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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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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. –  Florin Andrei May 13 '14 at 22:50
@FlorinAndrei Aren't all (except Mercury, who is just being rebellious) within ~3° of each other? Not perfect, but good enough for me. –  IQAndreas May 22 '14 at 22:00
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. –  harogaston May 23 '14 at 3:41

2 Answers 2

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.

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$x\equiv \theta_i (\mod l_i)$ 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. –  Gerald May 14 '14 at 1:14
@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. –  harogaston May 14 '14 at 2:25
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. –  LocalFluff May 14 '14 at 4:51
@LocalFluff That is true. This only provides answer to radial alignment configurations. Edited. –  harogaston May 14 '14 at 4:59

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.

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