Transits and occultations

Builds on: Sections 5.2, 6.1 and 6.4
Reading time: 30 min
Exercises: 4
Olympiad problems: 2

On 6 June 2012 Venus crossed the face of the Sun as a small black disc, taking more than six hours to do so. Eight years earlier, on 8 June 2004, it had made the same crossing, but the next one will not come until December 2117. In the eighteenth century astronomers sailed round the world to time the transits of 1761 and 1769 from distant places, because the small differences in timing gave the distance to the Sun. Mercury crosses the Sun more often, about thirteen times a century, always in early May or early November, most recently on 11 November 2019. In this section we work out why these events are so rare and so regular, and what a body passing in front of another can tell us.

6.5.1Transits

Before you readMake a guess first

Venus passes between Earth and the Sun every 584 days. Why is it seen crossing the Sun's face only twice in more than a century?

A transit is the passage of a smaller body across the disc of a larger one, here Mercury or Venus across the Sun. It can only happen at an inferior conjunction, and, as with the eclipses of Section 6.4, only when the conjunction falls close to a node of the planet's orbit.

The planet is much closer to us than the Sun, so its height above the ecliptic is magnified when seen from Earth. A planet of orbital radius $a$ , an angle $u$ round its orbit from a node, is $a \sin i \sin u$ above or below the plane of Earth's orbit, and at inferior conjunction we see it from a distance $1 - a$ . Its latitude seen from Earth is therefore about $a / (1 - a)$ times its latitude seen from the Sun. For Venus the factor is 2.61, and for Mercury it is 0.63.

ResultThe transit limit(6.12)

\frac{a}{1 - a} i \sin u_{lim} \cos ψ = ρ_{⊙} + ρ

$u_{lim}$	the largest distance of the inferior conjunction from a node, round the planet's orbit, that still gives a transit
$a$	the radius of the planet's orbit in au
$i$	the tilt of the planet's orbit to the ecliptic
$ψ$	the angle between the planet's path across the Sun and the ecliptic
$ρ_{⊙}, ρ$	the angular radii of the Sun and of the planet seen from Earth

The left side is the closest approach of the planet's centre to the Sun's centre, in the same angular units as the right side. The path is tilted to the ecliptic because the planet's latitude changes as it moves, by 8.8° for Venus and 9.1° for Mercury. With the Sun's radius in June for Venus and in November for Mercury, the limit is 1.8° for Venus and 3.6° for Mercury.

Earth must also be on the line of the planet's nodes, so transits can only happen on two dates of the year. The ascending node of Venus's orbit lies at heliocentric longitude 76.7°, so the Sun is in line with the nodes in early June and early December. For Mercury the nodes lie at 48.3° and 228.3°, which brings the Sun there in early May and early November. Mercury's orbit is quite eccentric, and in November Mercury crosses its node near perihelion, 0.31 au from the Sun, where its latitude seen from Earth is magnified less. Its transit limit is then about 4.8°, against 2.7° in May, so November transits are about twice as common.

For Venus the limit of 1.8° is so tight that the regularity of Section 6.1 takes over. Five synodic periods of Venus are 2.5 days shorter than eight years, and in that time Earth moves 2.4° round its orbit, so each inferior conjunction near a node is followed eight years later by another 2.4° further back. A window 3.6° wide can hold one or two of these but never three. At present it holds two each time, so transits of Venus come in pairs eight years apart, as in 2004 and 2012. The pairs repeat after 243 years, almost exactly 152 synodic periods, alternating between June and December with gaps of 105.5 and 121.5 years: $8 + 105.5 + 8 + 121.5 = 243$ .

Figure 6.7A transit across the Sun

Planet

Conjunction past the node

Watch

At conjunction the planet is
Closest approach
From contact I to contact IV

The figure shows the Sun's disc with ecliptic north up and east to the left, as in the sky, with the ecliptic drawn dotted. The planet crosses from east to west along the line, entering the disc at the first contact I and leaving it at the last contact IV, and C marks the closest approach to the Sun's centre. The slider sets how far round its orbit the planet is from the node at the moment of conjunction. The filled dot is the planet at the moment of conjunction, and running the transit carries it across. For Venus the dashed lines are the paths of the conjunctions eight years before and after, 2.4° further on and further back, so a setting where two paths cross the disc is a pair of transits. The settings can be changed to , , or .

The presets show the pair at work. With the conjunction 1.2° past the descending node, Venus crosses 11.0′ south of the Sun's centre, close to the 10.5′ of 2004. Eight years later the conjunction comes 1.2° before the node and the path is as far north, as in 2012. Eight years after that, 3.6° before the node, Venus passes 33.0′ north of the centre, well clear of the disc, whose radius is only 15.7′.

6.5.2How long a transit lasts

Against the Sun the planet moves west at $a / (1 - a)$ times the rate at which it gains on Earth round the Sun, which is $360 ° / S$ for the synodic period $S$ . For Venus this is 1.61° a day, or 4.0′ an hour, and for Mercury 4.9′ an hour. A transit then lasts as long as the planet takes to cover the chord of the Sun's disc along its path.

ResultThe length of a transit(6.13)

t = \frac{2 \sqrt{(ρ_{⊙} + ρ)^{2} - b^{2}}}{ω}

$t$	the time from the first contact to the last
$b$	the closest approach of the planet's centre to the Sun's centre
$ω$	the angular speed of the planet along its path against the Sun
$ρ_{⊙}, ρ$	the angular radii of the Sun and of the planet

Between the second and third contacts, when the planet's disc is wholly inside the Sun's, the time is the same with

ρ_{⊙} - ρ

in place of

ρ_{⊙} + ρ

. Seen from different places on Earth the planet is shifted slightly against the Sun, and Earth's rotation carries the observer during the transit, so the contact times differ by several minutes from place to place. The expeditions of 1761 and 1769 timed exactly these differences.

Example 6.9A central transit of Venus

How long would a transit of Venus last if Venus passed straight across the centre of the Sun in June? Take the Sun's angular radius as 15.75′, Venus's disc as 60″ across, and its speed along the path as 4.07′ an hour.

Solution

For a central transit $b = 0$ , so the chord is $2 (ρ_{⊙} + ρ) = 2 (15.75' + 0.50') = 32.5'$ . By (6.13) the transit lasts $32.5 / 4.07 = 8.0$ hours.

AnswerA central transit of Venus would last about 8.0 hours. The transits of 2004 and 2012 were off-centre and lasted a little over six hours.

Quick checkA central transit of Mercury

In November the Sun's angular radius is 16.2′. Mercury's disc is 11″ across at inferior conjunction and it moves along its path against the Sun at 5.0′ an hour. How long does a central transit of Mercury last?

hours

Example 6.10How far away is Venus at a transitfinish it

At a transit Venus is between Earth and the Sun. How far is it from Earth, and how wide is its disc then?

The start

Venus's orbit has a radius of 0.723 au, and at inferior conjunction it lies on the line from Earth to the Sun, so its distance from Earth is $1 - 0.723 = 0.277$ au.

Your stepNow turn this into kilometres, with 1 au =

1.496 \times 10^{8}

km.

× 10⁷ km

6.5.3Occultations

Before you readMake a guess first

An asteroid passes in front of a star. From one town the star vanishes for 3 s, and from another for 5 s. What does that tell us?

When the Moon passes in front of a star or a planet, the event is called an occultation. The Moon moves eastwards against the stars by about its own width every hour, 33′ an hour, so a star behind the middle of the Moon stays hidden for about 57 minutes as seen from Earth's centre, and longer from a place where the Moon is high, because the turning Earth carries the observer the same way as the Moon. Because the Moon has no air, a star disappears at the Moon's edge almost at once, and the precise moment fixes the Moon's position, and so its orbit, very accurately.

An asteroid occultation works the other way round. The asteroid is too small to see as a disc, but when it passes in front of a star, its shadow, which has the asteroid's own size and shape because the star is so far away, sweeps across Earth. Observers in a line across the shadow's path each measure how long the star is hidden. The shadow moves with the asteroid's velocity relative to Earth, measured across the line of sight, so the length of each chord across the asteroid is that speed times the time the star was hidden. Several chords together trace out the asteroid's outline.

Common errorsSection 6.5

The planet's latitude seen from Earth is not the same as seen from the Sun. For Venus it is magnified 2.61 times, because Venus is so much closer to Earth than to the Sun at conjunction.
A transit needs the inferior conjunction to fall near a node of the planet's orbit, not near a node of the Moon's orbit. The two have nothing to do with each other.
The speed of a planet across the Sun is not its orbital speed divided by its distance. What matters is its motion relative to Earth, which is only the difference between the two angular speeds round the Sun, magnified by $a / (1 - a)$ .
The planet does not cross the Sun from west to east. It is overtaking Earth on the inside, so against the Sun it moves backwards, from east to west.
In an asteroid occultation the speed of the shadow is the relative speed across the line of sight, not the full difference of the two orbital speeds. The part of the relative velocity along the line of sight does not move the shadow at all.

ExercisesSection 6.54 questions

6.5.1
The next transit of Venus comes in December 2117. In which year will the one after it be?
6.5.2
Why are transits of Mercury much more frequent than transits of Venus?
6.5.3
In which months can transits of Mercury happen?
6.5.4
An asteroid's shadow sweeps across Earth at 20 km/s, and an observer on the centre line of its path sees the star vanish for 4.0 s. How wide is the asteroid?
km

Olympiad problemsPast IOAA problems, restated2 problems

2010A transit of Venus seen from the North Pole10 points

Adapted from IOAA 2010 (Beijing), theory problem S14. We have restated it in our own words and kept the original data, describing in words the path that the original shows in a figure. The official solution turns the chord into an angle round the Sun and uses the planets' angular speeds round the Sun, which is the same method, and gets 14h 57m. Its factor cos ε, never defined, is cos 3.4° for Venus's inclination. It corrects for the tilt of the path, but seen from Earth the path is tilted 8.8° to the ecliptic, not 3.4°. The official solution takes the Sun's angular diameter as 32.0′ and Earth's radius as 6378 km, which we state. The solution was last checked on 23 October 2026.

An observer near Earth's North Pole watches a transit of Venus. The path of Venus's centre across the Sun is a straight line. It crosses the Sun's limb at the points A and B, which are 90° apart as seen from the Sun's centre O. The first contact, when Venus's disc first touches the Sun's from outside, is at 9:00 UT. Find the time of the last contact. Take Venus's sidereal period as 224.70 days, Earth's as 365.25 days, the radius of Venus's orbit as 0.723 au, Venus's radius as 0.949 times Earth's radius of 6378 km, and the Sun's angular diameter as 32.0′.

I solved it without the solution

AnswerWith the data given, the last contact comes about 5.97 hours after the first, at about 14:58 UT. Allowing for the tilt of Venus's path brings it to about 14:54 UT.

2015The size of the asteroid Julietta

Adapted from IOAA 2015 (Magelang), theory problem S3. We have restated it in our own words and kept the original data. The book gives only the weights 20, 25, 40 and 15 for the steps of its solution. The official solution takes the shadow's speed as Earth's speed minus Julietta's speed times the cosine of the angle at the Sun, 13.98 km/s, which gives 86.7 km. That leaves out the direction of the line of sight, and the speed across it is 9.79 km/s. Like the official solution, we ignore the rotation of Earth. The shadow moved west while Borobudur was carried east at 0.46 km/s, which would raise the size to about 64 km. The solution was last checked on 23 October 2026.

On 27 May 2015 the asteroid 1285 Julietta passed in front of the star δ Sagittarii, as seen from the Borobudur temple in Indonesia, which lay on the centre line of the asteroid's shadow. The star was hidden for 6.201 s. At the time Julietta was 3.076 au from the Sun and 2.156 au from Earth, near the far end of its orbit, whose semi-major axis is 2.9914 au. Take Earth's orbit as a circle and Julietta's orbit as lying in the plane of the ecliptic, with both bodies moving the same way round the Sun. Estimate Julietta's diameter.

I solved it without the solution

AnswerJulietta is about 61 km across.