less mass than Earth, water on the surface - enough for liquid water near the light-dark terminators, less water on the sun side, and some frozen water on the dark side - and an atmosphere strong enough to repel most EM radiation.

Given all that - else no one would ever have tried to colonize it...

And given that Epsilon Indi A's orbit around the star has no "wobble" except for its own elliptical orbit, and I think we could call it stable. I also decided to give it a very slight axial tilt - 4 to 5%, no more than that.

Libration would have a different perspective entirely, since unlike the Earth-Moon scenario, the observers would be on the smaller body and observing the larger. But it should be in play.

Based on Kepler's Third Law, one year would be about 54.07 days.

Given all that, it appears to me that any humans existing on the world near the terminator (which is the idea) would perceive that short, 54-day year.

But during that year, would the effects of libration and tilt create four short "seasons?"

Assuming that they're on (or very near) the terminator itself, it seems that they would experience...

Summer - when the red dwarf is entirely in the sky, but not too high,
Autumn - when Epsilon Indi is on the horizon, half above and half below,
Winter - when Epsilon Indi is fully below the horizon and the sky is dark,
Spring - when Epsilon Indi is back on the horizon, in the same position as Autumn.

...all of which would create "seasons" of around 13 days.

Estimating that the libration effect meant that the terminator actually shifted about 4 degrees on both the leading and trailing edges (totally an estimate unrelated to the axial tilt), and the tilt affected things as well...

I would assume that if one traveled north or south up the leading and trailing edges, one would see a fairly dramatic change of the "seasonal" appearance.

It seems to me that if one were at a higher latitude, their would appear to be more activity on the horizon during the Spring and Autumn, and during Summer, the red star would appear to move around the sky somewhat.

...and at the Equator, things would be much more stable.

...But, if one was merely to travel east or west at any latitude, either the summers or winters would change drastically, lengthening or shortening, depending on which edge and which hemisphere you were in.

Here's the big question: How Wrong Am I?

I've spent days working on this, using this forum, some others, and several other websites - including (ew) Wikipedia. I'm taking this seriously. Since red dwarfs are far more common than other types, I wanted someone to address this. But due to the radiation given off by the stars at closer orbits - which is necessary to support life - it would be difficult.

If you have an answer for me, or a way to correct, or any sort of "shut up and write detective fiction" advice, I'd love it.

I'm not a dummy, but I've just never been able to handle math or science with any real aptitude. For that, I'm hoping y'all can help out this hillbilly.

I thank for any response you offer.

2. Jul 15, 2011

### BobG

With a 4 degree tilt, you'll essentially have no seasons. And your equivalent of the Arctic Circle & Antarctic Circle will be at about 86 degree latitude, since the tilt also determines that.

If you're considering the small variations caused by a 4 degree tilt, you should also consider that the orbit is almost surely elliptical (in fact, orbital perturbations pretty much ensure every orbit is elliptical). Depending upon just how elliptical that orbit is will determine how much warmer the planet is at perigee compared to when it's at apogee.

In the case of Earth (with a near circular eccentricity of .0167), the tilted axis is the main reason for the seasons. However, reaching perigee when the Northern hemisphere is in winter makes the winters slightly less severe and reaching apogee in the summer makes summers slightly less extreme.

In any event, since the planet is tidally locked, you essentially have a winter (or night, whichever you want to call it) on the hemisphere opposite the star and a summer (or day) on the near hemisphere. If you're talking about seasons, the latitude won't come into play at all, except for the slight tilt meaning the poles themselves are sometimes all dark or all light. However, the angle that light hits the surface will mean it will be cooler near the poles than near the equator.

I think your habitable zone is just a band around the transition area between light and dark and you have no seasons worth worrying about.

3. Jul 15, 2011

### Nick Writes

Thanks, Bob!

I'm afraid I muddled up my previous question - as if that was possible - by referring to the position of the sun in the sky as "seasons."

Assuming the planet's elliptical orbit causes the perceived "wobble," that should make some change in where exactly the light is hitting, shouldn't it?

If a settlement vanishes into shadow, which would be similar to the shadowed edges of our moon once a month, conversely that would mean that Epsilon Indi vanished from the sky of the planet - correct? And when the settlement drifts into full light, wouldn't the red dwarf be higher in the sky, fully above the horizon?

Using that 54-day year, I was referring to "seasons" as referring to the star's position in the planet's sky: Spring and Autumn when it was on the horizon, Winter when it was below, and Summer when it was above.

Do your figures take that into account, or did I just *entirely* misread what you were saying?

Thanks!

4. Jul 15, 2011

### Janus

Staff Emeritus
You seem to be talking about two different effects here. One due to axial tilt which causes a North-South movement of the star, and the other due to libration which causes a East- West movement of the star.

Assuming you are at a position near the median terminator line, the effect these have on the stars position relative to the horizon depends on your latitude. at the poles, the tilt causes up-down motion and the libration East-West drift. At the equator, it is the libration that causes the up-down motion and the tilt causes North-South drift. Latitudes in between see a combined effect.

Another consideration is that the libration and axial tilt do not have to be in phase to each other. In other words, when the South pole is tilted full on to the star does not have to coincide to when the planet is at periapis or apoapis and can happen when the planet is at any point of its orbit.

5. Jul 15, 2011

### Nick Writes

Yes! Exactly, Janus!

I figured it would be easier to keep axial tilt to a minimum - primarily to make the math a bit easier, but also because it just seemed to me to make sense that a planet tidally locked to a red dwarf might simply *be* a bit more rigid in its movements.

I have *zero* science to back that up. Just a thought that that might be the case.

It was primarily the East-West movement I was considering. If we're talking about a settlement on the terminator, then the biggest movement of the sun (if we keep axial tilt to a minimum) is the east-west motion. But at that angle, if I'm not *completely* mistaken, I believe that that motion would be perceived as the red sun's up-down motion on the horizon.

It occurred to me that libration would be the dominant force on this world, and not axial tilt.

I admit that this is all conjecture. As far as I can tell, science has no example of the mechanics of how one object becomes tidally locked to another. We have the Moon and numerous bodies in our own solar system, but (again) as far as I can tell, we've never seen a body actually lock into place.

What does it look like? How long does it take? Does it suddenly *snap!* into place? Or does it slow over millennia until its reached its new speed? (Yes, I think it must.) And when it does eventually become locked, is there a shudder or wobble as it settles into its new position or does it just suddenly start working smoothly?

I can't imagine there's not a little bit more back-and-forth when it finally locks in place. I believe that as time passed, the lock would become more and more solid, but I can think of no example of anything in nature that would suddenly click in place - not when one is dealing with multiple orientations, orbital motion, the mass of those objects, and the movements of the tides.

Yes, I was hoping to have this particular world still in the "shuddering" phase - after it has become tidally-locked to Epsilon Indi, but before it was completely in synch with it.

Thank you.

Last edited: Jul 15, 2011
6. Jul 15, 2011

### Drakkith

Staff Emeritus
There is no sudden point in time that we can say that a body gets tidally locked. It is a gradual occurance over millions or billions of years. The "lock" becoming more solid over time is up in the air. Variances in the stars makeup and geometry could cause some slight movement back and forth I think.

For ease of writing, you could simply model it after the Moon. You don't have to explain every little detail and the history of it.

7. Jul 15, 2011

### Janus

Staff Emeritus
The only way that a body could be perfectly synced with the primary is if it were in a perfectly circular orbit, Otherwise there will be some libration. And since perfectly circular orbits are hard to come by, you are not likely to ever see a perfectly synced body.

8. Jul 16, 2011

### Nick Writes

Janus, I never considered a perfectly circular orbit. I want it to be like the majority of the universe's orbits - eliptical. I didn't think it would ever be perfectly in synch, but I did assume that at some point it might become a little more stable than the Moon is to the Earth.

Like I said earlier, my goal was to have this set during the time when it's still falling in synch - which I describe as "shuddering."

Drakkith, thank you, too. I'm still debating between this event taking millions or billions of years, but I tend to think it's probably closer to a million. Who knows?

As much as I'd love to ignore this and consider it backstory, it's not. I needed the setting for a current work to be set on a tidally-locked planet in close orbit to a red dwarf star. It was easy enough to research the atmosphere; convection; water; light, sky, and flora color; and all the rest. But to set it along the leading or following edge of a terminator... I could find nothing for it.

Because the setting for the story is part of it, I wanted to be able to actually picture how things would look and act, and like I've said, I wanted this to be as accurate as possible. Science's best estimate is that most worlds that we find that could support life would be around a red dwarf - the most common type. If the planet has evolved to the point of supporting life, it would need to be close in to the star, and almost certainly in a tidal-lock situation.

It's just my preference to have a wee dram of accuracy, instead of another "Galactic Colony" under a yellow sun, over green grass... or on a dark, frigid world that could never support life.

Thank you gentlemen!

9. Jul 16, 2011

### Janus

Staff Emeritus
For reasons that I'll cover later, as long as the planet remains in an elliptical orbit, it will undergo libration and this is as "stable" ad it is going to get.
There is a way to estimate this:

the formula is

$$T = \frac{.075 \omega a^6 m_p Q}{G M_s^2 k_2 R^3}$$

$$\omega$$ is the original rotational rate of the planet in radians/sec
a is the average star-planet distance
mp is the mass of the planet
Ms is the mass of the star
G is the universal gravitational constant

The other two factors are hard to determine but can be estimated by

Q= ~100

and

$$k_2 = ~\frac{1.5}{1+\frac{19u}{2 \rho g R}}$$

u is the rigidity of the planet
$$\rho$$ is its density
g is its surface gravity.

For an Earth-like planet Q/k2= ~ 125

So if we place an Earth-like planet 0.28 AU from Epsilon Indi and assuming its staring rotation rate is equal to the Earth's present one, it would take ~33 million years to reach tidal lock (the short time is due to the fact that the tidal locking time is strongly affected by orbital distance. Move the same planet out to 1 AU and it increases to ~ 40 billion years.)

Now let's see how this affects the "wobble" due to libration. But first, I want to address a couple of points in your first post.

You said:

What made you conclude the last part? The distance from the star has no effect on what size the planet needs to be in order to be close to Earth-like. In fact, in order to hold on to a significant atmosphere it will need to be approximately the size and mass of the Earth. Especially considering the following:

Putting your world close to the star increases its exposure to stellat wind which tends to strip the atmosphere away. You world is going to have to be on significant mass to hold on to an atmosphere.

Also:

Epsilon Indi is ~0.762 the mass of the Sun. By the Mass/ luminosity law, this makes it 0.38 times as luminous as the Sun. Since the light striking the surface of your world decreases by the square of the distance from the star, putting your planet 0.28 AU from the star increases the light intensity by some 13 times over what it would be at 1 AU. 13 times 0.38 results in ~5 times as much light striking the surface of your planet then strikes the Earth. This has a greater heating effect on your atmosphere, expanding it and making it more likely to be stripped away.

...Based on Kepler's Third Law, one year would be about 54.07 days.[/quote]

This would be true if Epsilon Indi was the same mass as the Sun( Kepler's law only works when you are comparing orbits around the same body or around bodies of equal mass), however, since it has less mass, the year at 0.28 AU would be closer to ~63 days.

Now let's look at how this works concerning tidal effects dampening libration.

Assuming a 5 degree libration ( a total peak to peak wobble of 5 degrees) this means that from the star's point of view, the planet rotates 5 degrees in one direction for ~ 31.5 days and then in the other for ~31.5 days. 5 degrees in 31.5 days gives an angular velocity of 0.000000032 rad/sec.

Plugging this into the formula for tidal locking gives 14,500 yrs to kill this rotation rate. However, the star doesn't have 14,500 yrs, it has 31.5 days, because after that, the planet starts rotating in the other direction.

The way to look at this is to consider why the planet exhibits libration in the first place. The planet, by nature, will rotate at a constant angular velocity. However, the angular velocity of its elliptical orbit changes over the course of the orbit. So what we end up with is a planet that at times is rotating faster than it orbits and sometimes rotates slower.

Tidal forces do try to make the planet rotate at the same speed that it orbits, but this means that half the time they are trying to slow the planet's rotation and half the time trying to speed it up. This ends up in there being no net effect on the rotation of the planet. the only slight dampening of the wobble is the little bit that it can alter the rotation in 31.5 days. But since it spends half an orbit pulling in one direction and half the orbit pulling in the other, this dampening is not accumulative and will not decrease the libration over time.

10. Jul 18, 2011

### Nick Writes

Thank you, Janus! I do appreciate it all. I'm trying to add a little science to my fiction - and I see I've got some more work to do.

(Sorry, I would've responded earlier, but I've been out-of-pocket for a couple days.)

I appreciate everyone's feedback, and for putting up with my complete lack of scientific understanding. You've actually made my life a lot easier. I'd tell you I'll come back next time I have an insurmountable question, but I'm afraid that would seem more a threat than a promise.

Thanks again!