Rimenight: Life on a Shepherd Moon

[Kobold King]

The forest frosts around you, but there's a spring in your step. The wood's getting colder by the hour, and already most of the region's larger wildlife has retreated to its dens. Or at least, so you hope; not many of Rimelight's nocturnal fauna have been catalogued yet.

 

Stopping by the shores of an iced over lake, you set your gaze skywards. A mere glance upward justifies this world's name; the sky itself almost seems to have frosted over. Arcs of magnificent icy rings dominate the heavens, appearing in celestial increments around the enormous turquoise globe in the center of the cosmic firmament.

 

Finding yourself beaming once again at the sight, you begin setting up your trusty telescope. There's a lot of ring-gazing to do before the sun rises in twenty more Earth-hours.

 

 

This is the vague conception I have for a new science fiction worldbuilding project: I am attempting to design a planetary system in which a functioning multicellular ecology could evolve on a shepherd moon. I think I could do some pretty nifty things with the setting, but I need a little guidance on the proper astrometrics of such a system.

 

A shepherd moon is a moon that orbits around the edge of a gas giant's ring system. They may also appear within gaps between the rings. A shepherd moon is defined by the gravitational effect it projects upon the ring particulates--it "herds" the dust and ice particles into sharp, stable shapes like those of Saturn's characteristic rings.

 

Shepherd moons orbit close enough to the rings of their planet that I imagine the view from one would be spectacular; unfortunately, there are a few astronomical issues I'm facing in this world construction.

 

First of all: the moon can't be located too close to the planet itself. The tidal forces of an object as large as a gas giant could literally destroy the moon every other week. If you don't believe me, check out the maps of Io--or the lack thereof. Jupiter's tidal forces, coupled with those of the nearby moons, is enough to bring the moon's magma to the surface and completely restructure the continents on a regular basis. It's hard to imagine carbon-based, water-filled life surviving in such an inhospitable habitat.

 

That means that the moon of Rimenight will need to be an outer shepherd, orbiting far enough away from the planet that the tidal forces don't tear the crust apart like an almond flour pizza crust that's been in the oven too long. But if I position the moon too far away, it could take well over a week to orbit its parent body--thus radically lengthening the day/night cycle. Weeks of day followed by weeks of night could make for a fascinating setting itself, but it isn't at all what I want for Rimenight. I want a day/night cycle no longer than three Earth days.

 

The proper distance from the gas giant would be bewildering enough for me to figure out--but since I want this to be a shepherd moon with a good view of the rings, I'll need to contend with the issue of the moon's "flock." Wikipedia cheerfully informs me that one of the primary mechanisms of a shepherd moon's herding is the accretion of rogue particles from the ring system.

 

Now, "accretion" is just fine for a rocky potato like the ones that pick up rocks as they orbit Saturn, but for a world with a complex ecosystem, "accretion" looks more like this:

 

 

I'm fine with a few extinction-level impact events on this world. As long as the major impacts are limited to once every thirty million years or so, I can play around with a fascinating Eocene-like ecosystem for this moon's bioregions. But my concern is that a shepherd moon, even one with a thick atmosphere, even one with other moons drawing away fire, would undergo such a massive and continuous pummeling that cellular life, much less multicellular life, might never get the chance the evolve.

 

So in summary:

 

• I need to know the habitable zone for a moon orbiting a Saturn-sized gas giant.
• I also need to know how common and how severe impact events would be over this moon's history.
• If the answer to the previous question is "pretty sparking bad", then I'd appreciate some guidance on how to lessen the hostility of the environment.

 

Thanks in advance! Anyone who posts comments on this thread gets a Rimenightian geologic feature named after him/her!

[Edgedancer]

So, I´m afraid that I lack the knowledge to be useful here but if it makes you feel better the idea sounds awesome.

[Voidus]

T = 2*pi sqrt(r^3 / GM) 
rearranging gives:
r^3=(T^2GM)/(4*pi^2)
According to google, Saturns mass is 568*10^24 kg
G of course is 6.67*10^-11
And for a maximum T of 25920 seconds:
r^3=(25920^2 * 6.67*10^-11 * 568*10^24)/(4*pi^2)
r^3= 6.45*10 ^ 23
r= 86,389,602 metres

Saturns radius is 58,000 km approx so it'd only be about 28,000 km above the surface at most with these limitations.

This was done mostly through google since I don't have my calculator to hand and I haven't done this kind of problem in quite a while, so there may be some errors.

Edit: Hmmm... given that Titan orbits saturn at over 1 million km and still only has an orbital period of 16 days I think I may have made a mistake somewhere given that it should only be 3 times further out.   

Also thinking about it, that was probably a way quicker way of doing this, the ratio is r^3:T^2 so if its period is 1/5th that of Titan it's distance should be about 1/3 so like 300,000 km I guess.

Edited January 31, 2015 by Voidus

[Seonid]

I haven't checked the math itself, but the form looks right for a Kepler's 3rd law problem, Voidus.

 

That orbit might be a problem, as it lies well within Saturn's Roche limit of 147,000 km. The big question is whether this is close enough to have the surface-destroying symptoms that we see on Io. I suspect that it is.

 

Just checked. Io orbits at 421,700 km from Jupiter's surface. This is a problem. Going to check the math now.

 

(Sorry, this is kind of stream-of-consciousness stuff, I thought it might be valuable to see the whole thought process - makes it easier to nitpick there)

 

Found a solution. 3 days is 259,200 seconds, not 25,920. Using Kepler's 3rd law with that value of T gives us 400,985 km, a much more reasonable orbit.

 

It's outside the Roche limit, and about as far as Io is from Jupiter (but Saturn has less mass by a factor of 1/3 - so we might not be too close to get the constant surface renewal).

 

This brings up a new problem. All of the rings we've so far observed (except for Saturn's E Ring) are within the Roche limit. However, the E Ring of Saturn (very thick and formed of microscopic rather than macroscopic particles) lies between the orbits of Mimas and Titan, outside that limit. Titan, in particular, is large enough to be a shepherd world with life on it (if we move its solar system position). It's big enough to have a thick atmosphere.

 

Under these circumstances, we also don't have to worry about impact from Ring particles, as the large particles are all orbiting significantly inside the orbit of the moon.

 

So, maybe not quite as spectacular as having a moon in between the rings, but much less risky. This might be able to solve your second problem. Titan's actual orbit is about 15 earth days, but we could rather easily imagine a situation where a Saturn-like body had a Titan-like moon with an orbital period of 3 days. (Mimas, the inner shepherd of the E ring, has an orbital period of .9 Earth days) We'd just have a smaller ring. We might have to check the resonance patterns to make sure we aren't violating something there, but I think we're on to a workable solution here.

[Voidus]

*Facepalm*
Knew it would be something simple.
Thanks for the assist Seonid!

[TwiLyghtSansSparkles]

This sounds like an awesome setting, and I am cursing my lamentable math skills for not allowing me to comment more intelligently, and Quota's tyranny for not allowing me to upvote it. 

[Seonid]

Hey, it was a better mistake than my constant mixing up of signs on simple algebra problems...

[BreathTaker]

*Drool leaks from mouth* *Snaps back in to consciousness.* What in the Storms just happened? All I saw was "Math stuff... math stuff... oh and some numbers having to do with math stuff." And my brain went, "Oh that's cool, I'm just going to take a nap." You guys are insane. On the other hand, KK this is an awesome idea. 

 

I did understand some of what you guys were saying but when the crazy mathing started I kind of zoned out. The concept is really cool and I really hope this works out for you KK. 

[Kobold King]

Thank you, everyone! I mean that. Thanks to those of you who've showed enthusiasm for the setting, and doubly thanks to the mathematical wizards who're making the setting feasible.

 

 

A note of confession: I don't understand any of the mathematics you've gone through, Seonid and Voidus. These kind of calculations are really not my strong suit. I can usually understand the issues involved in astrometrics, but seldom can I comprehend the equations and formulae used to solve them.

 

What I believe I have gathered--and please correct me if I'm misunderstanding anything--is that a Titan-sized moon orbiting at roughly 400, 985 kilometers from a Saturn-sized planet would be feasible, and that it'd orbit well out of the range of extinction-level Ring particulates. In fact the overwhelming majority of Ring particulates would exist between Rimenight and the host planet; assuming a synchronous rotation, I believe this means that the characteristic Rings of Rimenight's sky would only be visible from one side of the moon. A touch disappointing, but nothing serious.

 

If what I think I've gotten from your calculations is correct, then I've got a pretty solid foundation to start worldbuilding from. Once again, I cannot express my gratitude enough! If there's an interest, I'll keep this thread updated with details on Rimenight's flora, fauna, and possibly sapient aboriginals.

 

And of course, if you have anything to add or correct on the astrometry as I currently understand it, don't hesitate to speak up. Or if you want to take me up on the naming offer; there's still room for a Voidus Forest, a Seonid Sea, or practically any other feature someone might want to call dibs on.

[Seonid]

Any time! Physics is one of my majors at the university, and I love getting a chance to put it to use in...well, not real life, but you know what I mean. (TBH, half the reason I am a physics major is because I want to be able to write physically accurate sci-fi)

 

The math probably looks a little bit daunting, but what we did was use Kepler's 3rd Law. Detailed explanation inside the spoiler tag.

 

In the late 17th century (if I am remembering right), Kepler wrote down three equations that (still!) form the underpinnings of modern astronomy. The third equation was based on tabulated data of the orbits of all of the then-known planets, and related the orbital period (the time it takes to orbit the central body of the system) to the radius of the orbit, the mass of the central body, and some universal constants.

 

The common formulation is:

 

(T²/r³) = 4pi/(M*G), where T is the period in seconds, r the radius in meters, M the mass in kilograms, and G is Newton's gravitation constant (which most people just look up if they need to know it)

 

Voidus (and later me, when I checked it) just plugged in the mass of Saturn for M and 3 days (in seconds) for T, and the standard accepted values for pi and G. Then we just solved the equation for r, (which is rather simple if you are confident with algebra, and rather daunting if you're not). Step by step instructions in the spoiler:

 

To solve the problem for r, we multiply both sides by r³ and then divide by the quantity 4pi/(M*G)), which leaves us T²/(4pi/(M*G)) = r³

Then we just take the cube root (I just use my Windows calculator for that)

 

The Roche limit is the limit where a small body (such as a satellite or moon) would be torn apart by tidal forces rather than staying together. A particularly rigid body can exist temporarily inside of it, (we see this with Saturn, where the moons probably formed elsewhere and migrated within it), but nothing can form there.

 

That said, your layman's explanation/interpretation is exactly on-target.

 

I had a whole long investigation of the day/night cycle, but then I found that almost all moons are tidally locked, which means that only 1 side actually faces the planet at any given time, so the rotation around the planet would always equal the rotation around its axis. Which made all of my speculation useless. Ah well...

 

As far as naming things go, I'm good with whatever. Maybe use Seonid backwards, like the Dinoes River or Sea

Edited February 1, 2015 by Seonid

[Voidus]

Yeah I was going to say, I'm not sure exactly why but most are tidally locked. So unfortunately only half the planet would get to see the rings.

The 400,000 km was just the maximum distance the planet could be, that would give the full 3-earth-day cycle, anywhere between the Roche limit and that would be possible, although if it is inside the rings that raises some further problems that need solving.

I'm a bio/chem major so most of my physics comes from helping friends in other classes and my goal to read every book in the library. (I'm about 0.0000000001% done) but I've always loved physics.

[Seonid]

Now, if we were able to hypothesize a moon that wasn't tidally locked, but was in some non-1:1 spin-orbital resonance, we might be able to get somewhere with the whole moon seeing the rings. Mercury has this feature, actually. It has a 3:2 orbital resonance (meaning that each orbit around the Sun is 1.5 Mercury-days)

 

Something similar for this shepherd moon around its parent body would result in the whole moon getting to see the rings/planet (instead of having the planet and rings just 'hang' in the sky, which is what would happen for a tidally locked moon), but would create a complicated day/night cycle where for some parts of the day, the planet eclipses the sun (on a regularly recurring basis, unlike - for example - the chaotic pattern of the Earth's solar eclipses).

 

Actually, the gas giant eclipsing the sun is a problem for a tidally locked moon too. If you are on the 'inside,' the face that can see the planet, then during your night, you are facing away from the sun and at noon, the planet eclipses the sun. You would get a sunrise, but as the sun got brighter, the planet would start to block it. This could potentially be a problem for life on the planet-side (cold-wise, that is). One possibility might be to posit a higher spin-orbit resonance, such as 3:2 or even 5:3.

 

I'll give this some more thought, and come back to it.

[BreathTaker]

I for one  would love to see your progress in this story KK.

[Seonid]

(First off, sorry to totally dominate this discussion, Kobold)

 

Now that that's over with, I've given this the requisite amount of thought, and essentially decided that all of my concerns are resolvable. In astronomy terms, we want a 3:2 spin-orbit resonance (which means that the moon spins around its axis 3 time for every 2 times it goes around the planet) with a high eccentricity orbit (which means that it is more elliptical than circular - we are probably looking at an eccentricity of 0.21 or so, which is strongly elliptical) that is inclined compared to the ecliptic as well as to the plane of the planet's rings (say about 5 degrees from the ecliptic and about 20 or so from the rings - which implies a planetary tilt of either 25 or 15 degrees from the ecliptic - either would work).

 

In layman's terms, the moon would have an elliptical orbit around the planet, have 3 days for every 2 times it went around the planet, and the planet would occasionally and predictably but not constantly eclipse the sun. In addition, the rings would be visible from the whole moon (although only from about 1/2 at a time).

 

Essentially, the day/night cycle would be normal, and only interrupted by eclipses occasionally. The planet would rise and set like the sun, although on a different schedule. When the planet-rise coincided with sunset, you would see a spectacular view of the rings. Whether they were visibly during the day would depend on a lot of things, but it is possible.

[Kobold King]

No need to apologize, Seonid! I am immensely grateful for this help. If I could convert my gratitude into rocket fuel, I could personally power a small spacecraft to Barnard's Star within a typical human lifespan. That's how grateful I am.

 

I think I've got a pretty good idea of what the days and nights would be like now, thanks to you. Just a couple of small questions...

 

On a moon that was tidally locked to its planet, just how cold would the planetside get? Cold enough to prohibit complex life on that side of the moon, or could meteorological processes circulate enough warm air to the planetside to allow a few plants to grow, perhaps adapted for soaking up sunlight in the early morning?

 

Would it be safe to assume that the planet and other nearby moons would have massive effects on Rimenight's tides? I know pretty little about tides, but I want to know whether I can assume that the tides would be more severe and would operate on a more erratic schedule.

[Blaze1616]

I'm a chemical engineer, so my physics experience definitely does not include astrometry, but I'll try to help by making parallels with Earth.

 

On Earth, the moon and the sun are approximately the same size in the sky. Eclipses, then, are rather short lived. Even still, an eclipse can cause the temperature to drop upwards of 3 degrees Celsius, or 5 degrees Fahrenheit. The situation you are talking about is having a body much larger than our moon in the sky to eclipse an object that I am assuming is about the same size as our sun in the sky (you have to be in the hospitable zone of the star, after all). This should cause much longer eclipses, and during it there should be times where it is essentially night (that might be wrong though, as the light might be visible from the edges of the planet still, though probably fainter than the light around the moon during Earth eclipses, now that I think about it).

 

Regardless, it's going to be cold. Being tidally locked, the coldest part of the moon will be the point where the planet is smack in the middle of the sky (a straight line from Rimenight's surface to the planet would be perpendicular to the tangent line of Rimenight's surface). This is because the only rays from the sun to reach the land will be in the "morning" and "evening", and the rays will be Indirect and at low angles. For anyone reading this that does not know, there are two types of sun rays. Direct and Indirect. Direct rays are rays that are approximately perpendicular to the surface. Direct rays transfer the most heat. Indirect rays, then, are rays not perpendicular. These rays transfer less heat, and the more acute the angle to the surface, the less heat is transfered. As such, Antarctica and the Arctic are Earth's coldest points, and the areas near the equator are the hottest.

 

How cold that coldest point gets will depend on how much of the sky your planet consumes. Taking into account that one orbit of the planet is 3 days on Earth, this should mean that the side facing away from the planet has night for 1.5 days and sunlight for 1.5 days (is that right?). This would then also mean the side facing the planet will face the sun for 1.5 days (with the planet eclipsing the sun for some percentage of that), and away from the sun for 1.5 days as well. This means you should get ~36 hours of sunlght if it weren't for the planet in the way. So, if the planet covers 50% of the sky, that coldest section of Rimenight would get 9 hours of light before the eclipse, 18 hours of an eclipse, and 9 hours of light afterwards. This would then be followed by 36 hours of night to then repeat again.

 

So what percentage of your sky is being eaten by the planet? Well, the moon is 384,000 km from Earth, whereas Rimenight is 401,000 km from it's planet. That's about 96% of the distance. Close enough for a rough calculation. Earth's moon has a circumference of 10,917 km. Saturn, the planet you're basing your gas giant off of, if I've understood correctly, has a circumference of  378,675 km. That means your planet is 34.5 times larger than the moon. Now, in our night sky the moon is smallest when "smack in the middle" of the sky due to reasons scientists still seem to disagree on, according to Wikipedia. Now, I cannot find any information about what percent of our sky the moon consumes, but let us assume it's about 2% of the sky. So of Rimenight's sky, at this one point on the moon, about 70% is consumed by the planet, if my math is right. 

 

This means that your day/night cycle, starting at dawn, is ~5.4 hours of light, 25.2 hours of eclipse, ~5.4 hours of light, 36 hours of night, and repeat. Those 10.8 hours of light places the sun from the horizon to about where it is 1.8 hours after dawn, if it is daylight for 12 hours (as we've assumed a similar situation for Rimenight). Given this, I would argue that your sunlight angles are not much better than the Antarctic, but you're getting sunlight every day of the year, rather than the Antarctic's having weeks of no sun. I would also say, though, that during the eclipse the moon would lose a good amount of heat, much like it does at night.

 

So, after those rough calculations, my even rougher guesstimate on life in those areas would be to say it would resemble a tundra here on Earth. The plant life would have evolved to last on what precious sunlight it gets, and any fauna would be adapted for the cold. Cold water based fauna sould be fine too.

 

Now, obviously, as you move away from that coldest point, you'll have more and more daylight hours where the sun is at a more Direct angle, and eventually the planet will begin to disappear behind the curvature of the moon, until eventually you get a clear and open sky. So as you move further away, the climate will get warmer and will allow for more "regular" flora and fauna.

 

---------------------------------------------------------

 

That's what I have to offer. If anybody is reading this and thinking "what the heck is he talking about" feel free to rip it apart. Like I said, I don't know much about any of this stuff, so I just made parallels to Earth to try and help myself along. So Kobold, I'm sorry if this post is completely useless to you   .

[Seonid]

Alright. These ones I think that I can answer.

 

First, as I look back over the thread, I think that I overstated the eclipse case. In fact, we'd only have a light/heat problem for a tidally locked moon if and only if the moon orbits without any angle away from the ecliptic - that is, we only have daily eclipses if Rimenight orbits in the plane of the planet's motion around the sun. Since most planets are tilted, and shepherd moons orbit around the equatorial plane of the planet (like the rings must do, there's no way out of this one), for this to happen, Rimenight's parent planet would have to orbit with a vertical orientation relative to its star. If the planet were tilted, the nightly eclipses would be far less frequent (with the frequency decreasing as we increase the tilt, until we get an exactly zero chance at 90 degrees tilt)

 

For comparison, Earth is tilted at 23.4 degrees, and Saturn tilts at 26.7. Jupiter tilts at 3 degrees (which may or may not be enough to prevent nightly eclipses, I'd have to do a lot more work to figure out the exact minimum tilt we'd need). Uranus has a 97 degree tilt (causing its famous vertical rings - they still orbit the equatorial plane of Uranus, it's just the plane that's tilted). Neptune tilts at about 28 degrees, Mars is chaotic - varying between 0 and 60 degrees - likely due to the lack of a moon, Venus is 177 degrees (which means that it nearly points down relative to the earth's spin, in a rather funny circumstance) and Mercury is nearly zero.

 

An axial tilt (of the parent planet, not the moon) of between 20 and 30 degrees would certainly be enough to eliminate nightly eclipses (as we see with the Earth-moon system), and thereby avoid the heat problem. In fact, it seems that the more likely occurrence would be a tilted planet, rather than a non-tilted one (I am not aware of whether we have been able to observe the axial tilt of any exoplanets - so I just have the Solar System to provide data points)

 

Just in case, we can consider the case where Rimenight does orbit a planet with a very low axial tilt. In that case, let's examine the least illuminated point on the moon, the one that is closest to the planet. (I wish I had pictures) Detailed analysis follows in the spoiler tag:

 

Let us call time 0 the moment when the moon is exactly in between the planet and its sun. Since the rotation of the moon exactly equals its orbit around its planet, we can express the orbital position in terms of days. So, we will let t = 1 day be the point when the moon has returned to its original position.

 

Under these circumstances, the first time that the least illuminated point (hereafter called LIP) sees sunlight is at t = 1/4, one quarter of a cycle around the planet. (Try this with two balls and a light source!) The last time the LIP sees sunlight is at t = 3/4, three quarters of the cycle around the planet. So now we need to figure out at what point the planet eclipses the sun.

 

Aaand, I just did a lot of math that I don't have time to put down here. Essentially, I found out that the planet will block out all of the light for about 16.5 degrees of arc, or about 10% of the day.

 

I have no idea why this conflicts with Blaze's analysis above, and I've got to get to work. It might be because the moon takes up less than 2% of the sky (a quick back of the envelope calculation gives .01% of the sky covered by the moon. The gas giant has a radius 33 times that, so it has a (cross-sectional) area over 1000 times the moon's size. That translates to gas giant taking up about 3% of the sky from planet's surface.

 

I'll think more about how much that would affect life while I'm at work, and post again on my lunch break if I have something new to say.

 

Oh, and about the tide question - the planet would have zero effect on the tides (because of the tidal locking) but the other moons would certainly have a difficult-to-predict effect. The Sun would also have an effect (about half of the effect of the moon on Earth).

Edited February 5, 2015 by Seonid

[Blaze1616]

*cough* so when I wrote my response above I was in a bit of a rush to get out of work, so I submitted, logged off, and drove home. Approximately...oh, lets just go with 2 seconds...after I logged off it occurred to me how ridiculously high 2% is. It also occurred to me that all my calculations are based on Rimenight, it's parent planet, and the star are all in line with each other along one plane, which Seonid noted. A far more accurate guesstimate of the moon's percentage of the sky would be about .01%. This would then alter my calculations for the planet to be a measly 3% of the sky.

 

...Aaaand I just bothered to open Seonid's spoiler and see he came to the exact same conclusion. Well at least I know I'm not going crazy...

 

So with this 3%, which is a much more manageable and hospitable size, I would say that the coldest parts of the planet would be the "north pole" and the "south pole," where the sun's rays are the most acute angle, incredibly Indirect. These areas would probably be about as cold as Earth's poles, though I don't know for sure due to the increased day/night cycle.

 

Due to the increased day/night cycle (36 hours each), the sun would take longer to travel across the sky. This would mean surface temperatures (of the side facing away from the planet) would reach higher points than Earth, and at the same time the extended night would provide more time for heat dissipation.

 

A part of me wants to say that the "equitorial" section of the side facing away from the planet would get hot enough during the day, and cold enough at night, to be inhospitable, at least for human life, but I can't say for certain. To avoid the harsh heat of the sun, though, any kind of human-like civilization would probably avoid latitudes less than 15 degrees.

 

And I see I completely forgot to answer your tide question, which Seonid beat me to it. That's what I get for rushing. Sorry about all that!

[Seonid]

Quick question for you, Kobold. How large do you want your gas giant from the surface of Rimenight?

 

We can't get anything sky-devouring without being dangerously close, but we have a range of values to play with. You mentioned a 3-day as the max limit you wanted to have, so I'll just post a few representative values.

 

If Rimenight had a 24-hour day/night cycle, then we would be on the order of 200,000 km out (orbital radius), and the gas giant would have a 35 degree angular diameter. (In contrast, the moon has half a degree) To get a feel for this, go outside, point your arm straight up and point your other at the horizon. That's a 90 degree angular diameter. Now move your arm pointed at the horizon up until it's about 2/3 of the way to your vertical arm. That's a 30 degree angular diameter. With a 24-hour day/night cycle, the planet would look a little bigger than that.

 

With a 48-hour day/night cycle (like in the OP), we'd be roughly 300,000 km out (coincidentally...) and the angular diameter would be 20 degrees.

 

With a 72-hour day/night cycle, we'd be 400,000 km out (which is what we've been calculating based on now) and the angular diameter is 16 degrees, or about 32-33 times larger than the Moon as seen from Earth.

 

Just a few representative values. Also, if the moon orbits directly in the equatorial plane of the planet, the rings will be nearly invisible as you would be looking at them edge-on. You probably want a slightly inclined orbit so that the rings aren't edge on to Rimenight. Just a few degrees would be sufficient, but a larger inclination would give a more dramatic view of the rings.

 

Here is an interesting tool. I haven't tried it out yet, but I thought you might find it useful.

[Blaze1616]

Also keep in mind that the with large enough inclines, eclipses will happen less. Too large of an incline though will make all life impossible on Rimenight due to constant sunlight/night making the surface too hot/cold for any kind of life. As Seonid suggests, a small incline would be best, though I'm currently trying to figure out a "sweet spot" to get the best view of the rings, yet still be hospitable.

 

------------------------------------------------------------------

 

All right, so I've created a graph to help visualize this to help decide where to place Rimenight to get/not get daily eclipses. I apologize, but I hand-did the graph and scanned it to myself...then I realized my scan didn't show the effort I put in to color code it...so I took a picture with my phone as well. I will attach both below.

 

Oh no! I can't input an image from a file...HANG ON WHILE I FIGURE THIS OUT! *page should spin as I dash off to figure this out, like in the old Batman stuff, with the spin-ey music too*

 

Edit:

 

*page should spin with the music again as I dash back* HOLY DROPBOX KOBOLD KING! THE PUBLIC FOLDER IS PERFECT FOR THIS!

 

For COLOR!!!:

 

 

For crisper image but NO COLOR!!!:

 

https://dl.dropboxusercontent.com/u/109148420/NO%20COLOR!!!.pdf

 

So, I'm going to use the color image as I'm explaining things.

 

The horizontal line in the middle (the one connecting 0 and 180) is also the line that connects the center of the planet with the center of the star. As the image notes, Rp is the radius of the planet and Ro is the radius of the orbit.The blue circle (outer most circle, it's hard to see in the black & white image) is the orbit of Rimenight. The black circle (inner most circle) is the planet. The two red dots on the blue circle represent Rimenight.

 

The green line represents the "edge" of the light/heat from the star. This green line, then, is tangent to both the black line used to represent the planet, and the undrawn circle representing your star, way off to the left of the graph. This graph makes one big assumption though, that the green line is horizontal.  This is unlikely, and so theta is going to be smaller than what this graph shows. This graph is a good rough estimation though.

 

So what does theta mean? Any point of Rimenight that is above the angle theta will no longer get daily eclipses all year round. Instead, eclipse "periods" will occur twice a year, where for some nuimber of  days there will be an eclipse per day.

 

To make visualization of the graph easier to comprehend, the graph is a scale "model" if the following parameters are true:

Rp = 50,000 km (Saturn's radius is ~58,000 km)

Ro = 400,000 km (this means distance from the planet is ~350,000 km, making the day/night cycle equivalent to 2.5 Earth days)

Radius of Rimenight = ~6,000 km (same as Earth)

This makes the radius of each of the red, untraced circles on the graph 50,000 km larger than the circle it immediately envelops.

 

With those numbers, and the assumption I mentioned, the graph I made accurately shows theta to be ~7.5 degrees. Now, this number should be smaller, because the size of your star matters too. Unfortunately I cannot graph this as 1) my graph isn't big enough, and 2) It would be best if the much larger graph that would fix problem 1 had a second origin for drawing said sun. Picky, I know, but I don't have that, so mental picturing will have to suffice.

 

Our Sun has a radius of 695,800 km, and assuming your planet is about the same distance from it's star as Earth is from ours, that's 149,500,000 km away.  This could cause the green line (which is tangent to both the planet and the star) on my graph to decline down towards the 0 degree point on the blue orbit line. This would in turn decrease theta. I feel this is most probable, however the incline of the green line might be incredibly small to the point of being negligible, which would make my graph accurate.

 

So if you want to know an accurate theta, Kobold, I'll need the radius of your star, the radius of your planet, the radius of Rimenight, the distance of the planet from your star, and the distance of Rimenight from your planet. I would also then need to refresh on my very weak geometry skills (particularly because arc math was my weakest bit in geometry).

 

Edit 2: Sorry for all the little edits I've made across the day, I've just been fixing up the post for grammar and some typos and whatnot.

Edited February 6, 2015 by Blaze1616

[Seonid]

You're fine, Blaze. That's actually a rather brilliant solution to the eclipse issue. (Unless, of course, Kobold wants daily eclipses...but that's a different story).

 

Geometry's one of my weak points too. I far prefer calculus (I'm good at algebra, but I hate it).

[Blaze1616]

As you note, it depends on what Kobold wants, but either way theta is an important thing to note. Also keep in mind that if the incline of Rimenight's orbit is too steep, there will come a point where parts of the planet will get barraged by constant sunlight for parts of the year, and lack of sunlight the rest of it. Those inclines though would occur closer to 90 degrees than 0 degrees though, so keeping the number relatively small would not be an issue.

 

Also keep in mind, Kobold, that if Rimenight's orbit angle is greater than theta, the "north pole" would be similar to Earth's poles, and if all of Rimenight is above theta then the "south pole" would be the same as well.

 

As for math, I'm with you Seonid. I'm far better at calc, algebra, and differential equations, though that's because I haven't taken geometry since I was a freshman in high school. I've been doing calc and algebra every year since I was a sophomore in high school. I particularly like algebra though.