Quick note on terraforming Galilean moons

This post comes from an "Ask Tsana" comment.

Sam Keola asked:

Aloha from Hawai'i again Tsana! I have a hypothetical question. If in the very distant future we had the technology to terraform, would it be best to terraform Callisto and Ganymede or set up domed bases? Ganymede is suppose to have an ocean similar to Europa, but I'm not sure if that's "world wide". Your thoughts on terraforming!

The main problem with terraforming either of those moons is their gravity isn't large enough to keep any atmospheric gases for long after they're introduced. Ganymede, which is larger, has a surface gravity of close to a seventh of Earth's which is less than half of Mars's and Mars has difficulty keeping much of an atmosphere itself. Purely from that point of view, domes or something else sealed would be better.

Callisto.
Credit: Galileo Project, Voyager Project, JPL, NASA

Once you've decided to build something sealed, then it would be better for colonists to build on Ganymede, as opposed to the other Galilean moons, for a few reasons:

• It has the highest surface gravity, not by much but every little bit would prevent colonist's bodies from degrading. Actually, because the Galilean moons are less dense than Earth's moon, they have a lower surface gravity, despite being larger in volume. You're going to have low gravity-related heath problems in any case, however.
• It's not as close to Jupiter as Europa (and Io!) is. The phenomenon responsible for keeping Europa's interior liquid is tidal friction thanks to its proximity to Jupiter. It's the sort of thing that also makes the surface more unstable (prone to volcanoes -- not as much as Io, of course -- and quakes) and less hospitable to people. You can read more about it here.

On the other hand, if what you're doing is mining and the minerals etc you're interested in are found on both Ganymede and Callisto, then Callisto is the place to put your colony. It's gravity slightly lower and, more importantly, it's further from Jupiter, meaning that when you're exporting your rocks, there's less gravitational pull from Jupiter to overcome.

In terms of finding water to mine, all three moons in question (ie, not Io) have water on them, so that shouldn't be too much of a problem, especially if you're already planning to mine other things.

Of course, there are also reasons why Europa would be a desirable place for a colony, especially for scientific reasons, exploring it's subsurface ocean primary among them. There's a good chance there's microbial life there.

So there you have it, if you're going to colonise the larger Galilean moons, it's better to build a close structure on them rather than try to impart an atmosphere. It would be even harder than giving Earth's moon a permanent atmosphere.

Other Foreign Skies

This post is a response to a question I got on my Ask Tsana page.

Sam Keola asked:

Love the views of Jupiter from Ganymede and Io. How large would it appear from Europa or Callisto? And how large exactly would the sun appear? (I know tiny as hell, but another lovely picture would be amazing.)

The mathematical answer to that is explained in this old post. And my first set of Jupiter images (Io and Ganymede's skies) can be found here.

Jupiter

This time around, I used a different image of Jupiter so if you're wondering why it's rotated relative to the old pictures, that's why. For the Jovian images, I've used the same starting image because in the year since I last did this, I haven't managed to take a more suitable photo. Such is life.

The original photo with a full moon in Earth's sky.

So. Europa is the second Galilean moon out from Jupiter. It's made mostly of ice, is the smallest of the Galilean moons and might harbour life in its subsurface liquid ocean. The diameter of Jupiter as it would appear in the Europan sky is almost 24 full moons across. Remember that Europa's sky wouldn't actually look blue either since it doesn't have an atmosphere but I don't have a decent night skyline to work with. I'll do a night version eventually.

The size Jupiter would appear in Europa's sky. Or in Earth's sky if you swapped it with Europa.

You might be wondering whether Jupiter would actually be oriented the way it appears in these images. Well it depends. The direction the bands run relative to the moon's horizon would depend on where on the moon you were. Close to the equator, the bands would be vertical (although if Jupiter was high in the sky, it would be pretty difficult to tell. Perhaps better to say east-west). If you were near a pole, they'd be horizontal as in these images. And remember, the Galilean moons are all tidally locked, so Jupiter would never move, just change how much of it was illuminated by the sun.

And Callisto, the most distant of the Galilean moons. Callisto's Jupiter would appear "only" about 8.5 full moons across.

The size Jupiter would appear from Callisto. If Callisto had an Earth-like atmosphere and gum trees.

The Sun
 
The second part of Sam's question was how large would the sun appear from Jupiter. Well, on Earth, the sun and the moon appear to be approximately the same size (there's a little bit of a difference when the sun is at its closest and the moon at its furthest and vice versa). So the sun from Earth is about one full moon in diameter.

From Jupiter (or its moons) the sun would appear about 0.4 full moons across which is a little bit less than a sixth of the area of the sun as seen from Earth (remember, the moon and sun seen from Earth are on average the same size).

I cheated a little bit with these next two sun photos. They're actually two separate photos and I made the sun smaller in one of them. The reason the rest of the photo looks darker for the Jovian sun is because I was fiddling with settings on my camera. And if you're wondering why I chose sunsets, it's because those (and sunrises) are pretty much the only kinds of photos where the disc of the sun is properly visible.

Ordinary sunset on Earth:

Sunset. A little bit more than half the sun is below the horizon.

Sunset if Earth was at the same distance as Jupiter (but yet still warm enough to have liquid water. And plants. By the way, with these two, it's probably clearer if you click on the images to enlarge and compare the sun side by side.

A more diminutive sun, less than a sixth of the area of Earth's.

And there you have it. Photoshopped images (well, actually, I used Pixelmator) depicting the sizes of Jupiter and the sun from the Galilean moons and the Jovian system, respectively.

Foreign Skies: Daytime

When I wrote my first post on this blog, I wanted to include some photoshopped images of to-scale Jupiter hanging in the sky as it would above Ganymede and Io. At the time, I didn't have any good photos of the moon to compare with and paste over and I didn't want to steal something from Google Image Search so I put it off. I still don't have any good, cloudless photos of the moon at night with a suitably urban back drop. Instead, I decided to put my (very average) photoshopping skills to use and make a daytime Jupiter in the sky.

A caveat: Ganymede and Io both lack atmospheres that even remotely resemble Earth's. As a result, you wouldn't get a blue sky on either, even under a biodome of some sort (there wouldn't be enough air in the dome to have that effect). Instead the sky would be black outside of the giant orb of Jupiter (which, when full, would definitely be bright enough to make it hard to see stars).

So you shouldn't treat these images as "what the sky would look like on Ganymede/Io" more as an indication of how large Jupiter would be in the Earth's sky if you put it in Ganymede's/Io's place.

First! The original photo with the Earth's natural moon left in (taken with my phone, so it could be awesomer, but it served its purpose):

Unlike Earth, Ganymede and Io are unlikely ever have gum trees. Just saying. (Click to enlarge.)

For the remaining compositions, this APOD image is the shot of Jupiter I used. Credit to NASA, ESA, and E. Karkoschka (U. Arizona).

Next up, Jupiter as seen from Ganymede. I left the moon in for the first one, just so you can visually compare sizes better:

To scale, albeit physically unrealistic. I don't think Jupiter would quite be that colour either (particularly the edges wouldn't be so dark, but I couldn't fix it) and of course the lighting is all wrong. (Click to enlarge.)

Just Jupiter alone in the Ganymedian sky:

Jupiter in Earth's sky if Earth was in Ganymede's orbit around the gas giant. (Click to enlarge.)

And, last one, the size of Jupiter in Io's sky. Large doesn't really cover it.

Jupiter looming as though over the Io skyline. Of course, Io is even less likely than Ganymede to have apartment buildings on it, but shh! (Click to enlarge.)

And there you have it. Once I have a decent night skyline with the moon in it, I will repeat this but at night. It will be significantly awesomer. I just have to get some night photography in first.

EDIT: I have made some similar images for Jupiter from Europa and Callisto and the sun as seen from the Jovian system. See my new post here.

A couple of cool Io animations

So I was browsing around NASA's photo archive looking for a nice high-res image of Io and I came across this little movie of Jupiter-shine on Io. It's a time-lapse movie in which you can see the Io moving so that the sun is on its far side — it starts off behind and to the left which is why you can see a bright crescent at first. As the sun moves behind Io from the camera's point of view, Jupiter, which is behind the camera becomes more fully illuminated. The sunlight bouncing off Jupiter in turn more brightly illuminates Io's surface which is why the night side of Io gets darker as the crescent of Io gets smaller. Pretty cool, eh?

You can read more about planetshine in this post, which also talks about how much light you'd get from the sun at different distances.

Another cool little movie on the same site is this one, which shows some interesting happenings on Io's dark side. The bright spots away from the edges are volcanoes, whereas the two blue glows on the edges are similar to aurorae on Earth. Except those aren't the poles, but rather at the equator where charged particles are colliding with the tenuous wisps of Io's atmosphere. The thin atmosphere itself is only existent because it's constantly being replenished by volcanic gases. Nifty.

Living on a moon: How bright is the night?

Let's say you've stuck a colony on the moon of a gas giant. I've already talked about the unusual way in which the sun and the primary planet move (or don't move) across the sky. As you might recall, there will be times in the moon's orbit when, depending on where you are on its surface, the only natural illumination comes from its primary planet. the question this post addresses is: just how much illumination can we expect?
There are two things we need to know to work out how much illumination the primary is giving the moon:

1. How bright and far away is the sun?
2. How reflective is the primary?

EDIT: I've added in some comparisons with light bulbs thanks to Patty Jansen pointing out that the human eye can adapt to see in lighting conditions much dimmer than the sun

Star light, star bright?

The amount of light that reaches your planet-moon system from its sun will depend on what kind of star it is. Stars come in different sizes and different temperatures. Most stars lie on what is known as the Main Sequence. Two notable exceptions are red giants and white dwarfs. The main sequence refers to the band of stars running diagonally through the Hertzsprung-Russell Diagram (HR diagram, previous links to two different images). Is basically a plot of how much light a star gives out (it's magnitude or luminosity) against it's colour or temperature. Stars are then divided into types (O, B, A, F, G, K, M) based on colour/temperature. Giant stars (other than blue giants) lie above the main sequence and white dwarfs lie below it. The sun is a G type star with temperature 5800 K (on the surface, that is; it's much hoter on the inside). K means Kelvin and is the standard unit of temperature. To convert from Kelvin and Celcius, you need to subtract 273, so the sun is 5500ºC (with rounding).

Using a star's temperature we can work out how much energy, in the form of light, reaches our planet. The first step is to assume that the star is a black body. This might sound conter-intuitive since the last word you're likely to use to describe the sun is "black", but from a physics perspective, a black body is something that absorbs all incident light and emits light based on its temperature. Well, I say "light", but really I mean electromagnetic radiation.

The Stefan-Boltzmann law tells us how much energy a black body emits based on its temperature. When we're talking about stars, this is called the luminosity. The formula for calculating luminosity is:

A stars luminosity, given it's radius, R, and temperature, T. σ = 5.67 × 10^-8 is the Stefan-Boltzman constant and π = 3.14

The temperature has to be in Kelvin and the radius in meters to give luminosity in units of Watts (yes, like your light-bulbs) which is a measure of energy emitted per second. Radius and temperature are slightly trickier to come up with numbers for. If you're using a real star, you can just look it up on Wiki or Wolfram Alpha (Wiki even has a page listing the nearest stars to Earth). Otherwise you can make up a star with the characteristics you want such as temperature or class, then go to the second HR diagram I linked and look at the diagonal lines of radius. Whether you want a main sequence star, white dwarf or giant, this should give you an idea of radius (in units of the radius of the sun).

That's all well and good, but what we actually want to find is the light reaching a planet, not the total light emitted. Because stars emit light in all directions at once, their total energy output end up being diluted over an expanding sphere of light. Basically, not all the energy the sun produces hits our planet. It depends on how far away the planet is. This next formula will tell us how much energy hits the planet:

P is the energy per second hitting each square meter of the planet and D is the distance from the planet to its sun.

So P is the energy from the sun that hits a square meter of a planet which is D meters away from the sun. We're not quite there yet, but let's take a break and calculate some numbers. I'm going to work out the energy from the sun that hits the Earth/moon and Jupiter each second.

• Earth/moon are about 1.5 × 10¹¹ m from the sun. The sun's radius is 6.955 × 10⁸ m and its temperature is 5800 K. The energy hitting a square meter of the Earth or moon each second is 1400 Joules.
• Jupiter is 7.8 × 10¹¹ m from the sun. The energy hitting a square meter of Jupiter each second is 50 Joules, which is about 3.6% of the energy hitting the Earth. Jupiter's greater distance from the sun means that the sun's energy is about 30 times more spread out by the time it gets there. (As I calculate below, this is still about 14000 times brighter than the full moon as viewed from Earth.)

Planetshine

Light doesn't get completely absorbed by the planet, however. Some of it reflects back out into space and can illuminate other nearby objects. The property which determines how reflective something is (in this context) is called albedo. The average albedo of a planet is a number between 0 (non-reflective) and 1 (absolutely reflective), which represents the percentage of incident light that will be reflected.

In practice, it's fairly easy to implement albedo. The reflected energy is the incident energy multiplied by the albedo. Just multiply P above by albedo, A, and you get the power reflected off each square meter of planet. You can look up albedos for different planets/moons on Wiki and elsewhere. (But we all know Wiki's the easiest. It lists albedos in the summary box on the right of the relevant page. If more than one is given it's the Bond albedo, not the geometric albedo, that you want.)

What we actually care about, however, is how much of that reflected light goes on to reach the moon our colony is built on. In a way, we just reuse the equations I've already included above. Instead of putting L into the equation for P, use the P from the sun multiplied by albedo, radius becomes the radius of the planet, and distance is now the distance between planet and moon:

P is the energy per square meter per second hitting a moon, A is the albedo of the planet, R is the radius of their sun, r is the radius of the planet doing the reflecting, T is the temperature of their sun, d is the distance between planet and moon, D is the distance between planet/moon and sun. The last line is included because if you're using a real star, luminosity will probably be listed somewhere. Otherwise, the penultimate line is what you need to use.

OK, so this is getting increasingly more complicated looking, but remember that you only really have to do the last step. There rest are only there by way of explanation.

Now, one last thing before I calculate some more numbers. That last equation assumes that the primary planet appears full in the sky. If it's half full, you have to halve that number, if it's a quarter full you have to divide by four. Honestly? Just approximate.

• The moon has an albedo of 0.136. The energy the full moon is reflecting at the earth is 0.0037 W/m².
• For the purposes of comparison, a 100 Watt light bulb from 10 meters away has a brightness of 0.02 W/m².
• The Earth has an albedo of 0.306. The energy Earth reflects at the moon is 0.12 W/m². So because it's bigger and more reflective, the Earth as seen from the moon gives off about 32 times more energy per second. That means the full Earth in the lunar sky is roughly 32 times brighter than the full moon in Earth's sky and six times brighter than a 100 W light bulb.
• Jupiter has an albedo of 0.343. Ganymede is 1.1 × 10⁹ m away. The brightness of full Jupiter in Ganymedean sky is 0.07 W/m². That means Jupiter is almost twenty times brighter than the full moon. Not surprising given how big it is in the Ganymedean sky. A half-full Jupiter would be 10 times brighter than the full moon, a quarter-Jupiter about 5 times as bright and so-forth. The varying quantity here is what fraction of Jupiter's disk is illuminated (and that we're working under the assumption that Jupiter reflects evenly in all directions). A quarter-full Jupiter would be about as bright as a light bulb and a full Jupiter would be as bright as three and a half light bulbs 10 meters away.
• Io is 4.2 × 10⁸ m from Jupiter. The brightness of full Jupiter in Io's sky is 0.48. So Jupiter is shining a whopping 130 times brighter than the full moon. By comparison, the sun as seen from Io is only about 100 times brighter than Jupiter. Light-bulb-wise, Jupiter would be as bright as 24 100W light bulbs 10 meters away.
• For a bit of fun, the brightness of full Io (albedo 0.63) as seen from Ganymede varies from 1.2 × 10^-4 W/m² when it is at its closest point to Ganymede to 2.4 × 10^-5 W/m² when it is at its furthest. Neither of those are very bright, but it would still definitely be visible. It's about 0.6–3% the brightness of the full moon.
• And finally, let's say we put Jupiter at the same distance from the sun as Earth is. Now Ganymede would get around the same amount of energy from the sun per square meter as Earth does and Jupiter would be a lot, lot brighter. How bright? 1.9 W/m², which is 500 times more light that Earth gets from the moon and as bright as almost 100 light bulbs from a distance of 10 meters.

And there you have it. A method for approximating how much light you'd get reflected from a gas giant planet (or whatever planet/moon/asteroid you like). Unfortunately this post ended up being a little bit more complicated than I had initially anticipated (where complicated really means more maths), but it's a small price to pay for painstaking accuracy... Well, some semblance of accuracy, at any rate. There are a lot of approximations in the above (for example, the albedo varies for different types of terrain; so Earth's albedo is higher over clouds than over forest), but on average, it's close enough. Phew!

One last thing I came across after writing this post. I was looking for something else and came across this photo of Jupiter and Io. Notice how the line between Io's sun side and dark side (called the terminator) is very distinct and solid, whereas Jupiter has a bit more of a gradient going from light to dark? This is because Jupiter has an atmosphere (a very thick one, but the effect applies to Earth's atmosphere too) whereas Io's atmosphere is whispy and not really much to write home about. The atoms/molecules/particles in the atmosphere reflect light in all directions, allowing it to diffuse through a bit, giving us that gradient from light to dark. Io, on the other hand, only reflects light off its surface, leading to the solid terminator you can see in that image. Just something to think about when writing those realistic descriptive passages. ;-)

Update: I photoshopped some Jupiters into skies to give a size comparison with the full moon. You can see them here.

Living on a moon: Marking time

There is a certain class of exotic location often used in science fiction and that is the surface of a moon.

Most of what I'm going to say will apply to moons like Earth's but I'm going to focus on the moons of gas giant planets like Jupiter and Saturn partly because they're a little more interesting and partly because if you want to know about day and night on the moon, it's more trivial to look up.

Planets orbiting a star

First, let's talk about ordinary (Earth-like) planets orbiting a star. They will have a year defined by how long it takes them to do a complete orbit of their sun and a day defined by how long it takes them to spin on their axis. Actually, there are two possible definitions of a day:

• the solar day, which is how long it takes the planet to rotate all the way around so that the sun returns to the same place in the sky (or more accurately, until it returns to the same point above the planet. On Earth, the meridian passing through Greenwich and the middle of the Pacific ocean is the reference point we use).
• and the sidereal day, which is how long it takes the planet to rotate about its axis so that the stars return to the same position in the sky.

On Earth, a sidereal day is slightly shorter than a solar day (only 23.9 hours) and this will be true of any planet that spins in the same direction as it orbits. So the Earth, looking down on the north pole, spins anti-clockwise and orbits the sun anticlockwise. Such a planet is in a prograde orbit. This is true of all the planets except Uranus, which is sideways, and most of their moons. It is in general going to be true of all systems if they formed together (thanks to conservation of angular momentum) and if a planet isn't prograde (that is, if it's particularly lopsided like Uranus or if it's retrograde meaning spins or orbits in the opposite direction) then it probably has a more interesting history. In the case of Uranus, it is thought that some collision knocked it sideways a long time ago. For retrograde planets, where one each of orbit and rotation are clockwise and anticlockwise, the implication is that they did not form where they are found, but are interlopers from elsewhere. Or there could also have been a collision, but it would have to be a very large collision in exactly the right place. It's interesting to note that all the planets orbit in the same direction as the sun rotates. This is strong evidence that they all formed from the same nebula at roughly the same time.

Moons: Mostly tidally locked

OK, enough background. On to the moons. Let's assume we have a rocky moon orbiting a gas giant planet. All the interesting moons in our solar system (which is to say, the ones I checked and generally most or all of the big ones) are tidally locked with their primary, including Earth's moon. What does tidally locked actually mean?

I won't go into the details of the physics, but if a satellite is tidally locked with its primary, the same side will always face the primary. So on Earth, we always see the same side of the moon. If you go to the moon and land on the near side, Earth will always be in the same place in the sky (assuming you don't travel far from your landing place) varying only in how much of it is lit up by the sun. It's also possible to have planets tidally locked with their sun, but they have to be quite close to their sun for this to happen. Consequently, most of those planets wouldn't be habitable for humans, unless the star in question was a red dwarf, but that's a topic for another blog post.

Back to our rocky moon orbiting a gas giant. Since it's tidally locked, you will need to decide where you want to place your colony/city. Directly under the primary planet so that it always sits high in the sky? On the side of the planet which never sees the primary? These choices will depend a bit on your whim and a bit on the purpose of the colony. For the latter, if it's a research installation studying the primary or a mining installation skimming gas from the primary's atmosphere, it makes the most sense to build it directly below the primary. On the other hand, if the research installation is built for astronomy observations, you'd want to put it on the non-planet side so that light from the sun reflected from the primary interferes with your telescopes less.

Days and nights?

Once you've made that decision, you probably want to know how long days and nights will be on your moon. This is where it gets a bit tricky. I'm going to use Ganymede, one of the larger moons of Jupiter, as an example. Thanks to its synchronous orbit (another way of saying that it's tidally locked), a sidereal day on Ganymede is the same as it's orbital period. Orbital period is the general term for how long it takes to orbit all they way around Jupiter. (I'd prefer to say "Jovian day", but unfortunately that term refers to one of Jupiter's solar days. :-/ ) So unless it orbits very quickly, orbital period would not be a useful measure of time to base diurnal cycles on. And if it did have a fast enough orbit, it probably wouldn't be very habitable since that would imply that it was very close to the primary like Io (the innermost Galilean moon of Jupiter), leading to a host of problems like extreme volcanism and earthquakes. As I hope the crude sketch I did below helps illustrate, a solar day on Ganymede (that is, the length of time it takes for the sun to move all the way across the sky and come back to its starting point) is also the same length as an orbital period.*

Not to scale! Top right circle is the sun, orange circle is Jupiter with the lighter half the half that is illuminated by the sun and the darker brown half the dark side. The grey shadow is Jupiter eclipsing the sun and the rainbow circle is Ganymede, so coloured to illustrate that the same side is always pointing towards Jupiter. The thick black line shows its orbit around Jupiter and the light and dark semicircles inside Jupiter's orbit are to help guess how full/dark/crescent/gibbous Jupiter would appear in Ganymede's sky (if you're on the side of Ganymede facing Jupiter).

We can also use that diagram to work out how much of Jupiter would be lit up by the sun if we're on a the side of Ganymede facing Jupiter. It should also be noted that, unlike the Earth being lit up by the moon and human lights at night and hence being visible from the moon even when it's not lit up by the sun, the dark side of Jupiter would be completely dark. Against the black sky of Ganymede (and the sky would always be black, even during the day, since Ganymede has no atmosphere to scatter photons with) it would just look like a black hole in the stars. A black hole about 15 full moons across.

*Technically it would be slightly less thanks to Jupiter's orbit around the sun, but Jupiter is so far out from the sun and has such a large distance to travel that day to day we can ignore the small difference to the length of a Ganymedean solar day. If your gas giant is much closer to its star, it might become relevant, but this calculation is left as an exercise for the reader. ;-)

Time moves forward

Finally, it would be useful to work out how quickly Jupiter and the sun change in Ganymede's sky, especially if you're writing a story that involves spending longer than a day there. I will make this section more general so that you can use for any hypothetical moon orbiting an arbitrary gas giant.

What you need to know or decide is the orbital period, let's call it T,  a piece of paper with your own approximation of the diagram above (without all the different positions of Ganymede drawn in yet), and a protractor (or a really good eye for angles). For Ganymede, T = 7.15 (Earth) days. If you're making up a planet-moon system of similar size, it's probably best you're numbers don't deviate too much. I think I might make the proper physics you need to consider when making up planets the subject of a future blog post.

On your hand drawn diagram, choose a starting position for your planet and a location on the surface for your colony. I suggest putting your colony close to the equator because a) it will be more picturesque and b) Ganymede has some crazy magnetic fields and I suspect that radiation shielding would be easiest to achieving within about 30º latitude of the equator. This doesn't automatically apply to all moons in similar systems, but still, it can't hurt. Draw your moon in it's starting position and mark the location of your colony with a cross or something. Remember that looking down from above the north pole, the moon will probably be orbiting anticlockwise if it's in our solar system.

Next, you need to do a small piece of maths. Decide how much time you want to pass before you mention what the gas giant is looking like in the sky again. Call this time t. Make sure T and t are in the same units (convert them both to days or both to hours, whichever is more convenient, if they don't match). To work out how many degrees, d, of a circle the moon has moved in this time, you need to use the following equation:

d = 360*t/T

In one Earth day, Ganymede will move d = 360*1/7.15 = 50.3º which is a bit more than an eight of a circle. On the diagram above, that's a little bit more than the distance between two consecutive rainbow Ganymedes (ignoring the two close together in Jupiter's shadow). Since T is so small for Ganymede, this means that Jupiter and the sun change quite dramatically in the sky (Earth) day to (Earth) day. Depending on how your planet-moon system is set up, your mileage may vary.

Multiple moons

And a quick bonus calculation: if your planetary system has multiple moons your feel like caring about, you can do the above calculation for each of them, choose starting points and then see how far each one moves in the span of time you're interested in. This doesn't need to be very hard at all. In the jovian system, Io completes four orbits in the time it takes Ganymede to complete one and Callisto completes two in the same time. This convenient state of events is thanks to the physical principle of resonance. Resonance happens in all sorts of places in nature and celestial mechanics, including Saturn's rings and Mercury, so feel free to implement it with impunity.

Hopefully, I've given you enough information to convincingly set a story on a moon orbiting a gas giant planet. Well, in a colony at least, where you don't have to worry too much about external climate, so long as you stay away from Io.

Foreign Skies

Living on Earth, we are used to seeing the moon in the sky. Other planets, real or imagined, are unlikely to have moons identical to the Earth (well, ok, the imagined ones might). Other moons will look different and will take up different amounts of the sky. Today, I calculated how big various celestial bodies would appear from the surface of other celestial bodies and compared this with the size of the moon as seen from Earth.

In case you want to try this yourself, I will briefly explain how I did it. First I found the angular diameter of the object in the sky using the equation below. For objects that are much smaller than the distance they are away (so if radius divided by diameter is less than about 0.1), you can also use the small angle approximation, in which you can ignore the tan part and the angular diameter is approximately 2r/d.

Here, r is the radius of the object in the sky and d is the distance it is away. Both r and d should be in the same units,  that is, both in km (or miles). If you have your calculator set to degrees, then α is the angular diameter of the object in the sky.

 Now, because angular diameter doesn't really mean much on an intuitive level, I then divided this angle by the angular diameter of the full moon which is about 0.5 degrees. This gives me the number of full moons you could line up across the middle of the object. If you're interested in the area of sky covered by the object (in units of the full moon), then just square this number.

The Earth seen from the moon

The Earth is a lot bigger than the moon, so naturally it would look bigger if you were standing on the moon. In fact, the Earth seen from the moon is 3.6 times the size of the full moon. APOD have a nice photo for us of this scenario.

Living on Ganymede

Ganymede is the largest moon of Jupiter, and the third furthest away of the four Galilean moons. Ganymede is about three times as far from Jupiter as the moon is from Earth. However, Jupiter is very large (the largest planet in the solar system, in fact) and if visible from Ganymede's surface, would dominate the sky. As in happens, Ganymede is tidally locked with Jupiter, putting it in a synchronous orbit and meaning that the same face is always turned towards Jupiter. Unlike our moon rising and setting in the sky, it you stood on a part of Ganymede facing Jupiter, the gas giant planet would not move in the sky at all. Also, you could use Jupiter to navigate (even if it was directly above you, although it would be a bit trickier in that case) because the bands in its atmosphere run from east to west (here is another nice photo from APOD, which also shows Ganymede in the distance).

So, standing on the surface of Ganymede, how big would Jupiter look? Very big. Fifteen full moons across big. It's a bit hard to imagine and if I find a nice photo of a full moon above a city or other recognisable landmark, I will photoshop Jupiter in and post it here. Jupiter looks so big that you could spread 225 full moons over it and only just cover it up. Yeah. That big.

I also worked out how big Io, the closest Galilean moon to Jupiter, would look from Ganymede. Io is covered in volcanoes and sulfur and looks yellow. When Io is lined up between Ganymede and Jupiter, it will be 0.6 times the size of the full moon; a bit more than half. I initially decided that when Io was on the far side of Jupiter, even if it wasn't behind the parent planet, it would be too small to be very exciting, but doing a quick calculation I found that it would be about a third of the size of the full moon, so probably still big enough to be noticeable.

Phobos for a bit of fun

For a bit of fun, I also worked out how big Mars would look if you were standing on Phobos, one of its asteroid-like moons. Before doing this calculation, I never really appreciated how close in Mars' moons orbited. Firstly, Phobos is only about two Martian radii above the surface of Mars. This means that the small angle approximating definitely won't work when calculating Mars' apparent size. So how big does Mars look? Its angular diameter is a whopping 140º. That's the width of 280 full moons lined up next to each other. In fact, Mars would take up nearly the whole sky on Phobos. I think that's big enough to feel like you're falling towards it, particularly since Phobos has a very puny gravitational field. Scary.