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.

)

Determining the orbital period of Jupiter to a high level of accuracy

Updated March 11

This sketch represents a view from above of the Earth in its orbit (green), around the Sun (bright orange in the original sketch!), and an outer planet, Jupiter say, (pink/red).

The orbits are circular and are in the same plane.

At time 1, for an observer on the Earth, Jupiter is said to be at opposition, when at its closest to the Earth.

At time 2, 6 months later, the Earth has completed half an orbit. Being much further away from the Sun than is the Earth, Jupiter's angular velocity is much lower and it has completed only a small part of its orbit in that time.

At time 3, Jupiter is at opposition again and since it has moved along in its orbit, the Earth also has to move along its orbit for an extra distance (solid green line), in a time interval represented by the delta symbol, 𝛥, in green, between the points marked 1 and 3. Jupiter has now completed 𝛥/365 of an orbit in 365 + 𝛥 days.

In 2026, opposition was January 10th and in 2027 it will be February 11th. In that time, Jupiter will have completed 32/365 of an orbit in 397 days. 

From this, its orbital period in years is therefore 397/32 or 12.4.        (Method 1)

The accepted period is 11.86, so what happened? Mainly it's because I've assumed the orbits are circular and they are not - they are elliptical. Now, I'm going to try to eliminate that error, by considering a longer period of time when the eccentricities of the orbits of Jupiter and Earth will have been smoothed out, somewhat.

Let's look ahead to 2038, when opposition is on January 14th and Jupiter will have completed 369/365 ((365+4)/365) of an orbit in 12 and 4/365 years. That means it will have completed a full orbit in 365/369 of those 12 and 4/365 years, i.e. its orbital period is:

((12x365) + 4)/369 or 11.88 years.        (Method 2)

That's very close to the accepted figure. I've not taken into account the exact times of day of the consecutive oppositions (available here), nor that the Earth takes 365 and 1/4 days to orbit the Sun. If I do that, then I make it 11.85 years.        (Method 3)

A similar analysis using method 2 above for Saturn over July 15th 1990 to July 21st 2020 gives an orbital period of 29.45 years (actual=29.46) and for Mars over March 4th 2012 to April 9th 2014 gives 1.91 years (actual=1.88). Interestingly, the deviation from the accepted period is greater in the latter case (closer to us, shorter time interval) and smaller in the former (further from us, bigger time interval).

The Coherer - the simplest detector of radio waves?

 

Here's a simple circuit consisting of a power supply, an LED and series resistor, in series with a coherer made of some scrunched up aluminium foil balls inside a plastic beaker in contact with each other and a couple of aluminium foil strips wired to complete the circuit. 

Except that the circuit is only completed once there is a nearby source of radio waves, which somehow creates a low resistance pathway for the electrons in the aluminium balls. In demonstrations, this is usually achieved with a piezo-electric gas igniter. However, in this video, made perhaps 18 months ago, I instead touch one of the foil strips with a short length of wire which acts as an aerial to pick up radio waves from, IIRC, the electric motor of an adjacent freezer. The LED (top left) then lights up. If I tap the side of the beaker sharply, the pathway is disrupted and the LED turns off again.

An even better calendar?

Here's a modified version of that new calendar. There are now fewer changes (6),  March and September having the same number of days as they do now:

January	31	30	-1	-1
February	28	29	+1	0
March	31	31	0	0
April	30	31	+1	+1
May	31	31	0	+1
June	30	31	+1	+2
July	31	31	0	+2
August	31	31	0	+2
September	30	30	0	+2
October	31	30	-1	+1
November	30	30	0	+1
December	31	30	-1	0

Let's see what difference this makes to the dates for equinoxes and solstices: 

• first, we'll fix the 2025 Autumn Equinox at Sep 22 (i.e. unchanged)
• the 2025 Winter Solstice is now on Dec 22, one day later than before, as there are now 90, not 91, days in total in the three months of September, October, November
• the 2026 Spring Equinox, almost exactly 89 days later, i.e. 3 full months, is on Mar 22
• the 2026 Summer Solstice, nearly 93 days later, again 3 full months, is on Jun 22
• the 2026 Autumn Equinox, again nearly 93 days later, and again 3 full months, is now on Sep 22.

Wow - all these events, at least for 2025/6, are now on the 22nd of the month, so this mk2 version is even better! I find that quite surprising given the somewhat lopsided look with 2 months with fewer than 31 days at the start of the year and 4 such months at the end.