The Moon's orbit and eclipses

Builds on: Sections 1.6, 3.1 and 6.1
Reading time: 45 min
Exercises: 4
Olympiad problems: 2

On 12 August 2026 the Moon's shadow swept across Greenland, Iceland and northern Spain, and for up to two and a quarter minutes the Sun was hidden completely. Sixteen days later, on 28 August, the full Moon dipped partly into Earth's shadow. Earlier that year the same pairing had happened, with a ring of Sun around the Moon over Antarctica on 17 February and the Moon turning a deep red in Earth's shadow on 3 March. The Moon passes between Earth and the Sun at every new Moon, yet eclipses come only in these clusters about six months apart. In this section we see why, and why some solar eclipses are total while others leave a bright ring.

6.4.1The Moon's orbit

Before you readMake a guess first

The Moon passes between the Sun and Earth once a month. About how many solar eclipses are there in a year, seen from somewhere on Earth?

The Moon's orbit is tilted by 5.145° to the ecliptic, the plane of Earth's orbit. The two points where it crosses the ecliptic are the Moon's nodes. At the ascending node the Moon passes from south to north of the ecliptic, and at the descending node from north to south. The Sun's pull on the Moon makes the nodes slide backwards, westwards along the ecliptic, by 19.3° a year, so that they go once round in 18.6 years.

The orbit is also an ellipse, with a semi-major axis of 384 400 km and an eccentricity of 0.0549, so the Moon's distance swings between about 363 300 km at perigee and 405 500 km at apogee. The extremes reached over the years are 356 500 km and 406 700 km, which make the Moon's disc 33.5′ across at its largest and 29.4′ at its smallest. The Sun's disc ranges only from 31.4′ in July to 32.5′ in January, so the Moon can look a little larger or a little smaller than the Sun.

Because the nodes and the long axis of the orbit both move, the Moon has four different months, depending on what we count its return to. The month of 27.321662 days is its return to the same stars, and the synodic month of 29.530589 days (Section 6.1) is its return to the same phase. The draconic month of 27.212221 days is its return to the same node, shorter than the sidereal month because the node comes back to meet it. The anomalistic month of 27.554550 days is its return to perigee, longer because perigee moves forwards.

ResultThe draconic month(6.9)

\frac{1}{T_{dr}} = \frac{1}{T_{sid}} + \frac{1}{T_{node}}

$T_{dr}$	the draconic month, from node to the same node
$T_{sid}$	the sidereal month, 27.321662 days
$T_{node}$	the time the nodes take to go once round the ecliptic, 18.6129 years

This is the rule of (6.2) with a plus sign, because the node moves in the opposite direction to the Moon. The same rule with Earth's sidereal year in place of the sidereal month gives the eclipse year of 346.6 days, the time the Sun takes to return to the same node.

Quick checkThe Moon's mean apparent size

The Moon's radius is 1737.4 km. What is its apparent diameter at its mean distance of 384 400 km, in minutes of arc?

′

6.4.2Shadows in space

Before you readMake a guess first

Is the Moon's cone of full shadow long enough to reach Earth when the Moon is at its average distance?

A body lit by the Sun casts two kinds of shadow. Inside the umbra the whole of the Sun is hidden. Because the Sun is larger than the body, the umbra is a cone that narrows to a tip. Around it lies the penumbra, where only part of the Sun is hidden, and which widens away from the body. Beyond the tip of the umbra the body covers the middle of the Sun but leaves a ring around it, and this region is called the antumbra.

The length of the umbra follows from similar triangles. The cone's edges run from the edge of the Sun past the edge of the body. Over the distance $D$ from the Sun to the body they close in by $R_{⊙} - R$ , so they close the remaining $R$ in a further distance $L$ with $L / R = D / (R_{⊙} - R)$ .

ResultThe length of the umbra(6.10)

L = \frac{D R}{R_{⊙} - R}

$L$	the length of the umbra, from the body to the tip of the cone
$D$	the distance from the Sun to the body
$R$	the radius of the body
$R_{⊙}$	the radius of the Sun, 695 700 km

The half-width of the umbra a distance

s

behind the body is

R (1 - s / L)

. For the penumbra the edges cross between the Sun and the body, and its half-width grows as

R + s (R_{⊙} + R) / D

For the Moon at new Moon, with Earth 1 au from the Sun, the umbra is 373 600 km long. The ground under the Moon is $384 400 - 6371 = 378 000$ km away at the mean distance, so the tip stops about 4500 km short and the eclipse is annular. A total eclipse needs the Moon nearer than average. It is easiest in July, when Earth is farthest from the Sun and the umbra is longest, at 379 800 km. With the Moon at its closest, 356 500 km, the umbra then reaches 30 000 km past the ground and makes a band of totality about 270 km wide where the Sun is overhead, the widest the umbra itself can be. Where the shadow strikes the ground at a slant the band is stretched wider.

Figure 6.5The Moon's shadow

Moon from EarthEarth from the Sun

Length of the umbra
Moon to the ground below it
The eclipse is
Apparent diameters from the ground below the Moon

The figure shows the Moon M and the near side of Earth E with sunlight coming from the left, at the top to scale along the line and stretched 15 times across it. The darker cone is the umbra and the dashed lines mark the edges of the penumbra. The strip below magnifies the stretch near the ground, bracketed at the top, so that the tip T of the umbra can be seen. When T stops short of the ground the antumbra opens out beyond it and the eclipse is annular, and when T lies beyond the ground, drawn dashed inside Earth, the eclipse is total. The settings can be changed to , or .

Moving the Moon from 384 400 km to 356 500 km turns the eclipse from annular to total. With Earth 1 au from the Sun the change comes when the Moon is about 380 000 km from Earth's centre.

Earth's own umbra is 1.38 million km long, 3.6 times the Moon's distance. At the Moon it still has a radius of 4600 km, 2.6 times the Moon's radius, so the Moon can sit wholly inside it for well over an hour. Sunlight bent through Earth's atmosphere still reaches it, and because the air scatters blue light away, what gets through is red, which colours the totally eclipsed Moon.

6.4.3Eclipse seasons

An eclipse needs the Sun, Earth and the Moon to line up in three dimensions. At new Moon the Moon has the same ecliptic longitude as the Sun, but if it is far from a node it is too far north or south of the ecliptic to cover the Sun. If the new Moon falls a distance $u$ along the ecliptic from a node, the Moon's ecliptic latitude is $β$ , with

\sin β = \sin i \sin u,

where $i = 5.145 °$ is the tilt of its orbit.

How small $β$ must be depends on the sizes involved. For a solar eclipse seen from somewhere on Earth, the discs of the Sun and the Moon must touch, which needs the centre of the Moon within 0.525° of the centre of the Sun, the sum of their angular radii of 0.266° and 0.259°. Observers in different places see the Moon shifted against the Sun by up to its horizontal parallax of 0.950°, the angular radius of Earth seen from the Moon, so the limit grows to 1.47°. For the Moon to enter Earth's umbra, its centre must be within the umbra's angular radius of 0.686° plus its own radius, a limit of 0.94°. The distance from the node at which these limits are reached is called the ecliptic limit.

ResultThe ecliptic limits(6.11)

\sin u_{lim} = \frac{\sin β_{lim}}{\sin i}

$u_{lim}$	the ecliptic limit, the largest distance of the new or full Moon from a node that still gives an eclipse
$β_{lim}$	the largest ecliptic latitude of the Moon that still gives an eclipse, 1.47° for a solar eclipse and 0.94° for the Moon entering Earth's umbra
$i$	the tilt of the Moon's orbit, 5.145°

With mean distances these give 16.7° for a solar eclipse and 10.6° for an eclipse of the Moon by the umbra. The exact limits change by about a degree with the distances of the Moon and the Sun.

Example 6.7A new Moon 20° from the node

A new Moon falls 20° along the ecliptic from the Moon's ascending node. Is there a solar eclipse anywhere on Earth?

Solution

The Moon's latitude is $β = \arcsin (\sin 5.145 ° \sin 20 °) = 1.76 °$ north of the ecliptic. The limit for a solar eclipse is 1.47°, and the Moon is farther than that from the ecliptic, so its disc passes north of the Sun's disc as seen from every place on Earth.

AnswerThere is no eclipse. The new Moon is outside the solar ecliptic limit of 16.7°.

The Sun reaches a node once every half eclipse year, 173.3 days, and the weeks around each passage are an eclipse season. From one new Moon to the next the Sun moves 30.7° further from the node. The window for a solar eclipse runs from 16.7° before the node to 16.7° after it, 33.3° in all, which is wider than the 30.7° step. Some new Moon must fall inside it, so every eclipse season has at least one solar eclipse, and if a new Moon falls within 2.6° of the start of the window, the next one is still inside and the season has two. The window for an eclipse of the Moon by the umbra is only 21.2° wide, so some seasons have none. In the figure, sliding the nearest new Moon west of the node shows a second new Moon entering the window from the other side once the first is more than 14.0° before the node.

Figure 6.6Eclipse seasons

Eclipses

Nearest Moon from the node

Ecliptic limit
This eclipse season

The figure shows a strip of sky along the ecliptic, drawn dotted, around a node N of the Moon's orbit, with distances up stretched 5 times. The shaded band is how close to the ecliptic the Moon must be for an eclipse, so the heavy part of the Moon's path lies within the ecliptic limits. The markers on the ecliptic are the Sun at each new Moon of the season, or Earth's shadow at each full Moon, measured from the opposite node, and the Moon sits above or below each one on its path. The settings can be changed to , or .

The eclipses of 2026 fit this pattern. Each season brought a solar eclipse and a lunar eclipse about half a synodic month, 14.8 days, apart, and the two seasons were 176 days apart, close to half an eclipse year. A calendar year holds parts of two or three eclipse seasons, so it has at least two solar eclipses and at most five.

6.4.4The saros

Eclipses repeat after a period known since ancient times as the saros. It works because three of the Moon's months nearly fit into the same span of time. 223 synodic months take 6585.32 days, 242 draconic months take 6585.36 days and 239 anomalistic months take 6585.54 days. After one saros the Moon is again new or full, at nearly the same distance from the node and at nearly the same distance from Earth, so the eclipse repeats with nearly the same size and length. The saros also lasts almost exactly 19 eclipse years, so the Sun is back near the same node and the eclipse falls at nearly the same time of year.

A saros is 18 years and 10 or 11 days, depending on how many leap years it contains, and a third of a day. In that extra 7.7 hours Earth turns 116° further, so each eclipse in a series is seen about a third of the way round the world to the west of the last. After three saroses, 19 756 days, the eclipse returns to roughly the same longitude.

Example 6.8An eclipse of 2008 and its returnfinish it

A total eclipse of the Sun crossed northern Canada, Greenland, Russia and China on 1 August 2008. On what day of August 2026 did the next eclipse of the same saros fall?

The start

From 1 August 2008 to 1 August 2026 is 18 years containing the four leap days of 2012, 2016, 2020 and 2024, so it is $18 \times 365 + 4 = 6574$ days.

Your stepNow add the rest of the saros and find the date.

August 2026

Common errorsSection 6.4

The Moon's orbit is not in the plane of the ecliptic. If it were, there would be a solar eclipse at every new Moon and a lunar eclipse at every full Moon.
An annular eclipse is not a partial eclipse. The Moon is centred on the Sun, but its disc is too small to cover it, because the tip of the umbra stops short of the ground.
The draconic month is shorter than the sidereal month, not longer. The nodes move westwards, against the Moon's motion, so the Moon meets its node again before it has gone fully round against the stars.
The solar ecliptic limit is larger than the lunar one, even though Earth's shadow is much larger than the Moon's. A solar eclipse counts if it is seen from anywhere on Earth, and the spread of viewpoints across Earth adds the Moon's parallax of nearly a degree to the limit.
A saros does not bring an eclipse back to the same place. The extra third of a day shifts each eclipse about 116° to the west.

ExercisesSection 6.44 questions

6.4.1
With Earth 1 au from the Sun, how far from Earth's centre must the Moon be for its umbra just to reach the ground below it?
km
6.4.2
For a penumbral eclipse of the Moon, the Moon only needs to touch Earth's penumbra, whose angular radius at the Moon is the sum of the Moon's parallax, the Sun's parallax and the Sun's angular radius. Taking these as 0.950°, 0.002° and 0.266°, and the Moon's angular radius as 0.259°, find the ecliptic limit.
°
6.4.3
Why can a total eclipse of the Moon be seen from half the world, while a total eclipse of the Sun is seen only from a narrow band?
6.4.4
Which conditions together give the longest total eclipses of the Sun?

Olympiad problemsPast IOAA problems, restated2 problems

2008An annular eclipse on film20 points

Adapted from IOAA 2008 (Bandung), theory problem S11. We have restated it in our own words and kept the original data. The official solution takes the diameters of the Sun and the Moon as 1 392 000 km and 3476 km, which we state. The solution was last checked on 23 October 2026.

A photograph of the annular eclipse of 22 August 1998, seen from Dumai in Indonesia, was taken on 35 mm film through a telescope of aperture 10 cm and focal ratio 15. On the film the Sun's disc is 13.817 mm across and the Moon's is 13.235 mm. Taking the diameters of the Sun and the Moon as 1 392 000 km and 3476 km, find the distances of the Sun and the Moon from the observer, and the percentage of the Sun's disc covered by the Moon at mid-eclipse.

I solved it without the solution

AnswerThe Sun was $1.511 \times 10^{8}$ km away and the Moon about 394 000 km away, and the Moon covered 91.75% of the Sun's disc.

2010Annular eclipses against total ones10 points

Adapted from IOAA 2010 (Beijing), theory problem S15. We have restated it in our own words and kept the original data. Like the official solution, we take the Moon's distance to change steadily with time between perigee and apogee. Real counts over five thousand years give a ratio nearer 1.4 when hybrid eclipses are counted as annular. The solution was last checked on 23 October 2026.

The Moon's apparent diameter is on average slightly less than the Sun's, so annular eclipses are a little more common than total ones. The longest total eclipse lasts about 7.5 minutes and the longest annular eclipse about 12.5 minutes, both measured from second to third contact. Estimate the ratio of the number of annular eclipses to the number of total ones over a long time. Take Earth's orbit as a circle and the eccentricity of the Moon's orbit as 0.0549, and count hybrid eclipses as annular.

I solved it without the solution

AnswerThere are about 1.34 annular eclipses for every total one, roughly four to three.