The humanly habitable zone

If you want your little book people (otherwise known as characters) living on the surface of a planet that isn't Earth and without constant artificial support, you probably want the planet to be habitable. This is where the habitable zone (and hence this blog post) comes in.

Why is it a zone?

First let's talk about what we need from a planet. Earth is great; it gives us lots of handy life-sustaining conditions; air, water, the right amount of gravity, sunlight and radiation... Of course, the reason the Earth is so well-tuned to keeping us alive is that we evolved on Earth. It's not tuned to us, we're tuned to it. There is no reason that aliens couldn't evolve in very different conditions to those found on Earth. Even on Earth there are many forms of life such as extremophiles which live in conditions we humans couldn't survive in. There's also a good chance that life is possible on Europa or Titan (moons of Jupiter and Saturn respectively), but on the former it would have to be in a subsurface ocean and the latter has a very different atmosphere and composition to Earth, meaning that life couldn't be water-based.

Aliens are all well and good, but if we're interested in what conditions humans can live unsupported on a planet's surface, we need to be much more specific.

The things I mentioned earlier (air, water, heat, etc) depend on two things:

1. Planetary properties such as size and composition, and
2. Planetary location.

The first of those is most easily summarised as having to be similar to Earth to sustain unaided human life. The latter is the subject of this blog post.

You see, although the planet needs to be similar to Earth, its sun doesn't have to be that similar to our sun. Different stars output different amounts of light and energy (previously discussed in terms of how bright they are). If we were to pick Earth up and put it in orbit around a larger, hotter star then, if we didn't want to be burnt to a crisp while all our water boiled, we would have to put Earth in a further out orbit to compensate for the extra energy coming off the star. Similarly, if we put Earth around a smaller, cooler star, we'd need to put it in a closer orbit to stop it turning into a snowball.

The optimal region in which to put a habitable planet is called the habitable zone. It's a zone because there's no reason for life to have not evolved on Earth if it had been slightly closer or further away from the sun. I'll get back to life evolving on other planets at the end of the post, though.

There are a few different ways of defining what constitutes a habitable zone. Some definitions involve the region in which liquid water is possible, temperature-wise (since our biochemistry depends heavily on liquid water), other definitions go a bit further and include things like carbon cycles and the greenhouse effect (pdf link, sorry). For the purposes of the habitable zone calculations which follow, I'm going to use the definitions given in Selsis et al. (2007).

Before that, though, let's work out how warm a planet is based on how far from its sun it is. I am using a formula adapted from a Melbourne Uni 3rd year physics lab manual because it's simpler than the one given in Selsis et al. (2007), although they both give the same results. The approximate temperature of a planet, given a stellar temperature and radius and ignoring complex atmospheric effects (but assuming that an atmosphere exists) is:

T is the approximate temperature of the planet in Kelvin, T_* is the temperature of the star in Kelvin, A is the albedo of the planet (for reference, Earth's is 0.306, according to wiki), R_* is the radius of the star and d is the distance of the planet from the star.

The only tricky thing here is that both R_* and d have to be in the same units, so either both kilometres, or both AU etc.

According to Selsis et al. (2007), life is possible in the range where the temperature is between 277 K and 394 K. Rearranging the above equation then with these values and substituting R_* in annoying units for solar radii (the HR diagram I linked to a few weeks ago can help you estimate this much more easily than in km or AU), we find that the habitable zone for any main sequence star in AU is given by:

Quantities as above. R_☉ is solar radii (multiples of the radius of the sun) and d now must be in AU.

So all you have to do now is work out what kind of star you want, find out it's general properties (easily googleable if you're using a real star) and throw in the numbers. Unless you have plans to make your planet unusual, it's probably best to keep A as the Earth's albedo. That said, habitable distance will change with albedo so the only reason not to change it is because albedo reflects the planetary composition (and we probably want to keep a similar composition to Earth...).

For reference, Selsis et al. (2007) find the sun's habitable zone to be between 0.95 and 2.4. (Note that they use a different albedo for the Earth.) Wiki lists some other numbers which seem to vary mostly in the outer edge prediction.

I should also point out that all of this assumes that there aren't any other stars near our planet. Things get a bit more complicated with multiple stars around, something I will definitely address in a later post.

Evolving elsewhere

As well as the habitable zone, there is something known as the continuously habitable zone (or CHZ). This is the region around a star which remains habitable throughout the star's (main sequence) lifetime. (Main sequence lifetime because different stars end their main sequence lives in different ways, mist of which make the continued survival of planets complicated at best. More on this in a future post, I think.) The reason we need to worry about different times in a star's life is that stars tend to increase their rate of energy output as they age and use up fuel (because as more fuel is used up, hotter core temperatures are required to sustain fusion which leads to an increase in luminosity). The CHZ is narrower than the habitable zone calculated at any given time during the star's lifetime and it is possible that as the star's luminosity changes, the habitable zone could shift so that the planet moves into or out of it over (astronomical) time.

Of course, this doesn't matter so much if we just want to find a planet to throw humans at. Continuous habitability becomes more relevant when we're talking about life evolving. Evolution from scratch takes a long time, but human civilisation to date has lasted an insignificant amount of time, on an astronomical scale.

An interesting link and some housekeeping

Alien plantlife

Red Suns and Black Trees: Shedding a New Light on Alien Plants is a Universe Today article about how chlorophyll (the stuff that makes plants green) works well on Earth because of the colour of our sun. [Edit: link down as of posting because they're hosted on Amazon's servers... sorry. Edit 2: Here is an article from Astrobiology Magazine along the same lines, thanks to Patty Jansen.] On other planets, particularly those with different coloured suns, it's possible that the vegetation could evolve to be a different colour to better take advantage of the different light.

It reminded me of the red edge, which is a feature of the reflectivity of plant life. It has been proposed as a possible extraterrestrial biomarker (a way of determining if there is plant life like ours on other planets. I wonder if plants which don't use chlorophyll exhibit the same spectral properties in the infrared? That said, not overheating is a useful property, so on worlds with bluer suns, maybe the effect would be more extreme? This is pure speculation, of course.

Housekeeping

I have decided that I am going to aim to write one long, sciencey post a week and, for the time being, the plan is to have that post go up on Wednesdays. (That's Australian Wednesdays, for those of you in less progressive time-zones. ;-) ) I may also post some shorter blogs throughout the week, such as this one, but that's less guaranteed. Sometimes I may also write posts about the science (good or bad) in books I've read.

I know more frequent blog posts are meant to be a way of keeping readers interested, but this one proper post a week is all I can really commit to, time-wise, for the moment.

Gravity: Relatively general space

Two weeks ago, I talked about basic Newtonian gravity. Today I'll be talking about some of the contributions Einstein made.

Curvy

You may have heard in passing phrases like "space is curved" or "the curvature of spacetime", but what do these phrases actually mean? Einstein's theory of general relativity brought us the understanding that the force of gravity is the result of masses deforming (curving) the fabric of spacetime surrounding them. This is different to the other forces of nature which are quantised and mediated by force-carrier particles, and can be mathematically combined into a single force known as electroweak. Don't worry if that last sentence didn't make sense. The important thing is that our deepest understanding of gravity does not include quantum theories* but does involve classical geometry.

The most common metaphor used to describe the curvature of spacetime gives something like this:

Imagine the universe is an infinite rubber sheet (infinite because this is not the place to think about what's happening at the edges or even whether edges exist). Masses such as stars, planets and so forth are like ball-bearings stuck to the surface which, because of their mass, create dips in the sheet. Heavier things make deeper dips. Then, if you roll another ball bearing along the sheet (it doesn't actually have to be a lighter one, but it's easier to picture if it is). As it rolls past the other masses, it's path will be deflected by the dips so that it doesn't go in a straight line from the point of view of an external observer. If you were to draw a grid on the sheet before letting the masses stretch it, then rolled a very small mass very quickly along along it, it would follow the grid lines, even though the grid lines themselves are now stretched.

The very small, light mass I mentioned at the end would have to be a photon, a massless particle of light. Anything with a mass, even a small one, would get deflected off the grid lines by the masses because it would be travelling more slowly that the speed of light.

Of course, the universe isn't a two-dimensional rubber sheet and this metaphor isn't perfect. It's a little bit harder to picture in three dimensions, but the qualitative ideas are the same. And you can think of the grid lines as geodesics which mean they represent the shortest distance between two points and hence the path light takes. That's right, gravity curves the path light takes. The larger the mass, the greater the curvature. This is called gravitational lensing and is all sorts of useful in different areas of astrophysics.

Now, Newtonian gravity doesn't take the curvature of spacetime into account. Usually this doesn't matter much because even the sun only causes enough curvature for Mercury, the innermost planet, to really be effected. The general relativistic (GR) corrections for the other planets are small enough to be insignificant. GR becomes much more relevant around small and dense objects such as neutron stars and black holes.

* miscellaneous proposed but untested theories notwithstanding.

Black as black

Black holes are an interesting concept that falls out of general relativity. Chances are you've heard of them if you haven't lived in an internetless cave for the past hundred years. But what exactly are they?

A black hole is an extremely small and extremely massive object. It is so small and massive that it is denser than any form of matter we know of. We don't really know what sort of matter black holes are made of. This comes from the fact that we don't have a theory of quantum gravity as I mentioned earlier.

Escape velocity is the speed you need to go to escape a body's gravitational pull. To completely escape Earth's gravitational field, for example, you need to leave the Earth at about 11 km/s which is about forty thousand kilometres per hour. To escape the sun's gravity from the surface of the sun (let's ignore the fact that you'd be fried while you were there) you need to leave at about 618 km/s or more than two million kilometres per hour. To escape the sun's gravity from a distance of 1 AU (the distance between Earth and sun) you need a velocity of 42 km/s (150 thousand km/h) because the force of gravity drops off with the square of distance.

Because black holes are so dense, there is a point where the force of gravity is so strong that the escape velocity is equal to the speed of light. This is called the event horizon and from within the escape horizon nothing can escape the gravitational pull of the black hole (since nothing can move faster than the speed of light). This is where the "black" part of "black hole" comes from.

In practice, it would be very hard to escape the black hole long before you reached the event horizon, purely due to the energy needed to overcome gravity. And even before you reached the event horizon (also known as the Schwarzschild radius), lots of strange and interesting things start happening. Any signals you try to send out (to people further away from the black hole) would be redshifted as the wavelengths of light get "stretched out" by the extreme curvature of spacetime around the black hole. Notice how "spacetime" includes the word "time" as well as "space"? Time also gets stretched out near a black hole and passes more slowly (for complicated reasons I might explain in a future blog post). Not that you would necessarily notice if you were falling into a black hole because the tidal forces would be ripping you apart.

Tidal forces come from the difference in the force of gravity between the end of you/your spaceship closest to the black hole and the end further away. Gravity acts more strongly on the closer end, causing interesting (and painful) things to happen. This is the same principle which leads to moons being tidally locked in their orbits around their planets, but taken to an extreme scale. If the force of gravity on your feet is appreciably stronger than on your head, nothing pleasant will result from your feet being sucked into the black hole more quickly than your head.

In terms of where we can find black holes, that's a good question. The only black hole whose existence we're sure of is the supermassive black hole at the centre of our galaxy. We're also quite confident that most other galaxies also contain central black holes. The supermassive part relates to the fact that they are hundreds of thousands times more massive than the sun. In face, our supermassive black hole, Sagittarius A, is about four hundred thousand times the mass of the sun.

Theoretically, there should also be much smaller black holes around, only a few times the mass of the sun, ranging up to a hundred or so solar masses. These would come from very large stars which reached the end of their lifetimes, went up in a supernova and then collapsed into a (stellar) black hole. Even though there's no reason for these to not exist, we have yet to decisively detect any. The black part of the black hole makes that a little tricky. (The is some radiation coming off them, known as Hawking radiation, as well as some interesting things happing as things fall into black holes, which gives us some candidates, but as far as I know they're still only candidates.)

Barely scratching the surface

So that, very briefly, is what general relativity and black holes are about. You might have noticed that there weren't any equations in this post. (Gasp!) That is because most of maths that describes general relativity requires a maths or physics degree to understand. In my experience, general relativity is usually a graduate level subject (or Honours at most Australian universities).

Maths aside, I feel like I've barely scratched the surface of black holes [insert bad pun here], so I suspect another post focussing on black holes will happen some time in the future.

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.

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.

Space Elevators: Where to put them

The purpose of this post is not to exhaustively describe what a space elevator is and how it is intended to work. There are many other sources for that information. Instead, I intend to focus on what planets are suitable for building a space elevator on and which aren't and why.

WARNING: This post contains maths that requires a proper calculator.

Briefly: What is a space elevator?

A space elevator is a proposed system for (mostly) getting things into orbit without using rockets. The idea is that once it's established, the cost for going into orbit drops dramatically because you no longer have to use rockets (which are expensive and generally involve some disposable element).

Many different types of space elevators have been proposed, not all of which necessarily reach the Earth's surface. Instead they start part way up or outside of the atmosphere. This has the benefit of avoiding annoying atmospheric effects (wind, drag) and makes them lighter. But I'll leave researching what type of space elevator best fits into the world being created as an exercise for the reader. What I am going to talk about is what and where is it physically possible to place an elevator and more or less ignore engineering constraints (set it far enough in the future and, as Arthur C Clarke said, you can have technology advanced enough to seem magical).

There are two ways of making a space elevator stay up. You can have a rotationally supported one or a gravitationally supported one. In the case of lifting payloads from Earth into orbit, a space elevator would almost certainly be rotationally supported. One designed for getting things to and from the surface of the moon, however, would more likely be gravitationally supported. What are the differences?

Rotationally supported space elevator

The principle for a rotationally supported space elevator is to have the force of gravity pulling it down to Earth exactly opposed by the centrifugal force of it spinning around in its orbit. Since we also generally want it to remain above above the same point on the surface of the Earth (so we can connect the ground to the top of the elevator with a cable, for example), the centre of mass of the elevator needs to be at the height of geostationary orbit. Geostationary orbit just means that it orbits at the same speed that the surface of the Earth rotates and hence it remains above the same surface point of the Earth all the time. Also, this point has to be along the equator or the stationary part of geostationary won't work. For example, a lot of communication satellites live in geostationary orbits which makes it easier for receivers on Earth to talk to them since they are always in the same place. And in the case of TV broadcasting, you generally only want to broadcast at a particular region, so there is the benefit of always being able to do so.

A quick note on how to calculate the height of geostationary orbit, since I want this guide to be generally applicable, not just for Earth. For geostationary orbits, the acceleration due to Earth's gravity, g,  (or the gravity of whatever body we're interested in) needs to cancel out the centripetal acceleration, ac, from the circular motion of orbit.

So r is the distance from the centre of the planet in kilometres (just subtract the radius of the planet at the end to find the height above the ground if that's what you want to know), G = 6.67 × 10^-20 km³ kg^-1 s^-2 is Newton's gravitational constant, M is the mass of the planet in kilograms, T is the length of a day in seconds and π = 3.14 is a constant. We can now rearrange this:
 

And it becomes just a matter of plugging in the right numbers. We can now calculate that the height of geostationary orbit for Earth is about 36 000 km. Out of interest, for Mars, the height would be only 17 000 km thanks mostly to Mars' small mass.

You might have noticed earlier that I said the centre of gravity of the space elevator needs to be at geostationary orbit. This just means that half the mass of the elevator needs to be below the geostationary height (so, mostly this would be in the cable since thirty-six thousand kilometres of cable, even if it's made of carbon nanotubes, is a non-trivial mass), and half needs to be past the geostationary height. The latter "counterweight" could consist of something like a docking station, space craft manufacturing plant or whatever you like.


Gravitationally supported space elevator

You probably wouldn't use a gravitationally supported space elevator to lift things off the surface of the Earth into space, but it would be ideal for lifting payloads from the moon. Because the moon is in a synchronous orbit (where it takes the same amount of time to complete a rotation as it does an orbit around the Earth), it spins too slowly for a sensible rotationally supported elevator. Plugging the numbers into the equation above, I get a geostationary height above the surface of about 87 000 km, more than twice the height for Earth. The moon's great and all, but it's probably not worth the price of twice the length of the cable just for getting rocks back to Earth. Not to mention any gravitational effects of the Earth on the elevator. For another example, let's work out the geostationary height for Ganymede, the largest moon of Jupiter. Plugging in all the numbers, I get about 43 000 km above the surface of Ganymede. Sure, this isn't a much longer cable than for Earth, but at the distance you start getting annoying gravitational effects from Jupiter and other planets screwing you over. Basically, you couldn't make it stable.

The solution to this dilemma is not to try to make a rotationally supported space elevator, but to go for a gravitationally supported one instead. A gravitationally supported space elevator has its centre of mass at a Lagrange point, usually L1, which is sort of a gravitationally neutral location. So for the moon, the centre of mass would go at the point where the force of Earth's gravity is exactly balanced by the force of the moon's gravity. This point is called first Lagrange point of the Earth-moon system (after the guy who worked out the maths).

Thanks to the moon's synchronous orbit, the point on its surface that is closest to Earth doesn't change. Hence, a gravitationally balanced space elevator would automatically remain stationary relative to the surface of the moon, which is handy when you're running a cable between them.

How do we calculate how far from the moon the Lagrange point we're interested in? Well, we want the point where the acceleration due to each Earth and moon cancel out BUT we also have to consider the rotational acceleration due to the circular motion of orbit. (Remember, even though the space elevator is attached to the moon, the fact that it's suspended between moon and Earth means that it's also going around the Earth).

r is the distance from the centre of the moon to the Lagrange point, m is the mass of the moon, M is the mass of the planet, d is the distance between the planet and the moon and T is the time it takes for the moon to complete an orbit (hence the time taken for the space elevator to complete an orbit since it's attached to the moon). It's not actually possible to rearrange this into something nice. If you really need to do this for a general planet, I suggest going to WolframAlpha.com and typing in:

Solve[-((G m)/r^2) + (G M)/(d - r)^2 == ((2*3.14)/T)^2 (d - r), r]

But with the appropriate values in place of all the constants (in km and kg if you use the G I gave above). I personally did the same in Mathematica (which is also made by Wolfram and has the same maths engine as Wolfram Alpha). However, if you're interested in doing a calculation yourself, with some approximations to simplify things, this website (which I came to via Google) seems to do a reasonable job. It also explains the maths a bit more than I have.

So anyway, throwing appropriate numbers and making a computer solve it, here are a few results:

• For the moon, the distance from the surface to the first Lagrange point is about 56 000 km.
• For Ganymede it is about 29 000 km from the surface. 
• And because I feel like it, it's 8600 km for Io.

Clearly this is much more economical in terms of how much cable is used. And because you're taking advantage of the largest gravitational well in the vicinity, you don't have to worry too much about other effects mucking up your elevator. (OK, in the Jovian system you'd probably have some complications thanks to the other moons, so I'm not sure you'd necessarily want to go down that path, but it would work well for a gas giant with only one large moon and the rest small.)

You can also put a gravitationally supported space elevator on the far side of the moon at what is known as the second Lagrange point. This post is getting a bit too long to go into the details, but the height of such a space elevator would be approximately the same as if it was at the first Lagrange point so long as the satellite is much smaller than its primary. This isn't true of the moon (but is true of the Jovian moons) and it turns out that the height required for that space elevator's centre of gravity would be 67 000 km.

Summary

As we've learnt, there are a few considerations we need to take into account when placing a space elevator:

• Location
• Start and end points
• Type of body it services

The last consideration ends up informing the first two to a great extent. So if we have a planet orbiting its sun in a similar way to the Earth, we will use a rotationally supported space elevator which will have to be placed along the planet's (rotational) equator. If we have a moon in a synchronous orbit, we want to use a gravitationally supported space elevator which will be placed along the straight line connecting the moon's planet and the moon.

Everything else is just a matter of engineering. ;-)

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.