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EDG
2007-Sep-15, 04:55 AM
Usually, if you're on the surface a tidelocked planet orbiting a star in a perfectly circular orbit (and in a 1:1 spin-orbit resonance), you'd expect the sun to just be hanging in the same spot in the sky all the time (if you're on the side that can see it).

But if the planet's on an eccentric orbit, the star will move in the sky over the course of an orbit won't it? Obivously it'd get larger and smaller as the planet got to its perihelion and aphelion, but it should move laterally as well shouldn't it (I think that's down to the fact that the star is in one focus of the orbit)?

I was thinking that if you had the right circumstances on the ground then you could actually get false day and night cycles in the twilight zone as the star rose above and dipped below some mountains in the way during the orbit.

Taking a K4 V star (about the upper limit to where you can get a tidelocked planet in the habitable zone) with a luminosity of 0.16 sol and radius of 0.65 sol, the habitable zone should be at 0.4 AU. So the star's angular diameter 0.867 degrees, or almost exactly 52 arc seconds (1.62 times the angular diameter of the sun/full moon from earth).

So for this star at least it seems possible to be able to have the scenario I'm looking for - we'd only need the star to be able to move up and down by a degree or two over an orbit to be hidden behind mountains in the distance.

That being the case, is there a way to calculate exactly how far a primary star would move in the sky given a particular orbital eccentricity?

Delvo
2007-Sep-15, 10:55 AM
You wouldn't need mountains. The horizon will do the same thing.

The diurnal cycle would be very slow compared to ours, essentially the same period as the planet's year. And the solar angle would be low at its highest. Those are two things your diurnal zone would have in common with our polar regions.

The diurnal zone's width would depend not just on orbit eccentricity, but also on axial tilt.

EDG
2007-Sep-15, 04:15 PM
You wouldn't need mountains. The horizon will do the same thing.

True, but if you have mountains you can get a bit closer to the dayside in the twilight zone, where it's likely to be a bit warmer.


The diurnal cycle would be very slow compared to ours, essentially the same period as the planet's year.

In the case of the K4 V that'd be a problem, yes (though it'd only be the part in the mountains' shadow that'd be in darkness, so it won't get too cold there). For M V stars the orbital period can be pretty rapid since their habitable zone is that much closer to the star.


And the solar angle would be low at its highest.

...which means? (what's "solar angle"? And when what is at its highest?)



Those are two things your diurnal zone would have in common with our polar regions. The diurnal zone's width would depend not just on orbit eccentricity, but also on axial tilt.

Wait a sec... so we have a twilight zone where the sun is low enough in the sky that the temperatures are most habitable between day and night sides. But we also have a "diurnal zone" within that where the star can bob up and down below the horizon as well, is that what you mean?

Would the star actually go up and down though (relative to the horizon)? Could it move left-right in the sky (ie parallel to the horizon)? Wouldn't that depend on what latitude you are at within the diurnal zone on the planet?

As for axial tilt, I'm assuming a tilt of zero. If you're tide-locked then that should be zeroed out during the tidelocking process. That said, the eccentricity should also be zeroed out as well by it, but I think the influence of other planets in the system can cause that to still be non-zero.

Delvo
2007-Sep-15, 05:07 PM
...which means? (what's "solar angle"? And when what is at its highest?)I meant the angle by which the sun appears to be above the horizon if you're standing on the given part of the planet. In the polar regions of Earth, the sun never gets very high even though it's up for months, creating a months-long polar day. It just rolls around over the horizon in a circle every 24 hours. That's similar in a way to your planet's zone between the day side and the night side; if you stood there, the sun would stick close to the horizon (although it would always be in a roughly constant direction from you instead of doing circles around you), never getting far above it like it does at noon in Earth's temperate and tropical zones.


Wait a sec... so we have a twilight zone where the sun is low enough in the sky that the temperatures are most habitable between day and night sides. But we also have a "diurnal zone" within that where the star can bob up and down below the horizon as well, is that what you mean?I meant the zone where there's a diurnal cycle, getting bright then dark (or at least dimmer) then bright again, due to the sun going up and down a bit. Most of it would include most of what you're calling the twilight zone, and I didn't originally think of a distinction between the two names. But now that your question's made me think of it, I see a narrow strip on the day side where the sun is pretty low so it could possibly be called "twilight", but it never really sets because the lowest point it reaches is still above the horizon, making that land lack a diurnal cycle because the sun's always up. There could also be a counterpart on the night side where the sun never really rises, but sometimes it comes up close enough below the horizon to make the sky brighter, so you could argue over whether that's "occasional twilight" and whether or not it counts as a diurnal cycle.

Bottom line, because the sun does light the sky some even from below the horizon, here are the zones with functional descriptions instead of names:
1. Always night; sun never up or even close
2. Sun never up but sometimes close enough to light the sky from below like our dawn & dusk; dark otherwise
3. Sun wanders above and below horizon; full light when it's up, dimmer when it's down; probably not wide enough movement to allow full nightlike darkness
4. Sun always up, with the lowest point it reaches still being above the horizon; always day; counterpart to zone 2 where the sun's low enough in the sky, and to zone 1 where it's higher, but it would make little difference except greater/lesser overall brightness and shadowing


Would the star actually go up and down though (relative to the horizon)? Could it move left-right in the sky (ie parallel to the horizon)? Wouldn't that depend on what latitude you are at within the diurnal zone on the planet?If the planet librates relative to the star in the same way that the moon does relative to the Earth, then the star would appear to move in a small loop in one part of the planet's sky.

JMV
2007-Sep-15, 09:10 PM
That being the case, is there a way to calculate exactly how far a primary star would move in the sky given a particular orbital eccentricity?
I would assume it's the same as the difference between true anomaly and mean anomaly. If you plot the true anomaly as a function of time and subtract the linearly changing orientation of the planet (which I believe can be set to be equal to the mean anomaly because of the synchronous rotation), you should get a graph that oscillates around 0.

There's an example graph for a highly elliptical (e=0.7) orbit here (http://www.wolffdata.se/gps/gpshtml/anomalies_html_51409fd5.png). The center figure.

The problem is you can't get the true anomaly from time and other parameters that easily because one of the necessary equations (the one giving relation between mean anomaly and eccentric anomaly) can't be solved analytically. Read here (http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Position_as_a_fu nction_of_time) for more info about that.

EDG
2007-Sep-16, 01:00 AM
Thanks JMV. Ugh, so it's not remotely straightforward... :(

EDG
2007-Sep-16, 04:52 AM
Bottom line, because the sun does light the sky some even from below the horizon, here are the zones with functional descriptions instead of names:
1. Always night; sun never up or even close
2. Sun never up but sometimes close enough to light the sky from below like our dawn & dusk; dark otherwise
3. Sun wanders above and below horizon; full light when it's up, dimmer when it's down; probably not wide enough movement to allow full nightlike darkness
4. Sun always up, with the lowest point it reaches still being above the horizon; always day; counterpart to zone 2 where the sun's low enough in the sky, and to zone 1 where it's higher, but it would make little difference except greater/lesser overall brightness and shadowing

Yeah, that makes sense to me. The funky thing is that the "twilight zone" is basically a band that is equivalent to being at an earthlike (rotating) planet's poles, right? The star's energy is coming in at a really shallow angle all around the twilight zone.


If the planet librates relative to the star in the same way that the moon does relative to the Earth, then the star would appear to move in a small loop in one part of the planet's sky.

I'm having trouble visualising why the star would move in a circle/loop... Or even why it's moving at all. I think it's because the planet is locked to the empty focus of the elliptical orbit rather than the star, right? But if that's the case, why does the star trace a circle in the sky rather than a line? I'm assuming that the planet isn't on an inclined orbit.

Delvo
2007-Sep-16, 12:01 PM
It's because of the twisting effect you can see in the video here (http://en.wikipedia.org/wiki/Libration). Any point on the moon appears to move in a loop even though the moon doesn't move (although the path is stretched longer from the upper left to the lower right and skinnier from the upper right to the lower left). So the Earth, viewed from there, would appear to move in such a pattern relative to the lunar sky & horizon. It's a result of two different sources of libration in two different directions (lattitudinal and longitudinal).

EDG
2007-Sep-16, 04:21 PM
It's because of the twisting effect you can see in the video here (http://en.wikipedia.org/wiki/Libration).

But according to that page, the libration in latitude is because the moon's orbit is inclined. For my scenario I'm assuming it isn't.

neilzero
2007-Oct-07, 03:38 PM
We may be adding confusion by comparing the earth, moon, sun. Let's substitute a black hole for our Sun (negligible accretion disk/ten solar mass) The black hole provides essentually nothing but gravity.
For Earth substitute an MV star with 0.1 solarmass = very dim/ orbits 91 to 94 million miles in an almost circular orbit, For the moon substitute two Earth mass (average density double that of Earth) with a very thick atmosphere (water boils at 230 degrees f at sea level) tilted 5 degrees with respect the orbital plane of the MV star. It is tide locked to the MV star in an almost circular orbit with a radius from 1.1 to 1.2 million miles from the center of the MV star. If that is inside the Roache limit, assume the habital whatever has good adhesion and good cohession. Let's add a ten Jupiter mass planet in a circular orbit, 400 million miles from the black hole. Now we can expect about as much liberation as our moon? At selected locations near the sunlight terminator the MV star should produce day alternating with twilight, depending on mountains near the horizon. The planet = habital whatever is too hot for humans (average 130 degrees f) at the center of the day side, and averages zero c = 32 f at the center of the dark side. The thick greenhouse atmosphere keeps temperature variations moderate in spite of the 240? hour day and negligible energy reaching the center of the dark side, except by air currents. My guess is axial tilt is of no significance since the habital whatever is tide locked to the MV star. Am I on the right track? Neil

grant hutchison
2007-Oct-07, 04:43 PM
Am I on the right track?It's difficult to tell. You seem to have introduced a vast quantity of extraneous detail. Libration in longitude is caused simply by the mismatch between an object's constant angular velocity of rotation and its varying orbital angular velocity. The mismatch shows up, for an observer at the gravitational focus of the orbit, as a cyclical change in the central meridian of the orbiting object.
So the only thing we need to know is the shape of the object's orbit around its primary.

Grant Hutchison

tony873004
2007-Oct-07, 10:24 PM
The mismatch shows up, for an observer at the gravitational focus of the orbit, as a cyclical change in the central meridian of the orbiting object.
Interestingly, an observer in the empty focus of the ellipse would not see any change in the orbiting object's central meridian.
http://www.orbitsimulator.com/gravity/images/libration.GIF

grant hutchison
2007-Oct-07, 10:31 PM
Interestingly, an observer in the empty focus of the ellipse would not see any change in the orbiting object's central meridian.It's an approximate, rather than an exact, relationship; though pretty good for small values of eccentricity.
I gave a link to some of the maths here (http://www.bautforum.com/963936-post11.html).

Grant Hutchison

tony873004
2007-Oct-08, 05:25 AM
I was under the impression it worked for all values of e. Thanks for the math link.

hhEb09'1
2007-Oct-08, 07:18 AM
You wouldn't need mountains. The horizon will do the same thing.

The diurnal cycle would be very slow compared to ours, essentially the same period as the planet's year. Looking at tony873004's diagram, wouldn't the period be half?

And you could always take more mountains, with gaps, like a zoetrope (http://en.wikipedia.org/wiki/Zoetrope), to get any period that an ID desired! :)

Just not very dark...continually civil twilight, forget about nautical twilight or astronomical twilight.

grant hutchison
2007-Oct-08, 01:01 PM
I was under the impression it worked for all values of e.You can take a look at it in Celestia, with a bit of trickery.
First, I set up the gravitational focus of my orbiting body as a black hole in an *.stc file:

"Gravitational focus"
{
RA 0
Dec 0
Distance 5000
SpectralType "X"
AbsMag 0
}Then in an *.ssc I place a big planet in orbit, and trick an invisible object into sitting at the empty focus:
"Planet" "Gravitational focus"
{
Radius 1000000

Texture "white.jpg"
OverlayTexture "westgrid.png"

EllipticalOrbit {
Period 1
SemiMajorAxis 1
# Eccentricity 0.8
Eccentricity 0.2
ArgOfPericenter 180
}
}

"Empty focus" "Gravitational focus"
{
Class "invisible"

Radius 1

EllipticalOrbit {
Period 1e20 # keep it "stationary"
# SemiMajorAxis 1.6 # 2ae of "Planet"
SemiMajorAxis 0.4 # 2ae of "Planet"

}
}The SemiMajorAxis of the invisible object is always set to (2*SemiMajorAxis*Eccentricity) of the big planet.

I've set it up with two options: low eccentricity is active as shown, and you can uncomment the high eccentricity options while commenting out the low.
I'm using my lat/long grids (http://www.lns.cornell.edu/~seb/celestia/hutchison/grids.html) to show up the libration, but you can use any old texture you like.

With low eccentricity, tracking my planet from the gravitational focus, I see it librate in longitude by about 25 either way; migrating to the empty focus pretty much abolishes the libration. With high eccentricity, my planet librates by 100 either way as seen from the gravitational focus, and still by around 25 from the empty focus.

Grant Hutchison

tony873004
2007-Oct-08, 09:29 PM
Thanks for the Celestia code. You're right, at e=0.8 it's not perfect. Here's my Gravity Simulator attempt, photoshopping in the 2-tone planets, rotating them 30 degrees at each step, plotted every 1/12 of an orbital period:
http://www.orbitsimulator.com/BA/libration2.GIF

hhEb09'1
2007-Oct-08, 09:44 PM
Here's my Gravity Simulator attempt, photoshopping in the 2-tone planets, rotating them 30 degrees at each step, plotted every 1/12 of an orbital period:Seems to do a whoop-dee-do there at the far right :)

tony873004
2007-Oct-08, 10:45 PM
Seems to do a whoop-dee-do there at the far right :)

Whoop-dee-do? What whoop-dee do? I don't see anything...

Thanks for catching that. I edited the image. My blue and green were reversed on the far right image.

grant hutchison
2007-Nov-20, 09:59 AM
Just bumping this thread to point to an article in this month's Icarus (http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WGF-4PG8737-1&_user=10&_coverDate=12%2F01%2F2007&_rdoc=2&_fmt=summary&_orig=browse&_srch=doc-info(%23toc%236821%232007%23998079998%23674080%23F LA%23display%23Volume)&_cdi=6821&_sort=d&_docanchor=&_ct=19&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=9c6e60ff1be7d4df92fbaa4788215170), dealing with exactly this topic.

Grant Hutchison

neilzero
2007-Nov-21, 03:06 PM
If the mountains are shaped like the teeth of a comb, then horizontal liberation can produce day lenghts as short as a few hours. Near Las Crusis, New Mexico, USA, there is a group of mountains like this, with peaks over 11,000 feet. They are named The Organ Mountains because they look like organ pipes. Neil

JohnBStone
2007-Nov-22, 10:40 PM
If you have a close orbiting tide-locked librating planet in an eccentric orbit, wouldn't you get:

a) strong tidal heating forces, volcanoes, etc?

b) rapid circularizartion of the orbit due to the tides?