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Is there a perfect orbit?
October 16, 2009 4:56 PM   Subscribe

Is there a distance at which a satellite will orbit a body indefinitely? Is this distance a universal constant?

As I understand it, satellite bodies are generally getting closer or further away from the body they orbit. Our moon, for example, is getting further away.

This morning I was reading a great essay by Amy Leachin the Best American Essays 2009.

In the second paragraph, it is noted that unless you [a satellite] happen to roll onto a track precisely 18,254 miles above your planet, the law ejects you or dashes you down. One moon in our solar system has achieved synchronous orbit, being pledged forever to its planet--Pluto's moon Charon. The other 168 moons have not.

According to google, 18,154 miles = 29,216 kilometers.

The wikipedia article about Charon cites a distance of 17,536 ± 4 km to system barycenter, 19,571 ± 4 km to the center of Pluto
. The radius of Pluto is, apparently, 1,153 ± 10 km. 19,571-1,153=18,418, which is close to 18,254 (though not 'precisely'), but nowhere near 29,216.

Am I missing something obvious (I'm not very good at space, so I am pretty sure this must be the case) or is Leach using poetic license, or is it a bit of both?

I find it hard to believe that there would be a universal perfect orbit distance, as celestial bodies are of varying mass, and there are also external tidal forces acting on them. However, I also find it hard to believe that there are big balls of rock and ice in mutual orbit, and that these systems are, themselves, in orbit around a huge ball of gas.
posted by doublehappy to science & nature (28 answers total) 2 users marked this as a favorite
 
Assume for the minute that there is a perfect orbit. That would imply a place where things NEVER change.

Never is even a larger word in this sense than its usual sense. We are ever so slightly tugged across truly enormous distances, so it seems that there is no locale to which one could retreat that was free of influence of distant bodies. Of course, as far as governing effects are concerned, in the short term you can probably ignore infinitely small tugs, but you did say NEVER. Over a long enough period, even a small force integrates to a large number.

Since the universe is winding down, cooling off, and expanding, and since it is all pervasive, your scenario seems to have a bit of a problem. I can neither cite math or astrophysics, but the apparent paradox suggests I might not have to.
posted by FauxScot at 5:19 PM on October 16, 2009


What he's talking about is tidal effects. The "stable" orbit is synchronous, where the planet's rotation and the satellite orbital period are identical.

If the planet turns faster than the satellite and if the satellite is large enough to induce a non-negligible tidal bulge in the planet, then the tidal bulge will slightly lead the satellite, which causes a force on the satellite in the direction of its orbit. This adds energy, which moves it outwards. That's happening with Earth's moon, whose orbital radius is increasing now at a rate of a couple of meters per century.

If the planet turns slower than the satellite, and again if there's a non-negligible tidal bulge, then the tidal bulge trails the moon, pulling it backwards. That reduces the orbital energy of the satellite, making it move into a lower orbit. Eventually it'll either hit the planet or break up due to tidal stresses.

But the synchronous orbital radius is not a fixed distance. It depends on the mass of the planet, to a limited extent to the mass of the moon, and very critically on the rotational period of the planet. Generally speaking, the faster the planet turns, the shorter the radius to a synchronous orbit.

And if the moon is so small that it doesn't induce a substantial tidal bulge, then the process happens at a speed so slow that it can be ignored.
posted by Chocolate Pickle at 5:20 PM on October 16, 2009 [1 favorite]


...What she's talking about...
posted by Chocolate Pickle at 5:21 PM on October 16, 2009


With difficulty, I read the entire essay. Amy Leach turns science into poetry in a way that debases both science and poetry.

For a given planet, there is a "magic altitude" at which the period of a circular orbit will match the planet's day. However, it is not the same altitude for all planets. See synchronous orbit. This is a more stable orbit because it minimizes tidal drag. But, as FauxScot points out, nothing is forever. On preview, what CP said.
posted by drdanger at 5:22 PM on October 16, 2009 [1 favorite]


By the way, the problem is that even for a given planetary system, it'll change over time. Suppose Earth's moon was in synchronous orbit. (Well, it can't be because if it was that near it would break up. Make it smaller and put it in that orbit.)

The sun also induces a tidal bulge in the Earth. It happens not to be as large as the Moon's tides, but it's there, and it tends to slow the rotation of the Earth. As the Earth's rotation would drop, then the synchronous orbital radius would increase -- but nothing would move the moon upwards. On the contrary, it would start spiraling into the planet, slowed by a trailing tidal bulge.

So over the long term the system still wouldn't be stable.
posted by Chocolate Pickle at 5:23 PM on October 16, 2009


Hmm... Suppose the planet was tide-locked on the star. In that case its day and year are the same length, and those no longer change.

But in that case the synchronous orbital period for the moon would also have to be a year, and that means it would have to be so far from the planet, that it would fall out of planetary orbit entirely and go into orbit around the star. That's still no answer.
posted by Chocolate Pickle at 5:29 PM on October 16, 2009


I think I've found your problem:

"Amy Leach has an MFA in creative nonfiction" ≠ Science
posted by Aquaman at 5:46 PM on October 16, 2009 [1 favorite]


Yeah, but if you're quoting figures in an essay there surely must be some loose requisite truth. What I'm really asking is, ignoring external forces and the impending end of the universe, &c. is it feasible that there would be a distance that a satellite could orbit a planet and not get closer or further away, as Leach's passage suggests. And it looks like the answer is sort of.
posted by doublehappy at 6:13 PM on October 16, 2009


I think we're being too kind. This is not exactly some subtle error here that can be excused away:

The law is stringent about this; there are no clauses; and all moons are dutiful followers of the law. But, as all good followers of the law discover in the end, unless you happen to roll onto a track precisely 18,254 miles above your planet, the law ejects you or dashes you down.

18,254 miles is a magic distance? No, it isn't. This is false. Untrue. Wrong. In error. Inaccurate. Devoid of truth. Hokum. Bogus. BS.

Creative nonfiction, in the sense that it is a created fiction.

This is award winning? Does this sort of thing happen in Europe or Asia? I'm serious, I want to know.
posted by effugas at 6:18 PM on October 16, 2009 [3 favorites]


If you ignore the tidal and drag effects, all orbits that do not intersect the primary are stable. It's pretty much the definition of an orbit.

However, an orbit sufficiently close to a planet will have drag from the atmosphere and magnetosphere. That slows the orbit, dragging the satellite into progressively lower orbits.

Likewise, Chocolate Pickle has explained the tidal effects.

And all orbital calculations depend on the mass of the primary. So the altitude of geosync orbit is not a universal figure, as it varies by the mass of the primary (and the mass of the satellite, if that's significant). If you google "orbital calculator", there are dozens of web applications that will calculate all terms of an orbit for you.

As for the geosync figures the quoted author gives... well, that doesn't match at all for Earth, where geosync is roughly 35,000km. I'm inclined to believe that she confused kilometers with miles whens she wrote the essay, because then at least the figures match. As for the precision of the match... well, who knows who did her math, what measurements they used, and how it was conducted. Did she even get her sigfigs right?

Also: Aquaman nails it. An MFA in creative non-fiction leaves her lacking in all credibility for scientific figures. Maybe if she was a sci-fi author, I'd be inclined to believe she did and understood her homework, but even then, I dunno. Don't sweat it that her shit doesn't match up.
posted by Netzapper at 6:22 PM on October 16, 2009


doublehappy--

No. This is scientific people being charitable. There is _no_ such fixed distance, which is the whole point of the article. The distance is dependent on relative mass, volume, and spin rates, and will vary from moon to moon.

Put another way, this essay could not even remotely be written if there was any grasp of the science. The author might as well be about the solar system being an entertaining little mobile hanging over the crib of an infant diety, for all the scientific meaningfulness of the text.

(And yes, I'm intentionally trying to counterbalance all the hemming and hawing going on in the thread.)
posted by effugas at 6:23 PM on October 16, 2009 [2 favorites]


What I'm really asking is, ignoring external forces and the impending end of the universe, &c. is it feasible that there would be a distance that a satellite could orbit a planet and not get closer or further away, as Leach's passage suggests. And it looks like the answer is sort of.

Ignoring all other forces (including tidal) but simple gravity between the primary and the satellite, all orbits are stable if the orbital path does not intersect the primary. All circular orbits qualify for your "not get closer or farther away" requirement, as the altitude is constant for a circular orbit. But even an elliptical orbit is stable... it's just not a constant altitude.
posted by Netzapper at 6:25 PM on October 16, 2009 [1 favorite]


Ignoring tidal locking and geosynchronous orbits, it is possible that Ms Leachin was getting the Pluto-Charon configuration (in which, technically, neither are moons, as neither is a planet), confused with the concept of Lagrangian points, islands of stability that are possible in most any orbital three body problem. It should be noted that even these points are subject to permutation - L4 and L5 are the most stable, with, if I remember correctly, an average of 54,000 years stability for an object in orbit. (The upcoming James Webb telescope will be parked in L2).
posted by Bora Horza Gobuchul at 6:36 PM on October 16, 2009 [1 favorite]


It should be noted that the Lagrange Points are islands of stability in that they do not orbit. They're the points in an orbital system where you can remain stationary with respect to the primary.

But that's not what Ms Leachin was getting at. Ms Leachin is not qualified to write what she set out to write, and so wound up writing a whole steaming mound of horse shit--even confusing kilometers for miles in the Pluto-Charon altitude figure. She misunderstood something she heard at a cocktail party, decided it fit with some preconceived essay she had in mind, and ran with it. And now bright, sciency mefites are trying to square it with reality. But, really, it's horse shit.
posted by Netzapper at 6:56 PM on October 16, 2009


is it feasible that there would be a distance that a satellite could orbit a planet and not get closer or further away, as Leach's passage suggests. And it looks like the answer is sort of.

The answer is an unqualified "no". Not possible without drastically changing the way gravity works.
posted by Chocolate Pickle at 7:43 PM on October 16, 2009


I thought someone would have already explained how orbits work, but apparently not, so I'll give it a shot:

A satellite that is orbiting another body is actually "falling", just like everything else. Imagine you hold a gun exactly level on some huge open field, and then pull the trigger. As soon as it leaves the gun, it starts to fall towards the ground. However, because the Earth is (roughly) a sphere, the ground also curves away from the bullet. Normally this doesn't have much effect, because the bullet doesn't travel very far before it hits something or even before it hits the ground. Imagine that you had a very, very powerful gun, though. The bullet travels _really_ fast, so it can travel a long way before hitting the ground. There is a certain speed where over a given linear distance (that is, the "horizontal" distance away from you), the bullet drops vertically _exactly the same_ distance that the Earth curves away from the bullet. If you fire the bullet at this speed, then it just keeps on traveling forwards and falling down, but the Earth keeps curving away from it and the bullet orbits the Earth!

In real life, there is a ton of air resistance against the bullet, so it slows down rather quickly and can't stay in orbit. However, an astronaut in the space station doesn't experience (much) of this air resistance. Once you get going forwards at the right speed, the Earth will fall away from you at the same rate that you fall towards the Earth. That rate of falling, ie "the acceleration due to gravity", seems mostly constant for us walking-around-types near sea level, but it's actually dependent upon altitute - gravity is a little bit weaker on top of Mount Everest, for example. So for any given altitude, there is a magic number which is your fixed speed to remain in orbit. For the space station, this is approximately 5 miles per second.

That's all there is to orbits - there is no physical rule that will make an object get slightly closer to or further from an orbited body if it's not at some "perfect orbit" altitude. There is no such altitude. The moon is retreating from the Earth because of a ton of other effects - the pull of all the other bodies of the solar system, for example. These multi-body systems get really complicated, and the only way we can calculate what will happen is number crunching on computers - we don't have the formulas for a closed-form solution.

The article seems to be talking about a very special kind of orbit - geosynchronous. This is the orbit at which a satellite stays fixed over a particular spot on a planet. A huge percentage of Earth's man-made satellites are in geosynchronous orbit, so that each of them stays in constant line-of-sight of a particular area. Note that if a planet is not rotating, then it is impossible to be in geosynchronous orbit. If the planet is rotating really fast, then the satellite has to be moving very fast as well, which means a higher orbit. If the planet rotates slowly, then the satellite can be closer to the planet, but not so close that the orbit quickly degrades due to air resistance.

Obviously, then, there is no universal perfect orbit. It depends on the strength of gravity of a planet, which in turn is affected by the planet's mass. If the planet has no atmosphere, then lower orbits are possible, provided that the right speed is attained. And geosynchronous orbits depend directly on whether and at what speed the planet is rotating.

Award-winning essay indeed!
posted by RobotNinja at 7:47 PM on October 16, 2009 [4 favorites]


Fundamentally, it's a bad concept because gravitational forces are relative to the distance to the center of mass, not the distance to the surface of the planet.
posted by smackfu at 7:52 PM on October 16, 2009


If the planet is rotating really fast, then the satellite has to be moving very fast as well, which means a higher orbit. If the planet rotates slowly, then the satellite can be closer to the planet, but not so close that the orbit quickly degrades due to air resistance.

You got that backwards. Low orbits are fast. High orbits are slow. If the planet is rotating rapidly, a synchronous orbit will be lower.
posted by Chocolate Pickle at 8:14 PM on October 16, 2009 [1 favorite]


Am I missing something? Unless there's some odd positive feedback that isn't obvious from basic mechanics (and some have hinted at this, but I've never heard of it) orbits should self-stabilize. That is, a velocity that's too great for a given orbit will push a satellite away from the main body, but the wider orbit will correspond to a higher velocity, so -- unless the velocity is greater than escape velocity -- eventually an orbit will be reached that is "right" for the given velocity.

Obviously, the universe isn't Newtonian -- objects deform, systems aren't closed, the protons that constitute the bodies will eventually decay -- but is there any way that, under non-extraordinary conditions, the notion of the link is not just inaccurate but exactly the opposite of true?
posted by bjrubble at 10:15 PM on October 16, 2009 [1 favorite]


Amy Leach has an MFA in creative nonfiction from the University of Iowa

Could be worse; my sister has an MFA from the Iowa program, and she gives her kids homeopathic medicine. On the other hand, a very smart doctor friend of mine once spent a whole long car ride explaining to me why there could only be one stable orbital speed around the Earth, irrespective of altitude — I'd love to get him and Ms Leach in a room together.
posted by nicwolff at 10:34 PM on October 16, 2009


Man, there's lots of contradictory information in this thread. Hopefully the information I'm adding is no less correct than the rest.

In the simple situation you learn about in freshman physics, all orbits are stable. They may be elliptical (that is, the satellite/moon is closer to the primary during one part of its orbit than another) but they'll persist indefinitely without changing— there's a fairly straightforward conservation-of-energy argument for this, actually— and for a basic understanding of orbital mechanics this is good enough.

In the real universe there are other effects, which affect various satellites more or less: Doubtless there are others. Many of these effects will cause orbital precession as well.
posted by hattifattener at 11:36 PM on October 16, 2009 [1 favorite]


Am I missing something? Unless there's some odd positive feedback that isn't obvious from basic mechanics (and some have hinted at this, but I've never heard of it) orbits should self-stabilize. That is, a velocity that's too great for a given orbit will push a satellite away from the main body, but the wider orbit will correspond to a higher velocity, so -- unless the velocity is greater than escape velocity -- eventually an orbit will be reached that is "right" for the given velocity.

As chocolate pickle pointed out, you have it backwards. A wider orbit corresponds to a lower average speed. A better way to think of it is in terms of energy. More energy means a bigger orbit. The tidal drag adds energy to the moon which increases its orbital radius, but at the same time its orbital speed decreases. Think of it as pushing a car up hill. You have to add energy to push it up hill, but its speed decreases.
posted by JackFlash at 12:38 AM on October 17, 2009


I think the speed thing is being confounded.

If the orbit is geostationary, the further up you go, the faster the thing has to go to stay in track with the point on the ground.

But at the same time, there is the concept of stable orbit- that the object generally stays at the same altitude without needing to propel itself. The force of the object being pulled toward the ground is counteracted by the inertia of the object wanting to go in a straight line. Since the effects of gravity are lower the further out you go, the inertia of the satellite needs to be commensurately less.

(Like twirling a rock tied to a string over your head, where gravity is the string.)

So what she is saying is that for an orbit to be both geostationary and stable, there is a small band of altitude where that can only be possible for a given mass combination. If the thing were any lower, it would crash, or have to be going faster than geostationary. Any higher and it would shoot off into space, or be slower than geostationary.
posted by gjc at 6:10 AM on October 17, 2009


If mass and velocity remain the same, yes you could have equilibrium.
posted by blue_beetle at 7:10 AM on October 17, 2009


I came in to say basically the same thing that hattifattener said, except that there is no way I would have been cool enough to think of using the phrase "lithospheric drag" to describe the process of an orbit becoming an ex-orbit. That was funny.
posted by roystgnr at 8:49 AM on October 17, 2009


From the essay: So fast moons slow down and slow moons speed up, and only during excerpts of time do planetary dalliances appear permanent. Our moon through many excerpts--the Moon--is a slow moon. Thus it is speeding up, thus it is falling up, coming off like a wheel, at one and a half inches per year.

I see where some people in this thread have gotten the wrong idea. The statement above is exactly backwards. Fast moons don't slow down, they speed up as they move closer to the planet. Slow moons don't speed up, they slow down as they move farther away from the planet. This explains the orbital instability she is referring to -- the feedback is positive, not negative. If it worked backwards as she describes, all orbits would be stable regardless of tidal drag.


However, my favorite part of the essay is this breathless declaration, with her italics:

However, as the Earth is eighty-one times more massive than the Moon, the barycenter is eighty-one times closer to the Earth: thus the barycenter is inside the Earth.

Hey, I think I saw that movie.
posted by JackFlash at 11:23 AM on October 17, 2009


doublehappy: Yeah, but if you're quoting figures in an essay there surely must be some loose requisite truth. What I'm really asking is, ignoring external forces and the impending end of the universe, &c. is it feasible that there would be a distance that a satellite could orbit a planet and not get closer or further away, as Leach's passage suggests. And it looks like the answer is sort of.

The answer is yes although the author gets some of the reasons backwards as explained above. Pluto and Charon are a very special case in that they are the only two large objects in the solar system that we know of that are mutually locked in stable synchronous orbits.

As described above, in most cases tidal drag causes moons to move away from stable, synchronous orbits. Moons inside the synchronous orbit speed up and move closer to the planet while moons outside of the synchronous orbit move slow down and move away from the planet. So in both cases moons move away from the synchronous orbit. They are unstable.

It turns out that there is a small subset of starting conditions in which the opposite occurs, that the two objects move towards synchronization. It requires that the two objects be nearly the same size (Pluto and Charon are unusual in this respect), that they start out somewhat closer than the synchronous orbit (also unusual), and that the two have a narrow range of initial rotation speeds and tidal dissipation. These two papers (1 and 2) describe how under these special conditions, the two objects will slow down under tidal interaction and gradually move away from each other into synchronous orbits and become locked and stable once they reach that point.

It is not the case that Pluto and Charon magically happened to be created at exactly the right orbital distance, 18,254 miles, as the author implies. It is that Pluto and Charon started under unusual conditions that naturally evolved toward a stable, synchronous orbit. It may have taken a couple of billion years to get there, but once locked, the orbits became stable.

You might wonder that if most other moon orbits are unstable, why do they still exist. Why haven't all of the moons crashed into their planets in one case or escaped into the solar system in the other? The answer is basically time. The tidal effects are very small and the age of the solar system is not long enough for those terminal cases to happen.
posted by JackFlash at 12:37 PM on October 17, 2009


Why haven't all of the moons crashed into their planets in one case or escaped into the solar system in the other? The answer is basically time. The tidal effects are very small and the age of the solar system is not long enough for those terminal cases to happen.

We don't know that it hasn't happened. If Jupiter had swallowed a moon, how would we know?

Phobos is losing about 20 meters of orbital radius per century due to tidal drag and is expected to impact Mars in about 11 million years.
posted by Chocolate Pickle at 12:59 PM on October 17, 2009


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