Determining the oribital periods of the inner planets

Greated elongation etc. Source: wikimedia

Using a different approach, let's consider again two planets moving around the Sun in concentric circular orbits in the same plane.

Let the outer planet orbit the Sun at rate a (number of revolutions per time period).

Let the inner planet orbit the Sun at rate b (number of revolutions per time period).

At time t=0, let the planets be in the configuration shown such that viewed from the outer planet, the inner one is at greatest elongation (its greatest angular distance) from the Sun, in a specified direction. While in principle, we could have chosen inferior conjunction (when the inner planet is directly between the outer one and the Sun), in practice this is impossible to observe unless in the rare instance when the planet transits the Sun.

We can say that this configuration is repeated whenever:

(b - a).t(n) = n, where n is an integer > 0.

The time interval between the event at t(n) and the next, t(n+1), by definition is:

t(n+1)-t(n) = 1/(b - a)

This is independent of n. Let's call it 𝛥t.

Hence:

𝛥t = 1/(b - a)

So:

b/a = 1 + 1/(a.𝛥t)

Now, let's make this more concrete:

• the outer planet is Earth, so a is equal to 1 revolution per year
• the inner planet is Venus
• we'll choose the repeated event as the greatest elongation West, i.e. as observed in the morning sky - see table here.

From this table, we can see that an instance of greatest elongation West for Venus, observed from the Earth, occurred on the morning of March 22 2014 and then for a fifth time on almost the same day 8 years later on the morning of March 20 2022, so, 𝛥t is on average almost exactly 8/5 years.

Hence:

b/a = b ~ 1 + 5/8 = 13/8 (revolutions per year)

So, one orbit of Venus about the Sun takes 1/b, i.e. around 8/13 years or 224.8 days. The correct figure is 224.7 days, so we have achieved an accuracy of closer than 1 part in 1000.

Now consider Mercury in place of Venus. We can see there was an instance of greatest elongation West on March 14 2014 and then 22 instances later, another on March 6 2021. In this case:

b/a = b ~ 1 + 22/7 = 29/7 (revolutions per year)

Hence, one orbit of Mercury about the Sun takes 1/b, i.e. around 7/29 years or 88.2 days, which is close to the correct figure of 88.0 days, although with less accuracy than was the case for Venus. No doubt the relatively high inclination of its orbit to the ecliptic is responsible.

(We can adapt this method to the outer planets, with the inner planet now representing Earth and its rate of rotation, b, equal to 1. The rate of rotation of the outer planet, a, is given by:

a/b = 1 - 1/(b.𝛥t)

The time interval being measured is once more (the average) between one opposition and the next of the outer planet as seen from Earth.

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