In my previous post I referred to some difficulties encountered with the mechanics of *Aniara*‘s orbit with respect to the ring space station the crew is investigating. I mentioned how they could not simply park the ship 100 km away at the same altitude, paralleling the station’s orbit around the planet, without risking eventually colliding with the station.

Why not?

It has to do with a basic principle of orbiting a planetary body, whether it’s Earth or any other planet (such as Kepler 62f, for example). Any satellite or object in orbit describes an ellipse or circle as it moves through its orbit, and the center of that circle is the center of the planet itself. It’s not some point 100 km away from the planet’s center, it is *exactly* the planet’s center.

This means that any orbit that isn’t exactly above the equator must be at a particular angle of *inclination* away from the equator:

Granted, at only 100 km distance, the inclination would not be this extreme. Given that the orbit is at Kepler 62f’s geostationary altitude of 41,100 km, and therefore has a radius of 50,030 km (I’m assuming Kepler 62f’s diameter is 17,860 km, so yes, a bit larger than Earth), that gives the angle of inclination at 0.72°.

Perhaps an even more clear illustration of this effect, with plenty of technical astrodynamics terms thrown in for good measure:

So you see, that slightly north-inclined orbit would not parallel the equatorial orbit (which the station is following), nor even remain *north*-inclined for long. Instead, any such satellite (or ship) would spend half its orbit north of the equator, and half its orbit south of the equator (assuming the orbit is circular). If the orbit is at a different altitude, then this doesn’t represent any particular issue, but if it’s at the *same* altitude…

Yeah, that’s right. Since the station is a ring completely encircling the planet, *Aniara* would have made exactly half of an orbit before smashing into it. Not good for our crew, not good at all.

A far better option, then, is to choose a different altitude. So, instead, *Aniara* parks 100 km *above* the station, rather than beside it. This way, *Aniara* can maintain the same equatorial orbit and just hover there, right?

Well, almost. Not quite perfectly, as it turns out.

You see, every orbit has precisely one speed at which that orbit can be maintained. This can be an even harder concept to fully grasp than inclined angles circling around the center of the planet, but the best way to think of it is that the satellite is falling toward the planet at all times. It is called *free fall*, after all. The planet’s gravity is constantly pulling the satellite down. But, because the satellite has a tangential velocity vector…

Wait, what? Oh, hey, it’s been *years* since high school trigonometry, are you really going to lay that on me?

Ok, ok, I’ll back off just a little bit. So, the satellite is falling, but it’s not falling straight down, it’s falling at an angle, or rather in a high arc. And, it’s falling *really, really fast*. So fast, in fact, that it’s arc keeps on missing the planet and it falls *past* it. Except, as it falls past it, the planet’s gravity keeps trying to pull it down, so the arc keeps bending toward the planet, but never quite intersects the planet. The satellite falls past, and down, and past, and down, and so on, in a theoretically never-ending circle (or ellipse, more accurately).

The closer to the planet the satellite is, the more effect the gravity has on it, so it falls *faster*. But, because it falls faster, the arc always remains above the planet’s surface. In contrast, the farther from the planet the satellite is, the *slower* it moves.

Now here’s where things get really weird for most folks. Let’s say your spaceship is orbiting the planet, nice and stable like. Everything is great, yet you decide you really want to bring the ship down to a lower orbit. As if, for instance, you were descending to park closer to an alien space station you had just spent a few days observing from higher altitude.

You could just angle your ship nose down and fire the engines, powering your way down, but this would have the undesirable effect of changing the *shape* of your orbit. Where before perhaps you had been in a nicely circular orbit, now you would be in a much more highly eccentric elliptical orbit, which wouldn’t do at all if your intent was to park over the station with its own circular orbit.

Ok, so you can’t just power your way down there. Instead, your fire your engines in *retrograde*, meaning in the forward direction, to push back on your ship’s velocity vector. This has the desired effect of causing your ship to fall closer to the planet indeed, and nicely reduces the altitude of your orbit, but perhaps counter-intuitively it also causes your ship *to speed up*.

That’s right. Fire your engines in reverse, and you go faster. You descend, but you go faster.

By the same token, if you fire your engines *prograde*, meaning to push your ship forward, you will ascend to a higher orbit and go *slower*.

Well, that is, unless you fire them with enough thrust to overcome the planet’s escape velocity, thus leaving its orbit entirely, but that’s a different maneuver.

So what does this mean for *Aniara*, after descending, but parking 100 km higher than the alien station? She isn’t at the same altitude, so we know that she cannot orbit at the same speed. Indeed, she must be going slower. How much slower?

Orbital velocity for a circular orbit is defined by the equation √(GM/r), where G is the universal *gravitational constant*, M is the mass of the planet, and r is the radius of the orbit. There can be a lot of unknowns here, and often it’s easier to think of GM together (sometimes written as µ) as a given planet’s *standard gravitational parameter*. Kepler 62f’s exact gravitational parameter isn’t of supreme importance here, but rather from this we can determine that if the orbital velocity at 41,100 km altitude (50,030 km radius) is, say 3,000 m/s (which would be a reasonable velocity for that altitude for a planet of slightly higher mass than Earth, so let’s just go with that), then orbital velocity for an altitude 100 km higher would be 2,997 m/s.

So, *Aniara* drifts “backwards” relative to the station at a rate of 3 m/s, or roughly 10.8 km/h.

Astute readers will recall that this station has twelve tether counterweights (in various states of disrepair) extending upward for a considerable distance, certainly quite a lot higher than 100 km. How long before *Aniara* drifts backwards into one? Or more accurately, until one of them catches up to *Aniara* and knocks her out of her orbit (or worse)? Assuming the twelve tethers are equidistant around the globe, at this altitude that makes them more than 26,000 km apart from each other. At not quite 11 km/hr, *Aniara* has over 100 “Earth” days before this becomes a problem.

Depending on what happens, this could become a problem; possible plot point? Who knows? You’ll just have to wait and see. But it’s a reasonable assumption that the crew does not expect to maintain this same orbit for that long.

So why 41,100 km altitude for the station? What makes this altitude *geostationary*? Of course, I recognize that the word *geostationary* is inaccurate, as *geo* refers to our own planet Earth, and this isn’t Earth we’re orbiting now. But, *kepler62fstationary* just doesn’t sound right. Ok, semantics aside, remember how orbital velocity is a direct result of orbital altitude? For any given altitude, there is a defined orbital speed, and for any given speed, there is a defined orbital altitude. And it so happens that there is an altitude at which a satellite’s orbital speed is precisely matched to the angular momentum (I think I’m using the right term there — remember, long years since high school trigonometry?) so that the satellite is apparently *motionless* above the planet, or stationary. Around Earth, this altitude is approximately 36,000 km, but because Kepler 62f is a slightly bigger planet, the “geo”stationary altitude is slightly higher. I admit it, I just picked a number that seemed reasonable and went with 41,100 km. And yes, about 3,000 m/s would be an appropriate speed at this altitude for the presumed mass of this planet.

One final note about geostationary orbits: they are only possible with an inclination of 0°, or directly above the equator, and with a near-zero eccentricity, meaning practically circular orbit. Any other inclination, and the satellite will still revolve evenly with the planet’s surface, but it will appear to migrate north and south in a figure-8 pattern. This would be a *geosynchronous* orbit, but not geostationary.

One final, final note about ring stations with multiple tethers reaching all the way to the planet’s surface: I’m fairly certain that any satellites orbiting at a lower altitude would eventually find a way to smash themselves into one of the tethers. Thus, while it may take a while, installing these things would probably clear the low orbit space of all pre-existing objects, leaving clouds of extremely fast-moving, tiny, and *lethal* debris in their place (remember the film *Gravity*?). I need to research this point further, however.

Right. If your eyes haven’t glazed over yet, then you’re a good candidate for my geeky style of writing! And you should definitely read more from the authoritative sources: NASA’s Catalog of Earth Satellite Orbits for the easy overview, and then Terry Burlison’s Rendezvous and Docking: a User’s Guide for Non-Rocket Scientists for something deeper (he calls it “Part 1,” but I’ve never found a “Part 2,” sadly).

And if your eyes have glazed over, but you still managed to read this far, then you’re probably a great candidate to *beta read* my writing in order to rein me in a bit and get me to focus on the story and not all this technical stuff. ‘Cause obviously I can get myself way off-track with it all!

*header image credit: user:SpaceX-Imagery / pixabay.com*