Estimating aphelion and perihelion for Mercury and Venus

In an earlier post, I estimated the orbital radii for Mercury and Venus, by taking the sine of the average greatest elongation angles for each, using data for 2014 to 2029 from here.

In fact, I've been treating the orbits as being near-circular, but recently I've been wondering how to find their true shapes (which we know from Kepler and Newton are ellipses). Specifically, can I use the extremes of the greatest elongation angles to find accurate figures for the furthest (aphelion) and closest (perihelion) approaches of these planets to the Sun?

So, here is a summary for the inner planets. I've separated greatest elongation observations by East and West. The sines of the minimum and maximum angles within each set of observations provide estimates of perihelion and aphelion respectively and as can be seen are nearly exactly the same as the true figures, with exceptions to the exact figures highlighted with an *. This surprises me, because I'd assumed that I would have to find and make an allowance for the Earth's orbital eccentricity (loosely, non-circularity) for greater accuracy (which I have not done yet), especially so in the case of Venus, because the Earth's orbit is actually the more eccentric of the two planets.

Notice how eccentric Mercury's orbit is!

Mercury

• Greatest Elongation West
• 50 observations
• min. 17.9 (sine: 0.31)
• max. 27.8 (sine: 0.47)
• Greatest Elongation East 
• 51 observations
• min. 18.1 (sine: 0.31)
• max. 27.4 (sine: 0.46*)
• wikipedia: perihelion 0.31 AUs, aphelion 0.47 AUs

Venus

• Greatest Elongation West
• 10 observations
• min: 45.8 (sine: 0.72)
• max 47 (sine: 0.73)
• Greatest Elongation East
• 10 observations
• min. 45.4 (sine: 0.71*)
• max. 47.2 (sine: 0.73)
• wikipedia: perihelion 0.72 AUs, aphelion 0.73 AUs

The Astronomical Unit - improving on 8 light minutes

For almost as long as I can remember, I've known that:

• it takes light from the Sun about 8 minutes to reach us
• the Sun is about 93 million miles away
• light travels at about 186 thousand miles per second.

However, being on the trailing edge of things, it seems to have passed me by until just now that from the last two statements, we can see that in one half second, light travels one thousandth the distance to the Sun and so it takes 500 seconds for light to travel the full distance.

That's a nice easy round number to remember and is far more precise than the rather vague 8 minutes, which is out by twenty seconds.

Estimating the radius of the Moon (and the Earth-Moon distance)

Graphic to follow!

Many years ago I timed a lunar eclipse and attempted to work out the diameter of the Moon compared with the Earth's from that information. Unfortunately, I didn't keep a record, but was pleased to note that the result was very roughly in the right general area.

With an estimate of the size and knowing that the Moon is about half a degree of arc in diameter, one can then estimate the distance from the Earth to the Moon. This is what the Ancient Greeks would have done.

One of my assumptions was that diameter of the umbra was equal to that of the Earth. Now, I knew this to be incorrect, but I didn't bother to find out by how much.

(I also assumed that the centre of the Moon went through the centre of the umbra, also wishful thinking, but I'll leave that for now).

Here is a very belated attempt to calculate the size of the Earth's umbra at the Moon:

• let's think of a solar eclipse that is momentarily total
• let the Moon be at distance d from the Sun
• let the radii of the Sun and Moon be rₛ and rₘ respectively
• let the length of the Moon's umbra be lₘ
• we can say that lₘ = d.rₘ/(rₛ - rₘ)
• now think of a lunar eclipse.
• this time the Earth, radius rₑ, is at distance d from the Sun
• let the length of the Earth's umbra be lₑ
• we can say that lₑ = d.rₑ/(rₛ - rₑ)
• hence the ratio of the length of the Earth's umbra to that of the Moon is 

     lₑ/lₘ = (rₑ/rₘ).(rₛ - rₘ)/(rₛ - rₑ) ~ rₑ/rₘ ~ 3.7                    (since rₛ >> rₑ, rₛ >> rₘ)

Now, at distance lₑ, the size of the Earth's umbra is 0 and at distance lₘ (the Earth-Moon distance) the diameter of the Earth's umbra is therefore about (3.7 - 1)/3.7 or 0.7 times the diameter of the Earth. The difference is not insignificant and if we ignore it, we will over-estimate the size of the Moon and also the Earth-Moon distance. While I didn't take this into account, I'm sure that the Ancient Greeks would have done so in order to estimate the Earth-Moon distance, as they did so quite accurately, at around 60 Earth radii IIRC.

Measuring the Astronomical Unit - part ii

Updated 29 May 26

Introduction

Here's a summary of two (identical!) estimates of the AU made in 1672, based on telescopic observations of Mars at or near opposition. They are remarkably accurate for the time, being well within 10% of the accepted figure.

With these estimates, for the first time the terrestrial scale could be connected to the astronomical, via the radius of the Earth, the size of which had been known since the time of the Ancient Greeks. 

(True, the Earth-Moon distance had also been fairly accurately known, but that was a dead-end as a unit of measurement).

Summary

i. Flamsteed (Britain)

Method: diurnal parallax, i.e. non-simultaneous observations made at the same location

Location: Greenwich (51.5 N)

Baseline not known by me, but:

• if observations were made at 9 pm then 3 am (90° Earth rotation later), then 

        baseline = radius of circle of latitude x √ 2 = Rₑ x cos 51.5° x √ root 2: ~ 0.9 Rₑ

• if observations were made at 8 pm then 4 am (120° Earth rotation later), then

        baseline = radius of circle of latitude' x 2 x sin 60°  =  cos 51.5° x 2 x sin 60 ~ 1.1 Rₑ

(where Rₑ is the mean radius of the Earth)

Pros:

• no travel involved
• no need to determine longitude
• one observer - consistency of observation

Cons: 

• need to allow for relative movement of Mars between observations

Result: 92.5 % actual

ii. Cassini, Richer (France)

Method: 'standard' parallax, i.e. simultaneous observations at distant locations

Locations: Paris (49N,2E), Cayenne (5N, 52W), separation 63° of arc (from here).

Baseline: ~ 1 Rₑ 

Cons:

• (very) long-distance travel involved
• need to determine longitude accurately*
• two observers needed - inconsistency of observation?

Result: 92.5 % actual

* In 1668 Cassini himself had produced astronomical tables in order to do just that.

Sources: wikipedia plus links above.

Comments

These estimates of the AU are remarkable achievements for the time - a mere 60 or so years after the invention of the telescope and a similar number of years since Kepler's theories.

However, it is inconceivable that two independent estimates using different methods should be in exact agreement! 

Flamsteed's method is ingenious, but I wonder how easy it would have been to achieve in practice. It surely must have occurred to Cassini also.

Further investigation is needed! 

Further reading

Six stages in the history of the astronomical unit, David. W. Hughes

The Mars Parallax Project, John J. McCarthy Observatory

Measuring the Astronomical Unit - part i

Updated 29 May 26

2004 transit of the Sun by Venus, projected on a
wall through one side of 10x50 handeld binoculars.

Having seen that trying to estimate the size of the AU in terrestrial units by measuring the Moon's penumbra during a solar eclipse is a non-starter, let's look instead at the tried-and-tested method of parallax.

Consider Mars, for example, when it is at opposition and about half an AU from the Earth. Let's create a baseline of 5 thousand miles, say, measured through the Earth, with at each end, an experienced observer and telescope pointing at the planet. Now arrange for readings to be taken at an agreed instant when the baseline is perpendicular to a line between Earth and Mars and thereby maximizing parallax.

The method needs:

• accurate and precise local time-keeping - only possible with the invention of the long pendulum clock in the late 1600s
• an accurate and precise measurement of the difference in longitude to determine the baseline and time-difference - actually possible on land with the use of astronomical tables before the invention of the marine chronometer.

If Mars were say 5 million miles away, then the angle of parallax would be one thousandth of a radian. One radian is roughly 60 degrees or 60 x 60 minutes of arc. One thousandth of that is roughly three and one half minutes of arc. That's about ten times the angular size of Mars itself. Now if the observers can draw Mars against the background of stars with reasonable accuracy, the difference in position should be measurably accurate.

If Mars were instead say 50 million miles away, then the angle of parallax would now be one tenth of that and about the same as the angular size of Mars itself. Perhaps this would be too small to discern. However, if the planet were seen to obscure a star (an example of an occultation), from one location and to touch it at the other, then that would provide a measurement of parallax.

Even if the angle were too small to measure, it should at least be possible to establish a minimum size for the AU.

Venus is closer to us at conjunction, just 0.28 AU away, a little over half the distance to Mars at its closest. Its maximum angle of parallax is therefore about twice that of Mars and hence easier to measure. How to observe it in daylight? Just wait for when it transits the Sun! Then during the transit, our observers can record the progress of Venus across the disk of the Sun, noting the times as they do so. Afterwards, they can compare their findings, noting the different positions of Venus with the same 'timestamps', allowing for difference in longitude.  Armed with this information and the length of their baseline (and its orientation at each instant to the line between the Earth and Venus), they will now have an angle of parallax for Venus measured against the Sun's disk. However, this will be smaller than the true figure, had they been able to measure it against the fixed stars. The reason being that there will also be some parallax for the Sun - 28% of that measured for Venus. So, the true figure for the parallax of Venus will be the measured figure times 100/72. While it sounds easy in principle, I won't underestimate the difficulty of doing this in practice, especially before the invention of photography.

The Geometry of a Solar Eclipse

Updated 23 May 26

Geometry of a Solar Eclipse

The diagram shows a simplified representation of a particular type of solar eclipse in which the Sun is obscured for an instant only. If the Moon were closer, the eclipse would be total for a period of time and if it were further away, the eclipse would be annular.

At right is a red vertical bar representing the Sun, radius r₃; in the middle is a green vertical bar representing the Moon, radius r₂; at left is a black vertical bar representing the extent of the observed  partial eclipse (penumbra), radius r₁. 

Point 1 represents the location of an observer (on the Earth) of this momentary total eclipse. 

The distance between point 1 and (the centre of) the Moon is d₁ (known); the distance between (the centre of) the Moon and point 2 is d₂ (unknown) and the distance between point 2 and (the centre of) the Sun is d₃ (also unknown). The total distance between the point 1 and (the centre of) the Sun is therefore d₁ + d₂ + d₃.

Marked, but in the end not used, are angles θ (half the angle subtended by the Moon at point 1), and φ (half the angle subtended by the Sun at point 2).

From similar triangles:    

1.    (d₁ + d₂)/r₁ = d₂/r₂ = d₃/r₃

and:    

2.    d₁/r₂ = (d₁ + d₂ + d₃)/r₃.

From 1:

3.    d₂ = r₂.d₁/(r₁ - r₂).

From 2 & 3:

4.    d₃/(d₁ + d₂ + d₃) = d₂/d₁ =  r₂/(r₁ - r₂).

So: 

5.    (d₁ + d₂)/ (d₁ + d₂ + d₃) = 1 - d₃/(d₁ + d₂ + d₃) 

=  (r₁ - 2r₂)/(r₁ - r₂). 

(Note:  r₁ > 2r₂)

Hence, the ratio of the Earth-Moon distance to the Astronomical Unit is (minus the radius of the Earth):

6.    d₁/ (d₁ + d₂ + d₃) = ((r₁ - r₂)/r₁).(d₁ + d₂)/ (d₁ + d₂ + d₃)

= (r₁ - 2r₂)/r₁.

So, all we need to do is measure r₁, plug it into the above equation along with the known value for r₂ and we are done! However, there is a problem. Let's say the ratio in the above expression is 1/n, then:

7.    r₁ = 2r₂.(n - 1)/n. 

This means we must measure r₁ with an accuracy (much) greater than 1/n - an estimate will not do.

I had hoped that this could form the basis of a simple method to estimate the AU, but that does not seem to be the case. And I've not even considered yet just how to estimate r₁ from the timings of a solar eclipse on the spinning Earth. Not simple then and I can only think that this was never a practical method either for finding the size of the AU. However, for high n, r₁ to all intents and purpose equals 2r₂, so this is instead one way that the diameter of the Moon could be measured.

The Saturnian Keplerian system

Decades after the discovery of Jupiter's moons, Saturn was observed to have its own satellites: the first being Titan, the largest, in 1655.

Let's try to compute the Kepler value for the three largest, using the same method as for Jupiter, noting however, that accuracy and precision would likely have been low in the early days, so the figures below (especially its density) may not have been achievable until much later.

There is no orbital resonance with these, so the table shows the period in days.

(Rhea's orbit is actually about 4.5 Saturnian diameters. Were we able to estimate to this precision, then we should expect the Kepler value to be ~4.5).

We see that the values in the last column suggest a Kepler 'constant' for Saturn of ~ 4 and hence that Saturn, as well as the Sun and Jupiter, also has its own gravitational system obeying the same laws.

Converting this to Solar system units:

• as Saturn's angular diameter at opposition is ~ 1/3 minute of arc and it's 8.4 AUs away, it's diameter is ~ 0.0008 AUs
• we cube this:  ~ 5 x 10^(-10) and multiply our Kepler constant by it
• 1 day is 1/365 years, so we square this ~  7.5 x 10^(-6) and divide the above result by it

Hence, the average Kepler 'constant' for Saturn is ~ 0.0003.

We deduce that its gravitational pull is ~ 0.0003 times that of the Sun and ~ 0.3 times that of Jupiter. 

We suspect its mass is also in the same proportions. That being the case, as we know its diameter is ~ 0.8 of Jupiter's, its volume is therefore around half of Jupiter's and its density ~ 0.6 times Jupiter's, indicating a different composition.

Moon	Orbital radius (r) in Saturnian diameters	r^3	Period (t) days	t^2	r^3/t^2
I. Rhea	4	64	4.5	20	3.2
II. Titan	10	1000	16	256	4
III. Iapetus	30	27000	79	6241	4.3

The Jovian Keplerian system

Updated 11 May 26

The discovery of Jupiter's moons through the newly-invented telescope back in the early 1600s must have caused great excitement at the time and a wish to investigate further to see if the Jovian system follows Kepler's laws.

However, imagine trying to track these moons individually over many days and weeks in an attempt to measure the orbital radius and orbital period for each!

The orbital radius could be estimated crudely (being imprecise and prone to human error) using the diameter of the planet as the base unit, e.g. the innermost moon being found to orbit at around 3 Jovian diameters from the centre of the planet.

I've tried this myself on just a few occasions with binoculars and it was difficult! Doing this to an acceptable level of accuracy and precision in the early 17th century would have demanded the best optics of the time, a solid mount, good skies over many weeks and an experienced and patient observer. And then no doubt repeating the exercise again and again. 

The orbital period could be measured with more precision by timing the intervals between successive identical locations in a moon's orbit (e.g. maximum elongation), although we should bear in mind that this was before the invention of the pendulum clock.

(Would we have to allow for the movement of the Earth introducing errors due to parallax?)

See the table at the bottom of the page for an investigation. (The table had to be imported and in this case, if not placed there, messes up the font of the text following it!). The moons are numbered according to their proximity to Jupiter.

It turns out that the orbital periods of moons I, II and III are in the ratio 1:2:4, so I'm using the orbital period of moon I as the unit of time here.

The average estimate of the Jovian Kepler 'constant' (last column) is 26, but with a large range of values, no doubt due to the lack of precision in measuring orbital radii in whole Jovian diameters, which are then cubed. (In practice, there would be even greater variation with this method, because, unlike the case here, estimates would likely be inaccurate, as well as imprecise). However, these values are of a similar order of magnitude and a more precise and more accurate method should yield results that are more consistent*. Therefore, we tentatively assert that these moons do indeed obey Kepler's third law.

(* If instead we were to estimate accurately using the radius of Jupiter as our measuring unit, then we should find that the Jovian Kepler constant has an average of 202 with a proportionately smaller range of 182 to 216. However, I expect this advantage would have been easily removed by increased human error, in estimating, say, that Callisto orbits at 26 units from Jupiter! I imagine that the later invention of the reticule would have been revolutionary for making accurate and precise measurements). 

Moon	Orbital radius (r) in Jovian diameters	r^3	Period (t) as a multiple of that of moon I	t^2	r^3/t^2
I. Io	3	27	1	1	27
II. Europa	5	125	2	4	31
III.Ganymede	7	343	4	16	21
IV. Callisto	13	2197	9.4	89	25

Measuring the elongation of Venus with a watch?

Updated 3 May 26

As an alternative, I thought instead about just using the timings of the settings of the Sun and of Venus (I'm not a morning person!). Thinking that no doubt someone has already investigated this, after some digging, I came across this site with a method for such a measurement. However, there's no derivation of the method and it seems to assume that both will set at the same azimuth, which doesn't seem right. Anyway, I thought I'd check it with Stellarium and chose the greatest elongation East for Venus of 15 August 2026:

• Sun and Venus are about 45° apart, as we'd expect
• the Sun's declination (angular distance above the celestial equator) is around 14°
• that of Venus is around -5° (i.e. it's below the celestial equator), so very different from that of the Sun
• the Sun straddles the horizon, i.e. is half set at around 20.20 (BST?) at azimuth 294°
• Venus sets about 80 minutes later, at azimuth 261°, i.e. there's a whopping 33° difference!

If we use the formula, it says for a 45° separation, Venus sets 3 hours, not 1 hour, after the Sun, so the formula is indeed wrong - the Sun and Venus will set at the same point on the horizon only if they are at the same declination, which is not the case here nor in general. As a comparison, the star Denebola *is* at about the same declination as the Sun at this time, with an angular separation from it of 31°. The formula gives about 2 hours between the Sun and Denebola setting and Stellarium confirms that.

In addition to the time between the setting of the Sun and of Venus, then we'd need the difference in azimuths at setting or the declinations - and a new formula.

Verdict: What's the point?

Measuring the elongation of Venus

Updated 30 Apr 26

Given Mercury's proximity to the Sun, let's take it that measuring elongation will always be a challenge, so let's limit our investigation to Venus for now.

Measuring the elongation of Venus when it's distant from the Sun may seem easy enough in principle, if you have a clear view of the horizon -  just measure the direction on the horizon (azimuth) of the centre of the setting Sun and at the same time, record the azimuth of Venus and its altitude above the horizon and then determine the angular difference. However, given that the Sun will be bright and Venus much less so, perhaps it's going to be difficult in practice to observe both at the same time?

Wait then until the Sun has almost fully set, with just a remnant showing on the horizon, and Venus may be more obvious. Take measurements as before, but note the Sun's altitude is now one solar diameter (~ 0.5°) below the horizon, which we should probably take into account.

However,  refraction by the Earth's atmosphere will also make a difference to the measurements of azimuth of both Venus and of the Sun. At the moment we don't know how to quantify this effect, but hope that it will be similar in both. Let's not worry too much about this and consider our measurement an estimate only. One or two degrees in measuring greatest elongation is unlikely to make a massive difference to our estimate of the radius of the orbit of Venus, although it will of course have more significance for estimating that of Mercury.

Determining the orbital radius for Saturn

Updated 5 May 26

To complete the orbital radii of the classical planets, let's now find Saturn's:

rₛ ~ (dₑ + sin θ)/ ((dₑ/pₛ) + sin θ)

where pₛ is its orbital period, for which we'll give the value of 29.45 that we found earlier.

This time, let's double Δt to 4 days, because we know θ may be too small to measure otherwise. As a result, distance dₑ will now be 8π/365.25 AUs. However, as this is an approximation (chord vs. arc of a circle), we will have increased the error compared with a time of 2 days.

Considering the opposition of Saturn on October 4 2026 and using Stellarium to determine its position 48 hours before and 48 hours after, and this tool, θ, the angular distance Saturn is seen to have moved in those four days is approximately 0.32° of arc, slightly more than for Jupiter over 2 days, again hopefully measurable.

Using these values, rₛ ~  9.4 AUs, while the accepted figure is around 9.6 AUs, which perhaps we could have achieved by averaging the results from many opposition events.

Again, checking Kepler's third law:

• the cube of its orbital radius is 9.4³ ~ 831
• the square of its orbital period is 29.45² ~ 867
• the former divided by the latter is ~ 0.96

Note: in no case has the estimate of the Solar Kepler constant ever exceeded 1!

Determining the orbital radius for Jupiter

Updated 27 April 26

Adapting the formula that we used for Mars, the radius of Jupiter's orbit, rⱼ, is given by:

rⱼ ~ (dₑ + sin θ)/ ((dₑ/pⱼ) + sin θ)

where pⱼ is its orbital period, for which we'll give the value of 11.85 that we found earlier.

Having said earlier that for Jupiter etc, it would be better to consider a larger Δt, for now, I'm going to use exactly the same value as for Mars, i.e. 2 days, and see how it goes, so as before, distance dₑ is 4π/365.25 AUs.

Considering the opposition of Jupiter on January 10, 2026 and using Stellarium to determine its position 24 hours before and 24 hours after, and this tool, θ, the angular distance Jupiter is seen to have moved in those two days, is approximately 0.27° of arc. This is smaller than was the case with Mars, now about half the angular size of the full moon, but perhaps still measurable with reasonable accuracy using the naked eye, a good star chart and a quadrant, say?

Using these values, rⱼ ~  5.1 AUs, which is very close to the accepted figure of about 5.2 AUs, which we would likely have arrived at by averaging the results from many opposition events.

Again, checking Kepler's third law:

• the cube of its orbital radius is 5.1³ ~ 132.7
• the square of its orbital period is 11.85² ~ 140.4
• the former divided by the latter is ~ 0.94

If we instead use the accepted value for the orbital radius, then this ratio becomes ~1, as we should expect.

For Saturn, the angle θ over just two days will be smaller again, so in that case it really will make sense to increase the time interval Δt and hence the value of dₑ.

Determining the orbital radius for Mars

As mentioned in an earlier post, the (average) orbital radius for an inner planet can be readily determined from its (average) maximum elongations observed from the Earth. However, another way is needed for the outer planets and the sketched diagram below illustrates one such method.

As an approximation, we assume orbits are circular.

In a short interval of time, Δt, centred on the opposition of Mars (corresponding to position 2 in the diagram) seen from the Earth, the Earth will move a short distance, dₑ, in green, approximated as a short straight line segment of length rₑ.ѡₑ.Δt, where rₑ is the radius of the Earth's orbit and ѡₑ is its angular velocity.

Similarly, in the same interval of time, Mars will move a short distance, dₘ, in red, approximately equal to rₘ.ѡₘ.Δt, where rₘ is the radius of Mars' orbit and ѡₘ is its angular velocity, which we know (see earlier post) is around ѡₑ/1.9. Hence dₘ is about dₑ.rₘ/1.9 AUs.

Let's use the Mars opposition of January 16, 2025. It happened at 00:48 UT (position 2 in the diagram). Using Stellarium to determine its position 24 hours before (position 1) and 24 hours after this time (position 3), and this tool, from the Earth, Mars is seen to have moved approximately 0.8° of arc (0.014 radians) in those 2 days.

In this case, distance dₑ is 4π/365.25 AUs.

We see that the angular distance, let's call it θ, is the angle between the line joining the Earth and Mars 24 hours before opposition and the line joining the planets 24 hours after opposition.

These lines meet at a distant point, which is at distance l₁ from Mars and distance l₂ from the Earth. The distance between the Earth and Mars at opposition is therefore l₂ - l₁.

We can see that the distance travelled by the Earth in that time, dₑ, is approximately equal to l₂.sin θ.

Also that the distance travelled by Mars in that time, dₘ, is approximately equal to l₁.sin θ.

Putting this all together:

rₘ ~ (dₑ + sin θ)/ ((dₑ/1.9) + sin θ) ~ 1.5 AUs.

Being based around a single astronomical event (unlike the results determined for the inner planets), this result is fortuitously close to the the average of Mars' perihelion (1.38 AUs) and aphelion (1.66 AUs) and also to the semi-major axis of its orbit, according to wikipedia.

Let's check Kepler's third law for Mars as we did for the inner planets and the Earth:

• the cube of its orbital radius is 1.5³ = 3.375
• the square of its orbital period is 1.9² = 3.61
• the former divided by the latter is ~ 0.93

Actually, the accepted figure for the orbital period is 1.88 (years). Squaring this gives 3.53 and the ratio is now 0.95.

We could use the same method for the other outer planets, but as the angle, θ, would be smaller, and less easy to measure, it would be better to increase Δt, perhaps even extending it to the full period of retrograde motion. If we did so, however, the distance dₑ, at least, would need to be calculated as being the chord of a circle centred on the Sun.

Determining the oribital period of an inner planet

Updated 30 Mar 26

Greatest 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 inner plant be at greatest elongation (its greatest angular distance) from the Sun, in a specified direction, when viewed from the outer planet. 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 (near) impossible to observe other than 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 an outer planet

 Updated 22 Apr 26

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, (originally 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, after completing one orbit the Earth also has to travel 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).