Ignoring expansion of the universe, entropy, decaying orbits, and interference from any bodies colliding with or otherwise interfering with their orbits, will the eight planets known planets in our solar system ever align?

What is the "period" of the planets; how often would they align perfectly? And based on their current positions, how far off into the future is their next theoretical alignment?

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    $\begingroup$ In a strict sense - never. The orbits are not co-planar, they are not in the same plane. As such, an alignment in the proper sense can never occur, it's more of a media- and rumor-created notion. $\endgroup$ Commented May 13, 2014 at 22:50
  • $\begingroup$ @FlorinAndrei Aren't all (except Mercury, who is just being rebellious) within ~3° of each other? Not perfect, but good enough for me. $\endgroup$
    – IQAndreas
    Commented May 22, 2014 at 22:00
  • $\begingroup$ I posted an answer and would like to know if whether it does answer your question or you need a more precise one, so I could expand it. At least provide some feedback, I would appreciate it. $\endgroup$
    – harogaston
    Commented May 23, 2014 at 3:41
  • $\begingroup$ Never even if they were co-planar. $\endgroup$
    – Walter
    Commented Oct 5, 2015 at 8:07
  • $\begingroup$ Ignoring [...] interference from any bodies [...] interfering with their orbits -- this obviously includes the Sun, and without the Sun, the planets orbits are not well defined. Hence your question is unclear. $\endgroup$
    – Walter
    Commented Oct 5, 2015 at 8:40

5 Answers 5


This is a low accuracy - yet simple - answer

It allows you to calculate only radial alignment configuration of the planets.

If you would like an approximation, let's say, you approximate the position of the planets as hands in a clock, you could work the math out by something like this.

Assume $\theta_i$ is the initial angle for planet $i$ at time $t_0$ - measured from an arbitrary but fixed position, and $l_i$ is the length of the year - in days - for planet $i$.

Then it resumes to solving this system of equations:

$$ x \equiv \theta_i \left( \ mod\ l_i\right) $$

From here you would then simply apply the Chinese Remainder Theorem.

Finding the minimum x, will give you the angle that the planet that at $t_0$ had angle $\theta_i = 0$ would have travelled until an alignment configuration was reached. Asuming you choose Earth as the mentioned planet, then divide that angle by a complete revolution ($360^{o}$) and you will get the number of years for that configuration to be reached - from the $t_0$ configuration.

The different $\theta_i$ in degrees for all the planets on 01 Jan 2014 - you can use this as your $t_0$:

\begin{align} Mercury &\quad 285.55 \\Venus &\quad 94.13 \\Earth &\quad 100.46 \\Mars &\quad 155.60 \\Jupiter &\quad 104.92 \\Saturn &\quad 226.71 \\Uranus &\quad 11.93 \\Neptune &\quad 334.90 \end{align}


The different $l_i$ in days for all the planets:

\begin{align} Mercury &\quad 88 \\Venus &\quad 224.7 \\Earth &\quad 365.26 \\Mars &\quad 687 \\Jupiter &\quad 4332.6 \\Saturn &\quad 10759.2 \\Uranus &\quad 30685.4 \\Neptune &\quad 60189 \end{align}

Finally under an integer values aproximation and using this online solver for the system of equations the answer is $x = 4.0384877779832565 \times 10^{26}$ which divided by $360^{o}$ gives you roughly $$1.1218 \times 10^{24} \quad\text{years}$$

Edit 1

Just found this site you may like to play around with. It's an interactive flash application with the accurate position of the planets.

I also know that the all the information can be obtained from this NASA page and that is as accurate you can get, but it is just incomprehensible for me now. I will try to revise it later when I find time.

Also this book by Jean Meeus called Astronomical Algorithms covers all the fundamental euqations and formulas - it has nothing to do with programming algorithms though.

Edit 2

Seeing that you are a programmer, it could be worth for you checking out the NASA site I mentioned above, the data for all the planets can even be accessed via $\tt{telnet}$. Or this Sourceforge site where they have implementations for many of the equations described in the book also mentioned above.

  • 1
    $\begingroup$ $x\equiv \theta_i (\mod l_i)$ works the same in comments. I think, your approach is the best you can do without excessive simulations. All you need to do, is to insert the actual data; that has been the part, which made me hesitate to provide an answer. $\endgroup$
    – Gerald
    Commented May 14, 2014 at 1:14
  • 1
    $\begingroup$ @Gerald oh I thought equations markup didn't work in comments. Yes, I'm missing the data, most notably $\theta_i$. I will add the different $l_i$ information. $\endgroup$
    – harogaston
    Commented May 14, 2014 at 2:25
  • $\begingroup$ How could that solarsystemscope show the acurate relative positions of the planets, when their distances from the Sun are not correct? It might show each planets position relative to the Sun correctly in isolation and thus be good for this question, but not for finding conjunctions. $\endgroup$
    – LocalFluff
    Commented May 14, 2014 at 4:51
  • $\begingroup$ @LocalFluff That is true. This only provides answer to radial alignment configurations. Edited. $\endgroup$
    – harogaston
    Commented May 14, 2014 at 4:59
  • 3
    $\begingroup$ There are several blunders in this answer. First, using all digits in your tables (which implies converting to centidgrees and centidays) I actually get $x\approx1.698\times10^{42}$ (from the same online tool), which amounts to $1.29\times10^{33}$yr. I don't know how you obtained the lower value, but I strongly suspect you omited some digits. Secondly this shows that when adding more digits the solution tends to infinity: the correct answer is: radial alignment never occurs. Finally, assuming that the planets' orbits are following this simple motion is just wrong. $\endgroup$
    – Walter
    Commented Oct 5, 2015 at 8:24

The correct answer is 'never', for several reasons. First, as pointed out in Florin's comment, the planet's orbits are not co-planar and hence cannot possibly align, even if each planet could be placed arbitrarily in its orbital plane. Second, even pure radial alignment never happens because the planet's periods are incommensurable -- their ratios are not rational numbers. Finally, the planets' orbits evolve over timescales of millions of years, mainly due to their mutual gravitational pull. This evolution is (weakly) chaotic and thus unpredictable for very long times.

The wrong answer by harogaston essentially approximates the orbital periods by the nearest commensurable numbers, yielding a very long time (though he got that wrong by a factor of merely $10^{16}$).

A much more interesting question (and perhaps the one you were actually interested in) is how often do the 8 planets nearly align radially. Here, 'nearly' could simply mean 'to within $10^\circ$ as seen from the Sun'. At such an occasion, the mutual gravitational pull of the planets will align and hence result in stronger orbital changes than the average.


Any estimate of the common period of more than two planets (i.e., after how much time do they approximately align in heliocentric longitude again?) depends very strongly on how much deviation from perfect alignment is acceptable.

If the period of planet $i$ is $P_i$, and if the acceptable deviation in time is $b$ (in the same units as $P_i$), then the combined period $P$ of all $n$ planets is approximately $$P \approx \frac{\prod_i P_i}{b^{n-1}}$$ so reducing the acceptable deviation by a factor of 10 means increasing the common period by a factor of $10^{n-1}$, which for 8 planets is a factor of 10,000,000. So, it is meaningless to quote a common period if you don't also specify how much deviation was acceptable. When the acceptable deviation declines to 0 (to achieve "perfect alignment"), then the common period increases to infinity. This corresponds to several commenters' statements that there is no common period because the periods are not commensurate.

For the planets' periods listed by harogaston, $\prod_i P_i \approx 1.35\times10^6$ when the $P_i$ are measured in Julian years of 365.25 days each, so the common period in years is approximately $$P \approx \frac{1.35\times10^6}{b^7}$$ if $b$ is measured in years as well. If the periods are approximated to the nearest day, then $b \approx 0.00274$ years and $P \approx 1.2\times10^{24}$ years. If the periods are approximated to the nearest 0.01 day, then $b \approx 2.74\times10^{-5}$ and $P \approx 1.2\times10^{38}$ years.

The derivation of the above formula is as follows:

Approximate the planets' periods by multiples of a base unit $b$: $P_i \approx p_i b$ where $p_i$ is a whole number. Then the common period is at most equal to the product of all $p_i$. That product is still measured in units of $b$; we must multiply by $b$ to go back to the original units. So, the common period is approximately $$P \approx b \prod_i p_i \approx b \prod_i \frac{P_i}{b} = b \frac{\prod_i P_i}{b^n} = \frac{\prod_i P_i}{b^{n-1}}$$

The above derivation doesn't take into account that the $p_i$ might have common factors so that the alignment occurs sooner than $\prod_i p_i$ suggests. However, whether or not any two $p_i$ have common factors depends strongly on the chosen base period $b$, so it is effectively a random variable and does not affect the global dependence of $P$ on $b$.

If you express the acceptable deviation in terms of angle rather than time, then I expect you'll get answers that depend on the size of the acceptable deviation as strongly as for the above formula.

See http://aa.quae.nl/en/reken/periode.html for a graph of $P$ as a function of $b$ for all planets including Pluto.


Here is an estimate with acceptable deviation in terms of angle. We want all planets to be within a range of longitude of width $δ$ centered on the longitude of the first planet; the longitude of the first planet is free. We assume that all planets move in the same direction in coplanar circular orbits around the Sun.

Because the planets' periods are not commensurate, all combinations of longitudes of the planets occur with the same probability. The probability $q_i$ that at some specific moment of time the longitude of planet $i > 1$ is within the segment of width $δ$ centered on the longitude of planet 1 is equal to $$q_i = \frac{δ}{360°}$$

The probability $q$ that planets 2 through $n$ are all within that same segment of longitude centered on planet 1 is then $$q = \prod_{i=2}^n q_i = \left( \frac{δ}{360°} \right)^{n-1}$$

To translate that probability to an average period, we need to estimate for how much time all planets are aligned (to within $δ$) each time they are all aligned.

The first two planets to lose their mutual alignment are the fastest and slowest of the planets. If their synodic period is $P_*$, then they'll be in alignment for an interval $$A = P_* \frac{δ}{360°}$$ and then out of alignment for some time before coming into alignment again. So, each alignment of all planets lasts about an interval $A$, and all of those alignments together cover a fraction $q$ of all time. If the average period after which another alignment of all planets occurs is $P$, then we must have $qP = A$, so $$P = \frac{A}{q} = P_* \left( \frac{360°}{δ} \right)^{n-2}$$

If there are only two planets, then $P = P_*$ regardless of $δ$, which is as expected.

If there are many planets, then the fastest planet is a lot faster than the slowest one, so then $P_*$ is very nearly equal to the orbital period of the fastest planet.

Here, too, the estimate for the average time between successive alignments is very sensitive to the chosen deviation limit (if there are more than two planets involved), so it is meaningless to quote such a combined period if you don't also mention what deviation was allowed.

It is also important to remember that (if there are more than two planets) these (near-)alignments of all of them do not occur at regular intervals.

Now let's plug in some numbers. If you want all 8 planets to be aligned to within 1 degree of longitude, then the average time between two such alignments is roughly equal to $P = 360^6 = 2.2×10^{15}$ orbits of the fastest planet. For the Solar System, Mercury is the fastest planet, with a period of about 0.241 years, so then the average time between two alignments of all 8 planets to within 1 degree of longitude is about $5×10^{14}$ years.

If you are satisfied already with an alignment to within 10 degrees of longitude, then the average period between two such alignments is roughly equal to $P = 36^6 = 2.2×10^9$ orbits of Mercury, which is about 500 million years.

What is the best alignment that we can expect during the coming 1000 years? 1000 years are about 4150 orbits of Mercury, so $(360°/δ)^6 \approx 4150$, so $δ \approx 90°$. In an interval of 1000 years chosen at random, there is on average one alignment of all 8 planets to within a segment of 90°.


There is a much easier way to do this.

1) Look up the length of the solar year in earth days

2) multiply the length of the years like this: Mercury year * Venus year * Earth year * Martian year * Jovian year * Saturn year * Uranus year * Neptune year

3) Divide by 365 to get earth years.

And you have a time when they will align again longitudinally(meaning the angles will be different but from a top view they would form a line). It won't align at any higher of a frequency because some of these planets have a decimal number of earth days in their year.

  • 1
    $\begingroup$ 4) Realize that the number you got is much greater than the Lyapunov time of the solar system, and is thus meaningless. $\endgroup$
    – Mark
    Commented Oct 6, 2015 at 0:57

Technically the true way to find the period between alignment of all 8 planets is to find the LCM of all 8 of their year lengths.

LCM (88, 225, 365, 687, 4333, 10759, 30685, 60189) = 814252949520007202031000. I understand that this is a rough estimate since these are rounded to the nearest integer, but it gives a good idea of the number of days it would take.

814252949520007202031000/365 = 2230829998684951238441. That's how many years.

  • $\begingroup$ This seems to be the same method as described in Caters's answer. $\endgroup$
    – HDE 226868
    Commented Oct 4, 2015 at 19:58

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