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# Space Elevators Do Too Stay Put

In honor of Pixel-Stained Techno-Peasant Wretch Day, I'm showing some engineering work I did for free. This isn't anything I'd actually get paid for, at least not in this decade (or possibly century) but it does show how an engineer can analyze a practical problem.

Back in July, wrote a Pyramid article about an alternate history setting for GURPS Infinite Worlds. One colorful detail he included was a space elevator anchored at the equatorial island of Batam.

Fanboys being what they are, someone immediately took offense at this skyhook not being precisely on the equator and claimed it would wind up oscillating north and south. I pointed out the flaws in his logic and it got heated. This is when I wound up flaming someone for dissing centrifugal force. Eventually it was clear I'd have to run the numbers to end the back and forth.

The first step was how to model the elevator. For this discussion the only question was whether the elevator would hold a constant position relative to the equator or oscillate north-south. That lets us use a simple model: a single point mass attached by a weight-less string to Batam Island. If I had a finite-element-model simulation handy, I might use the more complex model of a series of point masses connected to each other with strings, but that would be a lot more work to do by equations, and require a bunch of assumptions about how the mass is distributed over the length of the skyhook.

The model only requires two dimensions, with the z-axis along the North Pole and x-axis extending from the center of the Earth through the equator at Batam's longitude. The elevator's counterweight weight isn't detailed in the article, so we'll assume it's far enough out to put the elevator's center of gravity at 70,000 km from the center of the Earth (r). That's where we put the point mass. The weightless string attaches it to Batam Island at 1.15 degrees North latitude (Rz, Rx). The mass has three forces acting on it: tension in the string (T), Earth's gravity (G), and centrifugal force (C). Since all three have a constant value they'll come to an equilibrium:

T + G + C = 0

It's easier to solve the math if we break out the equations for each axis, each force having an x and z component:

C = Tx + Gx
Tz = Gz

Note that centrifugal force is entirely normal to the Earth's axis and has no z component. Its magnitude is C = w r2, or 37 cm/s2. I'm neglecting the mass of the elevator here since it doesn't matter in the force balance. G is a function of the radius from the center of the Earth. At 70,000 km it's a modest 8.1 cm/s2.

Breaking G into its x and z components requires the value of a:

Gx = G cos a
Gz = G sin a

Since we don't know what a is, we'll have to iterate on it until we find a value that lets all the equations agree. We're still short a few more equations:

Tx = T cos e
Tz = T sin e

b = r tan a
e = atan( (Rz - b) / (r - Rx) ) (point for anyone who spots the simplifying assumption there)

That's enough to solve the problem. But to make the make a little easier we'll combine some of them:

T = (C - Gx) / cos e

Now we can plug in a value of a and see if Gz = Tz. If not, we try a new value. This can be tedious, but Excel's Goal Seek function doesn't mind. The results:

a = 0.083 degrees
b = 101.9 km

That puts the elevator's CG 26 km closer to the equator than the anchor point (Batam Island) but effectively over the equator, since that's a very small amount of latitude at that altitude. As C, G, and T are all constant forces the elevator will hold that position.

So, no, a space elevator anchored off the equator will not oscillate north-south. A satellite in geosynchronous orbit which is inclined to the equator will BECAUSE IT'S NOT FRIGGIN' ANCHORED! Sorry. Did I mention this was originally done for a flamewar?
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