Bunch of links, mostly outdated

First up, square Earth anyone? Forget realism, let's just have a think about what an Earthlike planet would be like if it were a cube. Puts me in mind of the planet builders in Hitchhikers' Guide to the Galaxy. Brought to you by Discovery News.

Second, the Planetary Habitability Laboratory talks about brightnesses of the various planets in the solar system and also of exoplanets. An interesting read, particularly if you enjoyed my old How Bright is the Night? post.

Martian moons eclipse the sun in these NASA photos from the Opportunity rover:

Credit: NASA/JPL/Cornell

Enceladus pics from Cassini. Enceladus is one of Saturn's moons, most famous for it's ice geysers.

Proposed space robot to cannibalise old satellites which have previously been boosted up to "graveyard" orbits. Many mentions of zombie satellites and grave robbing associated with this one ;-p . From New Scientist.

And finally, laser driven fusion in California. From New Scientist again.

Happy weekend, gentle readers!

A correction: You can't fall off Pan.

While writing a story which happens to be set on Pan, one of Saturn's moons, I realised that I had made an erroneous statement in a past blog post. Of course, I had to correct it.

In that post I made a passing comment that on Pan, the gravitational force of Saturn is greater than that of the moon itself. That statement was true. However, I went on to say that on the Saturn-side of Pan, you'd fall off because the gravity of Pan wasn't strong enough to overcome Saturn's gravity. This last part isn't quite true.

If you were on Pan, despite its weak gravity, you would be hurtling around Saturn at the same speed that Pan does, which means that you would automatically be going fast enough to stay in orbit (your centripetal acceleration would be balancing Saturn's acceleration due to gravity), no matter which side of Pan you were on. Admittedly, the low escape velocity (about 25 km/h) and the slight difference in gravitational pull from Saturn between the near and far ends would make it easy to fall off the planet and slowly spiral in towards Saturn, but you certainly wouldn't be falling upwards.

The more accurate statement I should've made was that if you were floating around in the vicinity of Pan's orbit and Pan came past you, its gravity would not be strong enough to pull you in over Saturn's gravity. No matter how close to it you were (even if you could touch the surface), if you were not already moving along with it, then Saturn's gravity would win out and you would fall towards Saturn, rather than towards Pan.

The original post has been amended to reflect the above correction.

Conquering the Horizon

And now for something completely different. The horizon; how far away is it? How different would it look on another planet? We're used to horizons on Earth but if we're writing a story set on an asteroid or on a small moon or planet the horizon will be closer because the planet's surface falls away more quickly.

Distance to the Horizon

When I talk about the distance to the horizon, what I mean is if you're somewhere flat, what's the furthest you can see (including with the aid of binoculars or a telescope etc)? On Earth a good example of this would be how far away the point where the sea meets the sky is, when you're standing on a jetty. Once we have trees and buildings in the way it can get a bit more complicated. Luckily, the sort of extraterrestrial locales where the horizon is going to be most different to Earth's are least likely to have a large abundance of trees. Convenient.

Working out how far away the horizon is takes a little bit of trigonometry. I've drawn a sketch below of all the relevant distances and whatnot.

R is the radius of the planet/moon, h is the height of your person (well, of their eyes) or if they're in a building, it can be how high up they are, d is the straight line distance to the horizon, s is the distance along the surface of the planet/moon and θ is an angle that will be useful in some calculations.

The important thing to that you need to know about your non-Earth planet is how big it is or, more specifically, it's radius which is labelled R in the image above. It's a reasonable assumption that you'll have at least a rough idea of how tall your characters are. If you don't, it doesn't really matter, you can just guess something close since there's not going to be much difference between a tall person's horizon and a short person's (the differences really come into play in non-horizon situations, such as crowds). For the purposes of my calculations later on, I'm going to set h = 1.7 meters. Because I can.

Now, that's a right angle between the line I've labelled d and the left hand radius line. Since we know R and h we can now use Pythagoras's Theorem to work out the distance d. Don't worry if you don't remember any maths, I'm just going to tell you the answer.

Chances are, your planet is significantly larger than your person, so you will usually be able to ignore the h² but not always (if on a small asteroid, for example). If in doubt, leave it in. It won't make your answer worse.

This is not an unhelpful result. However, I can't help but feel that when people stand in a tall tower and say things like "They're ten miles away but gaining ground!" they don't mean ten miles from their eyes, but ten miles from the bottom of the tower (if nothing else, they'd probably be estimating based on land marks and those are definitely relative to the ground distance).

So how do we find the distance along the ground, s? Unsurprisingly, with more maths. We use the fact that s = Rθ and then work out θ so we can substitute for it and not have to actually calculate it directly. Using the same triangle as before, we can find s in two different ways:

If you're wondering, cos and tan are trigonometric functions all scientific calculators (including the ones hiding in all your computers) can do. The ^-1 indicates that's it's the inverse of the function which you can usually access by pressing shift/2nd or something like that, depending on the calculator.

For planets/moons which are much, much larger than a person, s and d will be very close; it's the tiny, tricksy moons or asteroids are where it'll really make a difference.

So how far?

Some examples now for a person 170 cm tall and for a ten storey building (30 metres high):

•  On Earth, ignoring atmospheric effects which actually extend the apparent horizon thanks to bending light, the horizon is 4.7 km away. From a ten storey building it's 19.7 km.
• On the moon or Io, which are similar in size, a standing horizon is just under 2.5 km and the ten storey building horizon is about 10 km.
• Ganymede and Titan (moons of Jupiter and Saturn, respectively) are a bit bigger than those two, with standing and ten storey building horizons of 3 km and 12.5 km.
• Mars is about one and a half times the size of Ganymede and a bit more than half the size of  Earth. It has horizons 3.3 km and 14 km for standing and building respectively.
• Deimos, Mars's moon, is rounder than Mars's other moon, Phobos, but still not that round. If you stand on a fortuitously round bit, the horizon will be 140 meters away (that's right, metres not kilometres—Deimos is actually an oblong with dimensions only 15⨉12.2⨉10.4 km across. Its average radius is 6.2 km). If you somehow managed to put a 10 storey building on it... well you'd see about 600 meters away.
• Ceres is a large, round asteroid (or dwarf planet) in the asteroid belt. It was one of the bodies that, when Pluto's planethood was called into question, was up for being classified as a planet if Pluto got to stay. (If you're wondering, it is considerably larger than Deimos, with a radius of 471 km.) A person would see the horizon 1.3 km away (pretty close if you think about how far a kilometre looks when you're driving, for example) and a ten storey building would see the horizon drop off 5.3 km away.
• And speaking of Pluto, Pluto's largest moon (it also has two tiny ones), Charon, is a bit bigger than Ceres and has a standing horizon of 1.4 km and a building horizon of 6 km.

Seeing things beyond the horizon

The horizons I've talked about above are the limiting distances for seeing things on (or close to) the ground. Things are a little bit different if we want to work out from how far away we can start to see the top of a tall building, for example.

The diagram below shows that although my little stick figure can only see the ground up to d distance away, s/he can see the top of a building which is d+b distance away. Huzzah!

My drawing skillz know no bounds. Close up of previous diagram with a building of height H, that the stick figure can just start to see the top of, added in. The building is b distance further away than the ground point being cut off by the curvature of the planet.

So what if we want to work out how from how far away we start to see the top of a building/monument/spaceport/volcano? Easy. All we have to do is work out the distances to the horizon for both the person and the building/monument/spaceport/volcano and add them together. The distance, D, from which the person starts to see the top of the building/whatever is then approximately given by:

You may have noticed that if we want to work out the distance from which someone can start to see a ten storey building, all we have to do is add the horizons I worked out above together. How convenient! Just quickly, the distances at which the building will start looming out of the ground are:

• Earth: 25.5 km (assuming you can find an isolated ten storey building in the middle of a 25 km circle of flat ground...)
• Moon/Io: 12.5 km
• Ganymede/Titan: 15.5 km
• Mars: 17 km
• Deimos: 740 meters if you can manage an ideal situation (I strongly suspect that you can't, but these calculations do give you a good idea of how small Deimos is... the edge of Deimos would look so close!)
• Ceres: 6.6 km
• Charon: 7.4 km

There you have it. Note that with Earth being the largest rocky body in the solar system, it by far has the widest plains (or planes, if you prefer to be mathematical about it) around. The larger-but-not-as-bit-as-Earth planets and moons (Mars, the moon, Io, Ganymede and Titan) give us similar results for their horizons, which suggests to me that someone travelling between them wouldn't notice much of a difference. Our small-but-still-roundish bodies (Ceres, Charon) both give results about half that of the larger bodies (and hence a quarter that of Earth). I only included Deimos for fun, but it is important to remember that standing (or floating, as the case may be) on or near the surface of this moon (or any similarly sized asteroid, of course) would, visually, be a very different experience to any other body I discussed.

Tides and their locks

Gravity causes tides. On Earth, a planet with a whole bunch of wet stuff sloshing around on the surface, this leads to the sort of sea and ocean tides that most of you have probably encountered at some point.

Tidal interactions

Tides on Earth are mainly caused by the gravitational force of the moon pulling on the Earth. Water, unlike the rock making up most of the surface of the Earth, is able to move a little bit towards the moon in response. Obviously, it's not a massive effect—we're not talking about losing chunks of ocean into space—but it's significant for the sea level to rise a few meters in certain places. The sun also has a similar effect on the Earth so if we didn't have a moon, we'd still have some tides, just on a smaller scale and rather more regularly. As it is, the reason tides are so irregular is because moon and sun aren't in sync (this is also why lunar calendars and solar calendars are so different).

I should also add that while the water on the side of the Earth closest to the moon becomes deeper due to the gravitational pull of the moon, it is conservation of angular momentum which causes the water on the side of the Earth furthest from the moon to also bulge out. I won't go into the specifics unless someone asks in the comments, but the short version is that that opposite bulge of water is require to "balance out" the bulge formed by the gravitational pull of the moon.

OK, so the only thing that really sloshes around on the Earth is water, but what would happen if the moon was bigger or the Earth was closer to the sun and had no water? Or what if we had a rocky moon orbiting close to a gas giant? Well, instead of water sloshing about, it's possible that the gravitational pull of the large planet would pull on the moon strongly enough to deform rock. This is exactly what happens with Io and Europa, Jupiter's two innermost (Galilean) moons (although Europa is more ice than rock). The tidal tug of gravity on Io is what keeps its core molten and causes so many volcanoes on its surface. It's what keeps the interior of Europa liquid (or at least, what keeps some water in liquid form below the surface) and it's also what keeps Earth's core molten (because of our moon's tidal forces). Without these tidal interactions, there would have been more than enough time for these planets and moons to cool enough for their cores to solidify. In the case of Io, I believe the smaller gravitational tugs of the other Galilean satellites, particularly Europa and Ganymede, may also play a small part in its tidal heating.

The take home message is: tidal forces cause volcanoes. It's an important point to remember if you're situating your planet/moon close to its star/primary.

Locked with tides

Let's move away from the Earth and the Jovian satellites for a moment and think about a miscellaneous rocky planet, close to it's star. Like Mercury, for example. For a long time, it was thought that Mercury was tidally locked, meaning that the same side always faces towards the sun. It turns out this isn't quite true thanks to gravitational tugs from some of the other, bigger planets. So let's ignore the other planets. We have a star and a planet forms in place around it. I've mentioned before that conservation of momentum dictates which direction planets and moons will initially rotate and orbit. Any deviations from this will be a result of later collisions. So the planet will form orbiting in the same direction that its star rotates and also rotating in this same direction. (If you're a bit confused about how an orbit and a rotation can be in the same direction, stick your thumb out and curl your fingers around. Your thumb is pointing in the direction of angular velocity of something rotating in the direction your fingers are curling. You could make a looser curl with your fingers to represent an orbit and, so long as your thumb continued pointing in the same direction, that orbit would be in the same direction as the previous rotation.)

The angular momentum of the star-planet system has to be conserved. (Angular) Momentum is mass multiplied by (angular) velocity for each body and then summed. Since the mass of the system isn't going to change much (ignore comets and spare dust/gas that might accrete), then for the planet's rotation to change (that is, get faster or slower) one of the other angular velocities has to change to compensate. This is exactly what happens when a planet becomes tidally locked around its star or when a moon becomes tidally locked around its primary (example: the Galilean moons of Jupiter); rotational angular momentum is slowly converted into orbital angular momentum resulting in a slightly faster orbit but a slower period of rotation. Given enough time, the planet will become locked in a synchronous orbit (the same side always facing its sun). This is where the "tidal" part of "tidally locked" comes from.

Now, it's possible to work out how long this process takes or, for a given time frame, what the "tidal lock radius" is; that is, the distance from the star inside of which planets will be tidally locked after that time period has elapsed. The general equation for this, as given by von Bloh et al. (2007) (and Kastings (1993), pdf sorry) is:

This isn't the most helpful equation ever. And the units are confusing

Where P₀ is the original period of the planet, Q is a factor to do with the rotational properties of the planet and M_* is the mass of the star. As the caption says, this isn't super useful and they use somewhat baffling units. Subbing in the values they assume for P₀ and Q and assuming the same time they assume which is 4.5 Gyr (the G stands for giga—yes, just like in your computer—and means 4.5 billion years or 4 500 000 000 years), I get something close to:

See, isn't that nicer to deal with? And now we have r_T in AU and M_* in solar masses—yay!

So if you put in the mass of your star in units of solar masses (or the mass of your gas giant, but you still have to use solar masses) then you will learn the tidal lock radius in AU for systems which have had 4.5 Gyr in which to evolve. I think 4.5 Gyr was chosen because that's roughly the age of the Earth/solar system (actually, it's more like 4.6 Gyr, but close enough).

If you want to read more about orbital mechanics, I found a review written by the guy who originally worked this stuff out in 1977: Peale (1999) (pdf again, sorry). I haven't had the chance to read through all of it yet, and it's a bit heavy on the maths, but it looks interesting.

There is more that I want to say about tides, but I feel this post has gotten long enough so I'll leave it for a future blog. Coming up soon: the Roche limit and why Saturn has rings (and all the other gas giants too). Stay tuned!

Introduction to Gravity

I thought I'd do a series of posts on gravity because it's a huge topic and kind of important. I don't really want to do all of these in a row so if any of you have requests or suggestions for topics, please let me know in the comments. :-)



Newton's law of gravity



Almost everyone has heard the story of Newton sitting under an apple tree and being inspired towards understanding gravity thanks to a falling apple. You may have heard the version of that story where the apple falls on Newton's head but I recently read that it actually landed next to him. I am more inclined to believe this version of the story because I know that if an apple fell on my head I'd be too busy cursing the tree to have a flash of inspiration.

What was the revelation Newton had about gravity? Well basically, he gave us a simple, universal description of gravity. The same force that causes the apple to fall towards the ground also keeps the moon orbiting the Earth and the Earth orbiting the sun. The mathematical expression for the force of gravity between two objects that Newton left us with is:

F is the force of gravity between two objects of masses M and m kilograms, with their centres separated by distance r meters. G is the gravitational constant and is equal to 6.67 × 10^-11 km³ kg^-1 s^-2 .

However, in our day-to-day experiences, it is not forces, per se, that we are most aware of but accelerations. For example, if you are on a train moving at a constant speed (not accelerating, that is; not slowing down or speeding up), you can't tell how fast you are going solely from its motion. The only thing that would give it away is the bumping up and down thanks to uneven tracks. However, it's easy to tell when the train is slowing down or speeding up because you are either pushed backwards or forwards in your seat (depending on which way you are facing).

Although gravity is always pulling us towards the centre of the Earth, what we actually feel is the ground (or floor or chair or whatever) holding us up and preventing us from falling towards the centre of the Earth. It's gravity accelerating us into the floor that we experience as weight. If there was no floor and we were just falling, we would actually feel weightless (air resistance notwithstanding). The International Space Station is well within Earth's gravitational field but it is constantly falling, which is what makes the astronauts inside feel weightless. Luckily it's not just falling straight down; it also has a horizontal velocity which means that in the time it falls downwards a certain amount, the curvature of the Earth results in the ground also falling away by the same amount. This is called an orbit, and I will talk more about them in the next section.

Back to the equation above. How do we turn this into what we experience every day on Earth and then into what we would experience if we were living on another planet. The important quantity to calculate is g, the acceleration due to gravity. Every force (or sum of forces) can be described as an acceleration acting on a mass. F = ma is the famous equation and is the mathematical representation of Newton's second law (side fact: Newton's law of gravitation didn't get a number, his three laws refer to more basic laws of mechanics). To find acceleration due to gravity on the surface of a planet, we equate F = mg (rather than F = ma, since g is the symbol we use for acceleration due to gravity) with Newton's law of gravity:

Acceleration due to gravity, g, depends only on the mass of the body exerting a gravitational force and on the distance from the centre* of that body. All bodies, no matter what their mass, m, will accelerate at the same rate in the same gravitational field.

To find g at the surface of the Earth, you need to substitute in the mass of the Earth for M and the radius of the Earth for r. On Earth, g = 9.8 meters per second per second, which means that, if we ignore air resistance, something that is falling will gain 9.8 m/s of speed each second. It also means that the force with which we are constantly pushed into the ground/chair/bed is equal to our mass times 9.8.

On planets other than Earth which are smaller, larger, heavier or lighter, there are different accelerations due to gravity which we can find by throwing the right numbers into the equation above. Some examples from rocky† planets and moons in our solar system (where g is acceleration due to gravity on Earth's surface):

• Moon: 1.7 m/s² = 0.17 g (about a sixth of Earth's gravity, so you would feel a sixth as heavy.)
• Mars: 3.7 m/s² = 0.38 g (between a third and two fifths of Earth's gravity)
• Mercury: 3.8 m/s² = 0.39 g (coincidentally very close to Mars)
• Ganymede: 1.5 m/s² = 0.15 g (also about a sixth of Earth's gravity)

And because this is easily applicable to extrasolar planets — that is, planets outside of our solar system — I have also calculated some surface accelerations due to gravity for a few known exoplanets (links below are to Wiki, but I got my values from the Exoplanet iOS app; see below).

• Gliese 1214 b: 8.6 m/s² = 0.88 g (Just under nine tenths that of Earth. However, it's just inside its host star's habitable zone (the star is called Gliese 1214) which means it'll probably be too hot for human habitation. It could even have a runaway greenhouse effect like Venus. It's also possible that this planet has a very thick atmosphere making it more similar to a small gas giant like Neptune than to Earth.)
• CoRoT 7 b: 18.4 m/s² = 1.9 g (Just under twice Earth gravity so you would feel almost twice as heavy and, more vitally, your organs would all press down on each other twice as strongly. According to NASA (pdf, sorry), this isn't terribly sustainable in the long term for humans as we are now. Personally, I don't think it would take an awful lot of genetic engineering to fix this for us (we are talking science fiction, after all). The bigger problem with this planet is that it's much to close to its sun for our comfort or survival.)
• Kepler 11f: 3.4 m/s² = 0.35 g (About a third of Earth's gravity. Compare with Mars or Mercury. Unfortunately, it's also slightly too close to its star to be habitable. Incidentally, the whole Kepler 11 system is quite interesting with six confirmed planets so far.)

(If you have a hankering to include some real exoplanets in your story, I highly recommend this iOS app is an excellent resource. It is a frequently updated database of all the confirmed exoplanets that have been discovered, including their statistics (mass, distance from star, radius where available, whether it's in its star's habitable zone...) and you can even pan through and around a zoomable 3D map of the Milky Way. And when I say zoomable, I mean you can zoom right in to see the planets orbiting their stars at the correct distances and with the appropriate relative velocities. Even if you don't care about the specifics of the planets, that Milky Way map is worth the time it takes to click the free download link. For the record, I am in no way connected to this app, I just think it's awesome.)


* Technically, the distance from the centre of mass, but for round or roundish things like planets the centre of mass is generally the centre of the planet.
† Rocky because you can't stand on the surface of the gas giants. It's possible to calculate the acceleration due to gravity experienced by a hovering platform or similar, however.

Orbits

As we've established, gravity is what keeps things in orbit around other things. As such, we need to use what we know about gravity to work out how fast something has to orbit for different distances and masses of objects. In fact, Kepler had worked this out to some degree before Newton came along, but Kepler's third law was slightly less specific than could be calculated using Newton's law. Kepler realised that for orbits, the ratio between the cube of the semi-major axis and the square of the period was constant. The semi-major axis is the same as the distance to the larger body from the smaller (like r in the equations earlier) for circular orbits and half the length of the longest side of an ellipse (oval). the period, T, is the time taken to complete one orbit, so if we're talking about planets, then it's the length of a year. We can derive the constant part of Kepler's third law using Newton's law of gravity and the equation that describes centripetal force.

As it happens, I talked about a lot of the ingredients for working out how fast a planet should orbit around its star in my recent post about space elevators. The set of equations below starts by equating the centripetal force (the force required to keep something of mass m moving around in a circle with radius r and at speed v) with the gravitational force (centripetal on the left, gravitational on the right of the equals sign), then shows the derivation of Kepler's third law (the second last line) and finally gives the period, T, of a planet orbiting around a star of mass M at a distance r. Feel free to let your eyes glaze over if maths isn't your thing. You have been warned.

The final line gives us the period in seconds, which for most things isn't terribly helpful. To find out what the length of your planet's year is in days or Earth years you will need to divide your answer for the period by 86400 or 3.15 × 10⁷ respectively.

The reason I chose to rearrange Kepler's law to solve for the period rather than for the semi-major axis is because for science fictional purposes, the distance from the star is more likely to be fixed for plot purposes. For example, an human-inhabited planet has to be in the habitable zone. Exactly what the habitable zone is will be, I think, the subject of a future blog post.

I should also point out that these equations are completely applicable to man-made satellites or moons as well, you just need set the planet's mass to be M instead of the star's mass. Just remember that if your comms satellite is x km above the surface of the Earth/whatever planet, you have to add on the radius of the planet to find r to throw into the equations I've talked about today.

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.