Start of Summer

Updated 1 May 26

The Spring equinox this year was on the 20th of March. With the Summer solstice on the 21st of June, I estimate the mid-point - and surely the proper start of Summer -  will be about the 5th of May. However, the first of May seems a good enough time to celebrate it.

Traditionally, May Day is associated with fertility, so we have the symbolism of the May Pole and the May Queen in many European countries. In Beltane, there is a tradition of lighting bonfires, no doubt to symbolically to drive away the dark and the evil spirits (rather than to defiantly face them as done at Halloween).

In Germany (and elsewhere in Europe), the celebration has been co-opted by the Church, to celebrate Saint Walpurga, an Anglo-Saxon nun tasked with converting the Germanic tribes in the eighth century. There are bonfires the night before (Walpurgisnacht in Germany), indicating a common root with Beltane. Given Walpurga is known for having fought against witchcraft, she was an inspired choice for re-branding the event!

As noted in an earlier post, I've read that having four seasons in the year is a fairly recent innovation and that traditionally there were just Winter and Summer. (Ireland seems to be the exception, but some scholars have said that the first references to Imbolc were by Christian writers, casting doubt on an ancient four seasons tradition). However, if that is so, why celebrate or note the mid-points between solstices and equinoxes?

In addition, it seems there was a Germanic and Nordic custom of Summer starting with May Day and ending with Halloween. Perhaps this was nothing to do with seasonal mid-points, but because of the time it takes for earth and sea to heat up and to cool down again, although that seems to be at odds with the fire festivals, which suggest they were about daylight. Either way, mid-Summer and mid-Winter celebrations would still have been held on the solstices and not early August (Lammsas) and early February (Imbolc).

Perhaps, in the Germanic tradition, Summer and Autumn (albeit the latter un-named) were combined to form one large season with Winter and Spring (also un-named) the other? 

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 28 Apr 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

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.

Earth's gravitational pull vs. the Sun's

Updated 26 Apr 26

To compare the Earth's gravitational pull with the Sun's, let's look at the Moon's orbit about the Earth, but to keep it simple, we'll ignore its mass. (Assuming a similar density to the Earth's, its mass would be about 1/50th of it, and this approximation is fine for our purposes).

The Moon's orbital period is 27.3 days or 27.5/365.25 years.

Square that and it's about 5.6 x 10⁻³.

From memory, it's about 240 thousand miles away. Again, from memory, the Sun is about 93 million miles away, so the radius of the Moon's orbit about the Earth is about 2.6 x 10⁻³ AUs. 

(Actually, I've done this back-to-front and imagine that the size of the AU was first determined in relation to the Earth-Moon distance, something that had been measured fairly accurately by the Ancient Greeks).

Cubing that is about 17.1 x 10⁻⁹.

The ratio of this number to the square of the orbital period is about 3 x 10⁻⁶. This indicates the strength of the gravitational field of the Earth (from Newtonian Gravitational theory). Let's call it Γₑ.

The equivalent figure for the Sun (let's call it Γₛ)  is 1 (see my earlier post on Kepler's third law). So, the gravitational pull of the Sun is Γₛ/Γₑ, i.e. about 330,000, times that of the Earth. We know from Newton that this is also the ratio of the Sun's mass to that of the Earth.

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.