Sunday, March 20, 2011

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 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.


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. ;-)

Wednesday, March 16, 2011

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.


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