
#1
Dec213, 04:04 PM

P: 22

Is a solar system possible where there was constant planetary alignment (syzygy)
Let's make it the simplest scenario : just two planets orbiting a sun [the planets are different ] 



#2
Dec213, 04:16 PM

P: 566

Try applying Kepler's third law and see what happens.




#3
Dec213, 04:54 PM

P: 22

How about if one of the planets was a captured rogue ?, (i.e. not a native of that solar system) , would Kepler's third law still apply ? 



#4
Dec213, 05:04 PM

P: 566

Is constant planetary alignment (syzygy) possible ?
Indeed.
For two planets to orbit the star with the same periods, their distances from it have to be equal. There's no way around it. Two bodies can share the same orbit, though. Look up "threebody problem", and in particular the solutions to it called Lagragian points. Of the Lagrangian points, only L3 allows for all three bodies to be aligned in syzygy, with the two planets offset by 180°, and the star in the middle. Of course, you could not see one planet from the other, as it'd be obscured by the star. Additionally, L3 is unstable. Any, even a slightest, deviation from it will change the orbits in analytically unpredictable fashion. [edit]: I should probably clarify that the lagrangian points are solutions to the three body problem where one of the bodies is much less massive than the other two, so in general they do not apply to the situation with two planets and a star where masses are comparable. But if the two planets are of equal masses, they can still share the orbit in each other's L3 points(with the unstability caveat still applying). This, however, doesn't work for other Lagrangian points that otherwise allow a test particle(m<<M[sub]1[sub],M_{2}) to orbit in constant syzygy like L1 and L2. Also, it's worth noting that the concept of Lagrangian points as actual points applies to idealised scenario where orbits are circular. For elliptical orbits, the points are more like centres of areas around which the bodies librate. 



#5
Dec413, 04:43 PM

P: 1,271

As you know, the orbit of a planet traces an ellipse, of which the circle is a special case. What is necessary for two planets to have the same orbital period is that the semimajor axis of the ellipse for both has to be the same, and this will hold true regardless of the eccentricity of the orbit. For the earth, having an orbit which is nearly a circle, the distance of the earth to the sun can be considered to be the semimajor axis, and that the sun is at one focus ( or both foci ) of the earth orbit. The other planet would need an orbit of higher eccentricity, with the sun being again one focus of the orbit, but not the midpoint of the major axis. At times during the orbit the planet would be closer to the sun than the earth and at other times farther away. 



#6
Dec513, 12:56 PM

P: 61

"Constant" is a relative term. For celestial mechanics, where time is reckoned in billions or years, a constant syzygy for a threebody problem (any of the 5 Lagrange solutions to the gravity potential well problem) can only exist if 2 of the bodies are point masses. Otherwise tidal forces (gravity gradients) will deform the rotating bodies and dissipate kinetic energy into heat. The orbits will gradually decay as energy is reduced and angular momentum is conserved. That's is probably how Theia met Protoearth and gave birth to the Earth  Moon system, which continues to change  slowly  4 billion years later.




#7
Dec513, 01:03 PM

P: 63





#8
Dec713, 08:01 PM

P: 1,271

Perturbation of the orbits would occur and could be calculated in the same manner as that which what is done for the solar system planets. The planet with the more elliptical orbit would also suffer a perihelion advance due to relativistic effects than that of planet of the more circular orbit, but it could takes millions of year for a complete turn of the orbit, depending upon the eccentricity. 


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