Astronomy Answers: Common Periods of the Planets

# Astronomy AnswersCommon Periods of the Planets

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Suppose you have a number of planets that each orbit around the Sun in a fixed period, for example in 100/100, 188/100, and 1187/100 years for the Earth, Mars, and Jupiter (accurate to the nearest 1/100th of a year). After how much time do they all return to the same positions?

## 1. The Method

Every planet returns to the same position when a multiple of its period has passed, so the Earth returns to the same position after every multiple of 1 year (e.g., 1, 2, or 3 years), Mars returns to the same position after every multiple of 1.88 years (e.g., 1.88 or 3.76 or 5.64 years), and Jupiter after every multiple of 11.87 years (e.g., 11.87 or 23.74 or 35.61 years). They all return at the same time to their original position after a period that is a multiple of the periods of all planets. The least period after which they all return to their original position is the smallest of those multiples unequal to zero, and that is called the least common multiple (abbreviation to lcm hereafter) of all periods.

The method I describe below for calculating the lcm of a collection of numbers only works if all numbers are whole numbers. For periods of planets, this means that you must select a base period that evenly divides all of the periods, so that each period is a whole multiple of the base period. Small rounding errors may occur, and you can make those smaller by selecting a smaller base period. For our example, we can take a base period of 1/100 year. Then the periods of the planets are equal to 100, 188, and 1187.

Here is the first method to calculate the lcm of a collection of numbers:

1. Write each of the numbers as the product of numbers (factors) that are as small as possible. If you multiply all factors, then the original number must be produced.
2. For each of the factors, take the greatest count from each of the numbers and enter that into a separate list of factors.
3. The lcm of the numbers is equal to the product of the separate list of factors.

For example: 100 = 2×2×5×5 so it has two factors 2 and two factors 5. 188 = 2×2×47 so it has two factors 2 and one factor 47. 1187 has no divisors other than 1187. We now take the greatest count for each of the factors, so two factors 2, two factors 5, one factor 47, and one factor 1187. This yields 2×2×5×5×47×1187 = 5578900 base periods, or 55,789 years.

It takes quite some effort to find the divisors of numbers, so it is more convenient to use a method for which you don't need all divisors. For example, a method that uses the greatest common divisor.

The divisors of a number are the whole numbers that evenly divide that number (i.e., without leaving a remainder). The greatest common divisor (abbreviated to gcd) of a set of numbers is the greatest divisor that all of them have in common.

If you're dealing with just two numbers, then the product of those two numbers is equal to the product of the lcm and the gcd of those numbers, but this isn't the case in general if more than two numbers are involved. So, for two numbers, you can find the lcm by dividing the product of the two numbers by the gcd of the numbers.

For the calculation of the gcd of two numbers there is a nice method that does not require any searching:

1. Determine the remainder when you divide the greater number by the lesser one.
2. If that remainder is equal to zero, then the lesser number is equal to the gcd.
3. If the remainder is not equal to zero, then throw out the greater number. You now again have two numbers (the one that was the lesser one, and the remainder that you just calculated). Go back to step 1 with these two numbers.

For example, what is the gcd of 100 and 188? 188 divided by 100 yields 1 with a remainder of 88, which is not equal to zero, so we throw out the 188. Now we have 100 and 88. 100 divided by 88 yields 1 with a remainder of 12, so we throw out the 100. Now we have 88 and 12. 88 divided by 12 yields 7 with a remainder of 4, so we throw away the 88. Now we have 12 and 4. 12 divided by 4 yields 3 with a remainder of 0, so we are done and the gcd is equal to 4.

With this gcd we can now calculate the lcm of the two numbers. The lcm of 100 and 188 is equal to their product divided by the gcd, so 100×188/4 = 4700.

You can now determine the lcm of more than two numbers as follows:

1. Determine the lcm of the first two numbers.
2. Then determine the lcm of the previous lcm and the next number for each next number in turn.
3. The last lcm is the lcm of all numbers together.

You then find the lcm of 100, 188, and 1887 as follows: the lcm of 100 and 188 is 4700. The lcm of 4700 and 1187 is 5578900, so the lcm of 100, 188, and 1887 is equal to 5578900.

The lcm of a collection of numbers is at least as great as the greatest of those numbers, and is at most as great as the product of all of the numbers. If you calculate the lcm of a large collection of arbitrary numbers, then there is a good chance that the answer is so large that it doesn't fit in your calculator anymore. For example, the lcm of all numbers from 90 through 100 is itself a number with 16 digits. In such a case, the method with the individual factors (the first one discussed) can still be used. For example, here are the factorizations of the periods (to base 1/100th of a year) of all planets:

 24 =  2×2×2×3 61 =  61 100 =  2×2×5×5 188 =  2×2×47 1187 =  1187 2946 =  2×2×491 8401 =  31×271 16479 =  3×3×1831 24769 =  17×31×47

The lcm of all of those numbers is then equal to 2×2×2×3×3×5×5×17×31×47×61×271×491×1187×1831 = 786,503,499,835,444,655,400 and the common period (to 1/100th of a year) is then 7,865,034,998,354,446,554 years.

## 2. The Use

If the periods that are to be combined are not naturally multiples of a common base period, then the length of their combination period is very sensitive to the chosen base period, and has little use in practice.

In the worst case, the periods have no divisors in common at all, and then the lcm is equal to the product of the periods (measured in units of the base period). Suppose that we make the base period ten times smaller. Then the number that indicates the length of a planet's period becomes ten times larger, measured in units of the base period which is now ten times smaller. The product of all of the periods then gets larger by as many factors of ten as there are planets. At the end we transform the product back from a number of base periods to an actual time, by multiplying it with the base period, which has gotten ten times smaller, so then we loose one factor of ten again.

If you calculate the combination of $$n$$ periods and make the base period $$f$$ times smaller, then, in the worst case, the combination will get $$n - 1$$ factors of $$f$$ larger, or $$f^{n-1}$$ times larger. The combination of the periods of the nine planets discussed above will get larger by a factor of 100,000,000 every time that we reduce the base period by a factor of 10.

If $$P_0$$ is the product of the periods (without first dividing by a base period), then the combination period $$P$$ (calculated as indicated above) is equal to $$P_\text{max}$$ in the worst case:

\begin{equation} P_\text{max} = \frac{P_0}{b^{n-1}} \label{eq:p} \end{equation}

For the planets mentioned above, $$P_0$$ is about equal to 335 million (3.35 × 108). We used $$b = 0.01$$ years there, and we have $$n = 9$$ planets, so $$P_\text{max} = 3.35×10^{24}$$ years, and this gets 108 times larger every time we reduce $$b$$ by a factor of 10.

Below is a graph that shows the decimal logarithm of the combination period $$P$$ of the nine planets as a function of the chosen base period $$b$$ for three thousand randomly chosen base periods between 0.0001 and 0.1 years. The assumed periods of the planets were (measured in days):

87.96934; 224.70069; 365.256366; 686.9600; 4333.2867; 10756.1995; 30707.4896; 60223.3528; 90613.3055

In the graph, both $$b$$ and $$P$$ were measured in julian years "(a)". The numbers along the vertical axis roughly indicate "how many zeros" there are in the combination period. The dotted line shows the results of equation \ref{eq:p}, i.e., for the worst case. It is clear that $$P$$ follows that worst-case line: The smaller we choose $$b$$ to be, the larger $$P$$ gets. These results provide no evidence that there is more structure in the periods of the planets than in a random set of numbers.  languages: [en] [nl]

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Last updated: 2021-07-19