Astrodynamics

Three months ago I wrote about Orbital Mechanics, focusing on the ins and outs of how spaceships and satellites navigate their way around a planet’s orbital space. We got into the ins and outs of velocity vectors, inclination angles, and prograde vs retrograde thrust.

But what about when a spaceship leaves Earth — or whatever planet it happens to be in orbit around; Kepler 62f, perhaps? — and makes its way toward another planet in the Solar System. Mars, for instance. We’re all astute enough to realize that it’s not as simple as pointing the ship at Mars and firing the rockets. For one thing, of course, Mars is moving, as it is orbiting around the Sun just like Earth is, so even if a straight-line course were possible, we would need to “lead the target,” i.e. aim for the point in Mars’ orbit where Mars will be at the time we arrive.

Sphere of Influence

And, if that’s all it took, this would be a short post. But of course, it’s not as simple as that, as I’m sure you suspected by now. Once our spaceship leaves Earth’s sphere of influence, and until such time as we arrive in Mars’ sphere of influence, we aren’t just in some flat space with no gravitational pull on us. By sphere of influence, or SOI, I mean the region of space where the primary gravitational pull is from whatever celestial body “owns” that SOI. Close to Earth, or in orbit around Earth (even a very high orbit), the bulk of the gravity we feel is from Earth. And, if we don’t maintain sufficient velocity for the altitude of our orbit, then it is Earth into which we will fall. Likewise for Mars.

But in between Earth and Mars, we are still in an SOI — that of the Sun. So, once our spaceship leaves Earth’s SOI, we are primarily impacted by the Sun’s gravity, just as Earth and Mars are.

In fact, once we leave Earth’s SOI, we are still in orbit. It’s just that it’s no longer Earth that we’re orbiting, it’s the Sun.

Up, Down, and the Plane of the Ecliptic

If you’re like me, you probably grew up thinking of the Solar System as basically a flat plane (the plane of the ecliptic), with the Sun in the center and the planets in their various orbits sliding around that plane at a given distance from the Sun. Each of the planets has a north pole that points roughly in the same direction, “up” from the plane. Well, each of the planets except Uranus; Uranus is completely lying on its side, with an axial tilt of 98°, rolling through its orbit like a ball whereas the others are more like spinning tops (apologies to Fraser Cain of Universe Today for borrowing his imagery; I can’t think of a better way to describe it). Anyway, Uranus aside, this way of thinking about the Solar System and the plane of the ecliptic leads to the idea that the Solar System has something of an “up” that all the planets’ north poles somewhat point to, and a “down” that the various south poles point to.

This is not a useful way to think about space. Thinking like this is what leads to engineers constructing the Death Star, with such obvious design flaws that both the first and second versions suffered catastrophic failure with just a tiny nudge. Don’t be one of those engineers.

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image credit: Andy Langager / flickr.com via cc by-nc 2.0

Astronauts onboard the ISS don’t think of the general direction of Earth’s north pole as “up.” If anything, it’s sort of sideways. The effects of free fall notwithstanding, the ISS and everyone in it remain subject to Earth’s gravity at about 90% of what we feel on the ground — they just don’t notice it because they are falling all the time, just as I described in my previous post on this topic. If there’s a “down,” it’s toward the Earth’s surface, and “up” is away from Earth.

So likewise, when we are orbiting the Sun, and outside any planet’s SOI, it’s more useful to think of “down” as being toward the Sun, and “up” as away from the Sun. On a more technical level, the inner planets (Mercury and Venus) are in lower orbits, and the outer planets (Mars and beyond, though the belters of The Expanse might have a thing or two to say about referring to Mars as an “outer” planet) are in higher orbits.

Delta-V, Prograde and Retrograde

So, referring back to the earlier post, you’ll recall that it’s not possible (or at least not feasible without a very powerful torchship, but that’s beyond the current discussion) to travel from a lower orbit to a higher one by simply aiming the spaceship “up” and firing the rockets, and that likewise we cannot travel from higher orbit to lower orbit by aiming the spaceship “down” and firing rockets. Well, we could, but it wouldn’t give us the desired result.

Instead, we need to fire our rockets either prograde (to raise our orbit) or retrograde (to reduce or lower our orbit). Either way, what we want to achieve is the appropriate delta-v, or change in velocity (often expressed as Δv), required to match orbits with our target.

To get to Mars, once we have left Earth’s SOI (achieved an Earth orbit high enough that the Sun’s gravitational influence takes over), we don’t point our spaceship toward Mars at all. Instead, we point it forward along the direction of Earth’s orbit around the Sun, i.e. in a prograde direction, and fire the rockets. We are already orbiting the Sun at the same velocity as Earth, which just like the orbital velocity of a satellite around Earth is defined by the same equation, √(GM/r), that you surely recall from Orbital Mechanics. Without going into the equation’s details, the Earth moves along its orbit at a rate of approximately 108,000 kilometers per hour, which is fast.

So our spaceship already has this much velocity around the Sun. Mars, in contrast, orbits at an average of 86,760 km/hr (you’ll recall that objects in higher orbit move slower). What we need to do is accelerate prograde, expending delta-v to raise our orbit until it matches that of Mars.

Hohmann Transfer

In Orbital Mechanics I didn’t go quite far enough into the details of just how raising or lowering an orbit works, other than discussing about thrusting prograde or retrograde. However, the principle remains exactly the same, whether in orbit around the Earth or orbit around the Sun.

When we accelerate prograde, the orbit as a whole doesn’t just expand outward to the higher altitude. Instead, it takes our current mostly circular orbit around the Sun and reshapes it into an elliptical orbit instead, with the perihelion, or closest point to the Sun, being the distance (altitude) of our starting orbit, i.e. Earth’s orbit. As we apply thrust, either more powerfully or for a longer period of time, our orbit becomes more and more elliptical, with the aphelion raising higher and higher, or farther and farther out.

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image credit: user:MenteMagica / wikimedia.org

The perihelion of our orbit doesn’t change, however. Eventually, after we have accelerated for a long enough time, our aphelion matches the orbit of Mars, but our perihelion remains at the orbit of Earth. We are now in what’s known as a Hohmann transfer orbit, and if we do nothing else, our spaceship will cycle endlessly back and forth between Mars’ orbit and Earth’s orbit. That, of course, could be quite useful if our intent is to set up some kind of shuttle or cargo transport back and forth, assuming we can match up the times of aphelion and perihelion to when Mars and Earth will be in the same space as our transport craft.

But we aren’t setting up a cargo shuttle. We want to get to Mars, and we want to stay there (for now). We could, upon arriving at aphelion (and thus the orbit of Mars), once again burn prograde. By burning at aphelion instead of perihelion, we won’t be further raising our aphelion; instead we’ll be raising our perihelion. This has the effect of (slowly) reducing the elliptical eccentricity of our orbit, i.e. of circularizing it. And indeed, this is precisely how we would set a satellite into, say, geosynchronous orbit around the Earth.

But, this would negate all the efficiencies of the Hohmann transfer, and require a lot of burning, a lot of propellant, and a lot of time (unless we use a continuous thrust engine, such as an ion drive; more about this later).

Instead, if we are clever and time our launch window so that our arrival at Mars is close to when it will be 180° around the Sun from where Earth was at launch time, then we arrive with least travel time (and least delta-v), and the last thing we want to do is now waste all that by raising our perihelion. Instead, by arriving at just the right time, we will insert our spacecraft into Mars’ SOI (sphere of influence, you’ll recall) and decelerate to slow the craft down enough that it is captured by Mars’ gravity.

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image credit: NASA/JPL

Now we’re in Mars orbit! From this point, all the same rules apply about adjusting our orbit around Mars as did for adjusting it around Earth, or for that matter around the Sun. Or around Kepler 62f if that’s what we’re talking about.

Return to Earth

So now it’s time to go home. How is it different? It isn’t, really. In fact, it’s just the same as any orbital maneuver aimed at reducing altitude. We need to reduce our altitude from 228 million kilometers to 150 million kilometers. To do this, we burn retrograde, against the direction of Mars’ orbit. This has the reverse effect from before, in that it reduces the altitude (from the Sun) of our perihelion until it matches Earth’s orbit. We aren’t actually orbiting the Sun in the reverse direction now — we aren’t burning nearly hard enough for that to happen — we’re just pushing back against our orbit so that we start falling in toward the Sun.

Again, we want to time it with a launch window, one in which it will take us just about one-half of a revolution around the Sun to arrive at Earth’s orbital altitude just as Earth arrives to the same spot. And, upon reaching Earth’s SOI, we need to decelerate so that we are captured by Earth’s gravity, and just like that, we’re home! Well, in orbit around home, anyway, but we know what to do from here.

Interplanetary Transfer Summary

When rockets launch from the surface of Earth to reach orbit, or beyond, they usually do so in an easterly direction, and from as close to the equator as practicable, in order to take advantage of the velocity already imparted upon them by the Earth’s rotation. In other words, to reduce how much delta-v needs to be expended to obtain the velocity required for orbit.

When spacecraft leave Earth to go to Mars, they take advantage of the velocity already imparted upon them by Earth’s revolution around the Sun. Burn prograde, and you ascend to higher orbit, or to the outer planets. Burn retrograde, and you descend to lower orbit, or to the inner planets. It’s really no different, whether in Earth orbit, or interplanetary.

But what happens if our spacecraft leaves the Solar System entirely?

Interstellar Orbits

Leaving out for now questions of faster-than-light travel, Alcubierre warp engines, and other such exotic drives, interstellar maneuvers are no different from interplanetary maneuvers. It’s just a question of scale. Just as the Earth and the other planets in our Solar System revolve around the Sun, the Sun and the hundreds of billions of other stars that make up the Milky Way galaxy all revolve around the center of the galaxy. Just as each planet has an orbit around the Sun, each star has an orbit around the galactic center.

And just as there is a path, and a sequence of burns, that describe an efficient Hohmann transfer route between two planets, there is likewise a Hohmann transfer route between any two stars within the galaxy. There are even more or less ideal launch windows, though given the timescale of stellar orbits, we don’t have as much choice about them; we pretty much have to go when we’re ready to go and accept the less-than-ideal configuration.

Interplanetary and interstellar movement is really the same thing; it’s just a matter of scale. A rather large matter of scale, to be sure, but conceptually no different.

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image credit: NASA/Goddard

So when we’re ready to travel between the stars, if our means remains conventional — whether with a short-burn, high-thrust reaction drive, or a continuous-burn, low-thrust drive, or a continuous-burn, high-thrust torch drive — then the way in which we plot our course doesn’t change. We burn prograde or retrograde about the galactic center, and we utilize our home star’s orbital velocity to impart a delta-v advantage upon ourselves.

There are no straight lines. And our spacecraft is always, always in orbit around something.


header image credit: NASA/JPL

Orbital Mechanics

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:

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image credit: Robert Simmon / NASA

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:

Orbit1
image credit: user:Lasunncty / Wikipedia

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.

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image credit: NASA

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!

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image credit: NASA

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