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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}km

^{3}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.

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

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

Just to be pedantic: GPS satellites orbit considerably lower than geostationary orbit. Each satellite completes one orbit in just under twelve hours. The result is that in addition to their distance from us, the precise time is needed in order to calculate one's location from the satellites' signals. This is why the satellites' signals include the exact time.

ReplyDeleteThanks Oolon. That's what I get for not double checking that fact. Changed post to be about comms satellites instead. :-)

ReplyDelete