To expand upon that, Kepler’s Second Law of Planetary Motion demonstrates that at different points in an (elliptical) orbit the body will be moving at different speeds; hence, Earth and Htrae would be lagging and leading each other at different locations in orbit. Htrae wouldn’t necessarily be visible from Earth–it would be more or less still behind the Sun–but we’d see gravitational anomolies stemming from the extra body.
There are no hidden planets on the other side of the Sun populated by fascist mind-eating aliens, I’m afraid. However, this guy claims that “the Sun’s centrifugal and magnetic forces hold each satellite in a precise NEUTRAL zone orbit around the Sun.” Uh huh. Let me grab my tinfoil beret.
Sam Hell technically asked whether a planet in the same orbital path as us would be entirely hidden from view by the Sun. The answer is no, as others have argued. The point that Chronos is trying to make is that you could have a planet orbiting the Sun that was hidden from Earth at all times. To do so, it would be in an orbit exactly the same size as Earth’s orbit, but that orbit wouldn’t be in the same location. Thus the two planets would not technically be in the same orbit. Of course we would see graviatational anomalies from Other Earth pulling on the other planets and comets and asteroids and such.
Why couldn’t the two planets remain 180 degrees apart stably when there are two asteroid clusters on Jupiter’s orbital path 60 degrees (or maybe 30 degrees, I can’t remember) in front of and behind the planet? The asteroid clusters don’t seem to lack stability. What’s the difference?
Apples and oranges, me lad, pineapples and papayas.
First of all, the asteroids you mention, commonly referred to as the Trojan Asteroids, are bodies that lie in the stable libration or Lagrange points, which are 60 degrees fore and aft of the orbital path of Jupiter (the L4 and L5 points). The reason stuff tends to collect here is that these are points of gravitational equilibrium with respect to the Sun and Jupiter; in other words, they are a special, degenerate case of the 3 (or n) body problem which has a discrete and stable solution. (The L1, L2, and L3 points are in quasiequilibrium–that is, they are temporarily stable but tend to shift with local dynamics and thus don’t collect permanent clouds of stuff.) In other words, these things are stable because any tendency to drift away is counteracted by a negative response.
Planets in opposition to a main body, however, are in a metastable state; if they are, as Chronos suggests, in complementary elliptical orbits such that they are always directly opposite to one another (imagine two identical elliptical orbits with colinear major axes, each with one focus centered on the Sun and the other foci in opposition to one another), they will always be in oppositional phase. However, if one of them gets even just a little out of tune, or is not coplanar, or is affected by the influence of another body, the system rapidly becomes unstable, with the ellipse shifting around so that the major axes are no longer lined up and then planets are not across from each other; then, this falls into a nondegenerate condition of the three body problem with the barycenter (the common mass center of the three objects) somewhere in between and bobbing and weaving around like the fox being pursued by hounds.
You have to work through the math to really understand why this is true, but because this is a nonlinear problem you have to have a good grasp of perturbation theory as well as vector calculus and phase (or state) space transformations. On the off-chance that your interest extends beyond the dilettante level I recommend checking out Fundamentals of Astrodynamics and Applications, 2nd Ed. by David Vallado, pgs 116-130, which offers a comprehensive introduction to libration points. If you actually want to calculate out the influences, I suggest boning up on MATLAB or Scilab and plan to spend a lot of late nights and weekends illuminated by the glow of a computer screen. But heck, if you’re going to go to that kind of effort, you might was well go ahead and get the degree so you can apply for a job at JPL. (Not that they’re hiring right now. )
Stranger already gave an explanation for this, complete with a fox hunting metaphor. But I wanted to add a little something.
The specific gravitational dynamics problem that Lagrange studied, and demonstrated was stable, considers three masses: M1, M2, and M3, with M1 >> M2 >> M3, and with M2 in a circular orbit around M1. (Or really, with M1 and M2 in circular orbits around each other; but M1’s motion is going to be much smaller.) Given these conditions, the tiny mass M3 can have a stable circular orbit around M1, the same one as M2, but either ahead or behind M2 by 60 degrees.
In other words you can’t just park yourself anywhere in a planet’s circular orbit and expect to stay put, relative to the planet. Only in the vicinity of the two Lagrange points just described is there an attractive “force” (a valley in the surface of potential energy) tending to keep you in place. This is what has happened to the Trojan asteroids, sharing Jupiter’s orbit. They fell into those valleys.
Moreover, the Lagrange points of stability are thrown out the window when M3 becomes comparable to M2. In this thread we’ve been contemplating an alternate Earth, which presumably would have the same mass, not the dinky mass of an asteroid. Perhaps if the alternate Earth were a hollow styrofoam ball, it wouldn’t come right up to us and collide, but would instead settle into one of Earth’s Lagrange points. That’s the best we can do I think.
Interesting thought – in one of Niven’s “Known Space” novels he has the Puppeteers engineer a “[something] rosette”, which seems to be a set of six masses in orbit around a central las mass (star). The six planerts apparently form a hexagon in the single circular orbit. He refers to this situation as if it is both an establised theoretical possibility and a stable one. I haven’t looked into it any farther, but I can see that for certain ranges of masses and distances such an arrangement might be stable.
If so, in that case, you could have a situation in which you had planets out of sight on opposite sides of a sun.
Of course, it’s a completely artificial situation. A “Trojan” situation is easily formed naturally, but seven masses of just the right size don’t just happen to fall together like that.
The term Niven used was “Kemplerian rosette”, but that doesn’t get any hits on Google. “Keplerian rosette”, however, does.
Even if two or more of the masses are comparable, there will still be stable points corresponding to L4 and L5, incidentally; they just won’t quite form an equilateral triangle. So one could have an Earth-twin leading or trailing the Earth in its orbit, just not by exactly 60 degrees (it’s a Comp-level problem to figure out how much it would lead or trail, so I’m not going to calculate it here).
If I bracket the terms so that I’m searching for “Keplerian Rosette” or “Kemplerian Rosettte”, I don’t get any hits on Google, Google Scholar, or Yahoo.
Not bracketing them gets me hits, but of course none of them for the combined terms.
Maybe Niven made it up. Or was quoting a very obscure work.
You’re thinking of a Klemlerer Rosette, proposed by astronomer William Kemplerer (see Klemperer, W. B., 1962, “Some Properties of Rosette Configurations of Gravitating Bodies in Homographic Equilibrium”, Astr. J. 67, 162-167.), who proposed that a configuration of an even number of bodies of alternating masses opposite of one another, or an number of equal mass bodies, all rotating at the vertices of a regular polygon inscribed on a circular orbit about a common barycenter would be stable. (There is no central star in the Puppeteer Rosette, and indeed, a central mass would create an unstable pole on the right side of the complex plane in state space.) This assumes a circular (and unperturbed) orbit of similar sized masses which is unlikely with planetary-sized bodies but which has application with smaller bodies (very small moons, asteroids, planetary ring constituants) as well as cosmic-sized objects like superclusters.
In essense, a rosette is a special case of the n-body problem which is similar to the Lagrange solutions for multiple bodies in a circular orbit. And if I may offer a slight correction to Bytegeist’s post, the Lagrange 3-body solutions of the Jacobian integral apply regardless of relative sizes of the masses; at M1 = M2 = M3 they still orbit at 60 degrees to each other (but at angles and distances relative to the barycenter dependent on the mass ratios). We often speak of the Earth orbiting the Sun and the Sun orbiting the Moon, but (as I’m sure you’re aware) correctly modeling of the behavior requires each body to orbit their common center of mass. The Trojan (L4 and L5) points (which are actually the loci of a scimitar-shaped region of stability) are stable because of their equilateral geometry, which forms at the intersection of geodesic curves about the two major bodies. Because of this, they resist perturbance and accumulate debris. Single bodies in direct opposition of a central mass (L3), however, are at best metastable, even if they are of equivilent masses, and any perturbance will send the system into positive feedback, with each of the secondary bodies pulling the other successively out of position.
sigh That’s a Klemperer rosette. Must learn to copyedit before sending off to the electronic void.
For the case of M2 = M3, the two secondary masses reside at each others complementary L4 or L5 point. Just a system would have some additional “zeros” (secondary points of true equilibrium) lying on the line bisecting the angle between them both nearer and further from the primary in lieu of the L1 and L2 colinear points, and there are other geometrical trivialities which would be of endless fun to Buckminster Fuller enthusiasts.
Well, for all three masses equal, one obviously must have an equilateral triangle by symmetry. But I’ve done the case where M1 = M2 >> M3, and it wasn’t equilateral. I’ve never done the case M1 >> M2 = M3, but I would imagine that it would also not be equilateral (though, of course, without having done the calculation, I might be wrong).
And I don’t know about you, but I don’t often speak of the Sun orbiting the Moon ;).
I haven’t worked the Lagrange L4 and L5 problem myself, but my recollection from Symon’s Mechanics was that you didn’t get a solution if the third mass was too large relative to the second, equilateral configuration or not. That doesn’t seem to square with what people are writing here.
Hal Clenment thought so, too. He used this as a plot device in one of his short stories – when the third mass was too great it didn’t stay in a stable orbit. He probably didn’t do the calculations himself, and I’ll bet he didn’t use Symon, but it must’ve been something similar.
There were actually TWO planets orbiting each other, on the other side of the sun from Earth, but the Lexx blew them up a few years ago. We didn’t know.
I cracked the book and worked through a few sets of parameters just for fun. (Yeah, I’m a riot at parties.) First of all, going through the assumpions, I note that for the Lagrange solutions (or the “Circular Restricted Three Body Problem”) it is implicitly assumed that M3 is negligible, so I have to recant on my claim that the relative masses are immaterial. However, for all solutions M1=M2>>M3 and M1>M2>>M3 I’m getting them in an equilateral triangle regardless of mass, with only their cocircular radius varying with respect to the total mass of the system. (For M1>>M2=M3 I’d assume that the primary dominates any behavior and could care less about how M2 and M3 relate to one another; technically there are stability foci but unless M2 and M3 happen to be in close proximity they’re not going to influence each other much with respect to the primary.) I’m not sure where or how we’re differing, but in poking around a bit every reference I’m finding says that they lie in an equilateral configuration, so either I’m making an assumption somewhere that isn’t true for situations in between Lagrange orbits and rosettes, or we’re talking past each other.
However, doing a little reading beyond my orbital mechanics text (which is mostly concerned with such prosaic stuff as satellites and moon rockets yawn) I find that the solutions aren’t stable unless unless M1/M2 > ~25. I haven’t modelled it out in state space and pumped it through MATLAB 'cause I’ve cheated the taxpayers enough for today, but here’s a neat little Java applet that shows bodies in a circular equilateral configuration (Example 1 & 2) and in a periodic elliptical orbit (Example 7 & 8), with Example 9 showing a satellite in the Earth-Moon L4 point. In this applet it shows a satellite in the L4 point of the Earth-Moon system with an exaggerated elliptical orbit.
You would if you’d only acknowledge the Truth of Lunacentricity. All hail Eros!
I take that back; going back to the generalized 3-body equations of motion with M1>>M2=M3 I still get an equilateral configuration, plus (graphical SWAG) two points of secondary relative equilibrium and five points of quasiequilibrium. But I don’t know how stable the overall system is; I’d hazard a guess that it isn’t stable (or at best, it marginally stable) for any values of M1 and M2.
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