We've all heard the statement that on the 21st of December, the planets in the solar system will "align" from the point of view of the Earth. I assume this means that they would all be in the same spot in the sky if we looked from here. The theory says that the alignment of the planets will somehow exert some influence on the Earth which would bring varying levels of catastrophe, depending on who you ask.

Now, it has been said many times that this will not actually happen, and that even if it happened there would be no effect on the Earth whatsoever. I know that, and that's not the question.

What I'm wondering here if it is actually possible for the planets to align in this way, regardless of whether it'll actually happen. As far as I know, the planets' orbits aren't all in the same plane, so it doesn't seem even theoretically possible, i.e., there's no straight line passing through the orbits of all the planets. Am I right?

  • $\begingroup$ I was under the impression that the planets do orbit on roughly the same plane. Similar to how Saturn's rings are flattened into one plane of orbit. Pluto was the rouge planet that had an orbit that deviated drastically from this plane, and therefore lost its right to be a planet. So let us even approximate it as one plane. Is there still a way which they would all line up? Or is the variation in their orbital periods too much to have a lining up? It might never happen even if they are on the same plane. $\endgroup$ – Todd R Dec 13 '12 at 2:01
  • $\begingroup$ Really? While I must admit I don't know where I got it from, I really thought the orbits were on different planes. $\endgroup$ – Javier Dec 13 '12 at 2:57
  • $\begingroup$ Some of the planes are "tilted", but in general they share roughly the same plane. This makes sense to me, because planets which orbit different planes would always see some degree of attraction between each other. Although minuscule compared to their attraction towards the Sun, this attraction would be non-uniform, always tending ever-so-slightly to the plane of their neighbors'. Over billions of years, maybe this attracts them all towards the same plane? (p.s. I'm a computer programmer, not an astrophysicist. So I might be WAY off. Like ASTRONOMICALLY off! haha GET it??) :-) $\endgroup$ – loneboat Dec 13 '12 at 3:22
  • $\begingroup$ Check this out: physicsforums.com/showthread.php?t=417310 , specifically the answers by member "mikelepore". $\endgroup$ – loneboat Dec 13 '12 at 3:28
  • $\begingroup$ I believe a planetary alignment generally means that they appear to line up across the sky to some reasonable approximation. $\endgroup$ – dmckee Dec 13 '12 at 5:18

First, Mercury "aligns" with the ecliptic plane only twice in its "year", when it comes from above to below and vice versa.

Luckily for our calculations, Pluto is not a planet any longer, because it would completely rain on our parade with its 248 Earth years of orbital period and another two points within it that it crosses the plane again. Getting Pluto and Mercury aligned alone would take millennia.

Now, what do we count as "aligned"? This is a very vague term because it doesn't state any tolerances. If you mean discs of the planets overlapping, just forget it, their own minor deviations from the ecliptic plane will suffice that it will never ever happen. Let us assume a tolerance of one earth day of their movement. This is fairly generous, in case of Mercury it's over 4% tolerance of its total orbit radius, which considering their size on the sky is quite a lot - in case of all planets the distance traveled over one earth day far exceeds their diameter. So, we're not taking a total alignment, just one night where they are closest to each other, a pretty loose approximation.

Now, we pick the day the rest of the planets are on the plane as Mercury, so let us simply take the 2 in 88 days of its orbital period and continue dividing by orbital periods of other planets.

1 in (44 * 225 * 365 * 687 * 4332 * 10759 * 30799 * 60190) days. That is one day in $5.8 \cdot10^{23}$ years. The age of the universe is $1.375 \cdot 10^{10}$ years.

It means planets would align for one day in 42 trillion times the age of the universe.

I think it's a good enough approximation to say it is not possible, period.

Feel free to divide by 365, if you don't want aligned with the Sun but only with Earth. (one constraint removed.) It really doesn't change the conclusion.

  • $\begingroup$ Is it really as easy as simply multiplying the periods? Imagine alignment of 3 planets with the sun, in the same plane, with perfectly circular orbits, and periods 1, 2 and 3. If you imagine a cube of side $2\pi$, the position of all 3 planets is determined by a single point, and alignment by points in the diagonal joining $(0,0,0)$ with $(2\pi,2\pi,2\pi)$. Because the ratios of the orbits are rational numbers, the trajectory through the cube will be a non-space filling line. I am not so sure that this line intersects that diagonal always... $\endgroup$ – Jaime Dec 13 '12 at 18:10
  • $\begingroup$ @Jaime: Imagine two celestial bodies (my case: Sun and Mercury, or the "divide by 360" case - Mercury and Earth) Draw a line through them. No matter where the line is on given day, the chance any given planet lies within a day distance of the line is 1/[orbital period of that planet]. Chance of multiple planets being on that line at given time is a simple product of these. There might be an order or two of magnitude error in my calculations but seriously, whether it's 10^23 or 10^18 years is moot. $\endgroup$ – SF. Dec 13 '12 at 18:36
  • $\begingroup$ @Jaime: Oh, wait. I see what you mean: each planet aligns with the line when it's on the same side of the Sun, and on the opposite too, twice per its orbital period. This exactly doubles the chance in case of each of them. So, my calculation is off by 2^7 times, or else divide my result by 128. Still, 10^21, or with alignment with Earth, 10^18 years... $\endgroup$ – SF. Dec 14 '12 at 12:18

All the planets except Mercury (7 degrees off) and Pluto (17 degrees off) are on the ecliptic plane. So a perfect alignment is not possible. I'm including Pluto as a planet out of habit.

  • 1
    $\begingroup$ BUT PLUTO'S NOT A PLANET! ;-) Poor Pluto. We miss you. $\endgroup$ – loneboat Dec 13 '12 at 3:28
  • $\begingroup$ Well what if we just forget about Mercury and Pluto altogether? They are both the hardest to see (I assume) due to their extreme proximity and distance to the Sun anyways. Would it be possible for the remaining planets to be aligned all along a single line pointing out from the sun? Will this alignment ever happen? If so, that would be a pretty spectacular event to observe. $\endgroup$ – Todd R Dec 13 '12 at 3:44
  • $\begingroup$ I disagree. Even if the planes are different, the planes still intersect and precess, so it's theoretically possible for all to be aligned. $\endgroup$ – gerrit Dec 13 '12 at 9:53

Theoretically, no, as the plane of orbit of each planet is tilted slightly in relation to other planets. However, if we ignore the intra-orbital plane tilt, then the chance of all planets (eight now excluding Pluto) to be at the right ascension with respect to sun at same time would be once in 180 trillion years.

For an exact alignment, when all the planets are inclined with respect to the ecliptic, we must factor the line-of-nodes recession into calculations, that makes the chance once in 86 billion trillion trillion trillion years (86 followed by 45 zeros).

The odds strongly favour the fact that an exact planetary alignment will never occur within the lifetime of the solar system that now has only about 12 billion years left.

— S P S Jain, Greater Noida


The orbital planes are all different. However, the orbital planes do intersect, and the orientation of the orbital planes precesses slowly. Therefore, it is mathematically possible that at some moment $t$, all orbital plane intersections would be at the same angle and all planets would be at this position within their orbital plane. One could do the calculations, but I'd expect that this state is so unusual that the expected time to wait for it is longer than the expected lifetime of the universe.


protected by Qmechanic Oct 4 '15 at 15:07

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