Precession and nutation

Builds on: Sections 1.3, 1.6 and 2.4
Reading time: 35 min
Exercises: 4
Olympiad problems: 2

Polaris has not always been the pole star. When the pyramids of Giza were built, about 4500 years ago, the north celestial pole lay close to Thuban, a faint star in Draco, which came within 0.2° of it around 2830 BC. Polaris will be closest to the pole around the year 2100, and in about 12 000 years the pole will lie some 5° from Vega. The stars have hardly moved, but Earth's axis has turned, slowly sweeping out a cone over about 25 800 years. In this section we see why the axis moves, what that does to the coordinates of the stars and to the calendar, and how to correct a catalogue position for it.

2.7.1The wandering pole

Before you readMake a guess first

Which star was closest to the north celestial pole when the pyramids of Giza were built, about 4500 years ago?

Earth is not quite a sphere. Its spin makes it bulge at the equator, so that the equatorial diameter is 43 km larger than the polar one. The Sun and the Moon pull on this bulge, and because the bulge is tilted to the plane of Earth's orbit by 23.44°, their pull tries to turn the equator into that plane. A spinning body does not simply tip over when it is twisted like this. It behaves like a spinning top, whose axis swings slowly round in a cone instead of falling. Earth's axis keeps its tilt of about 23.4° to the pole of the ecliptic and swings round it once in about 25 800 years. This slow turning of the axis is called precession.

As the axis turns, the celestial poles move round circles of radius 23.4° centred on the poles of the ecliptic, and the celestial equator moves with them. The ecliptic stays almost fixed, so the two points where the equator crosses it, the equinoxes, slide westward along the ecliptic. This is the precession of the equinoxes, discovered by Hipparchus in the second century BC by comparing his star positions with ones measured about 150 years earlier.

ResultPrecession of the equinoxes(2.15)

\begin{aligned} p & = {50.29}^{″} per year \\ T_{prec} & = \frac{360 °}{p} \approx 25 800 years \end{aligned}

$p$	rate at which the equinoxes move westward along the ecliptic
$T_{prec}$	time for one full turn of the axis

Measured from the moving equinox, the ecliptic longitude of every star grows by

p

each year, while its ecliptic latitude stays the same.

Figure 2.8The pole's path among the starsinteractive

Year

Bright star nearest the pole
Equinoxes moved along the ecliptic

The figure shows the northern sky around the north ecliptic pole, with stars down to magnitude 4.6 from a catalogue, placed by their ecliptic coordinates, which precession does not change. The north celestial pole P runs round the dotted circle, 23.4° from the ecliptic pole, once in about 25 800 years. The small ticks on the circle mark the pole in the years listed below. The slider sets the year, and the table names the bright star nearest the pole. Around the pole was close to Thuban, it is 0.6° from Polaris, around it will be near Alderamin in Cepheus, and around it will be 5° from Vega.

Precession has three consequences that olympiad problems use. First, the coordinates of every star change, so a catalogue gives positions for a stated date, called its epoch, such as J2000.0 in this course. Second, the tropical year is shorter than the sidereal year (Section 2.4), because the equinox moves 50.29″ westward each year to meet the Sun, and the Sun covers that angle in about 20 minutes. Third, the signs of the zodiac, which astrologers still count from the March equinox, no longer match the constellations they are named after. The March equinox, once in Aries, now lies in Pisces.

Quick checkUse the result above

How many years do the equinoxes take to move 1° along the ecliptic?

years

2.7.2Changes in right ascension and declination

Right ascension and declination are measured from the moving equator and equinox, so both change. Turning a star's ecliptic longitude by $p$ and working out the effect on its equatorial coordinates gives the yearly changes.

ResultYearly change of right ascension and declination(2.16)

\begin{aligned} Δ α & = p (\cos ε + \sin ε \sin α \tan δ) \\ Δ δ & = p \sin ε \cos α \end{aligned}

$Δ α, Δ δ$	changes in right ascension and declination in one year, in the same angular unit as $p$
$ε$	obliquity of the ecliptic, 23.44°

With

p = {50.29}^{″}

p \cos ε = {46.14}^{″}

, or 3.076 s of right ascension, and

p \sin ε = {20.00}^{″}

. The formulas are good for a few decades. Near the celestial poles, where

\tan δ

is large, the change in right ascension grows large and the formulas should only be used over short times.

Show the derivation

Precession turns the sky about the pole of the ecliptic by $p$ a year. A star at right ascension $α$ and declination $δ$ has the unit vector $(\cos δ \cos α, \cos δ \sin α, \sin δ)$ in the equatorial frame. The pole of the ecliptic has the unit vector $k = (0, - \sin ε, \cos ε)$ , and a small turn by $p$ about it moves a star by $p k \times r$ . Working out the cross product and comparing it with the small changes $d r$ produced by changes $Δ α$ and $Δ δ$ gives the two lines of (2.16).

Example 2.14Sirius from 2000 to 2026

Sirius has right ascension 101.29° and declination −16.72° for J2000.0. By how much do its coordinates change by 2026?

Solution

In (2.16) the second term in the bracket is $0.3978 \times 0.9806 \times (- 0.3004) = - 0.1172$ , so the change in right ascension each year is

\begin{aligned} Δ α & = {50.29}^{″} \times (0.9175 - 0.1172) \\ = {50.29}^{″} \times 0.8003 = {40.25}^{″}, \end{aligned}

which is 2.683 s of time. The change in declination is $Δ δ = {20.00}^{″} \times \cos 101.29 ° = - {3.92}^{″}$ .

AnswerIn 26 years the right ascension grows by

26 \times 2.683 = 69.8

s, about 1 min 10 s, and the declination falls by

26 \times {3.92}^{″} = 102^{″}

, about 1.7′. Section 2.1 used a right ascension of Sirius updated in this way.

2.7.3Nutation

Before you readMake a guess first

The Moon's orbit is tilted by 5° to the ecliptic, and the line where it crosses the ecliptic turns round once every 18.6 years. What would you expect this to do to the precession of Earth's axis?

The pull of the Moon and the Sun on the bulge is not steady. The Moon's orbit is tilted by 5° to the ecliptic, and its nodes, where it crosses the ecliptic, go round once in 18.6 years, so the direction of the Moon's pull changes with the same period. On top of its slow precession, the axis therefore nods back and forth by up to 9.2″ in obliquity and 17.2″ in longitude every 18.6 years, with smaller terms of half a year and half a month from the Sun and the Moon. This nodding is called nutation, and it was found by James Bradley in 1748. A position corrected only for precession is called a mean position, and one corrected for nutation too is a true position. Nutation changes coordinates by less than 20″, so olympiad problems usually leave it out.

Common errorsSection 2.7

Precession moves the equinoxes westward along the ecliptic, so the ecliptic longitudes of the stars grow. Their ecliptic latitudes do not change.
The coordinates in a catalogue belong to its epoch. Over 26 years the right ascension of a star near the equator changes by more than a minute of time.
The formulas for the yearly changes break down close to the celestial poles, where $\tan δ$ is large.
The tropical year, not the sidereal year, keeps the calendar in step with the seasons, because the seasons follow the equinoxes.

ExercisesSection 2.74 questions

2.7.1
By how many degrees did the equinoxes move along the ecliptic between 2000 and 2026?
degrees
2.7.2
Which coordinate of a star does precession leave unchanged?
2.7.3
Spica has right ascension 201.3°. By how many arcseconds does its declination change each year?
arcseconds
2.7.4
The Sun moves along the ecliptic at 360° in a sidereal year of 365.2564 days. How many minutes does it take to cover the 50.29″ by which the equinox moves each year?
minutes

Olympiad problemsPast IOAA problems, restated2 problems

2011Sirius and Canopus from Kraków10 points

Adapted from IOAA 2011 (Katowice), theory problem 12. We have restated it in our own words and kept the original data. The solution was last checked on 17 October 2026.

Because of precession, the part of the sky that can be seen from a given place changes over time. Is it possible that at some time Sirius will not rise as seen from Kraków, while Canopus will rise and set? Assume that Earth's axis sweeps out a cone of angle 47°, so that it keeps a tilt of 23.5°. Kraków is at latitude 50.1° N, and the present coordinates of the stars are 6h 45m, −16°43′ for Sirius and 6h 24m, −52°42′ for Canopus.

I solved it without the solution

AnswerNo. Precession will at times take Sirius out of sight from Kraków, but Canopus can never rise there, whatever the date.

2025Makar-Sankranti10 points

Adapted from IOAA 2025 (Mumbai), theory problem 2. We have restated it in our own words and kept the original data, adding the length of the sidereal year. The solution was last checked on 17 October 2026.

The Indian festival of Makar-Sankranti is celebrated when the Sun appears to enter the zodiacal region of Capricorn, measured against the stars. It now falls around 14 January. Long ago it fell on the winter solstice, which we take to be 21 December. Take the precession of the equinoxes as 1° every 71.6 years and the sidereal year as 365.2564 days.

(a) In what year did the festival last fall on the winter solstice? [3 points]

(b) In 2006 the Sun entered Capricorn at 11:50:13 local time in Mumbai on 14 January. On what date and at what time did it enter Capricorn in 2013? [3 points]

(c) The festival is celebrated on the date of the first sunset after the Sun enters Capricorn. Taking sunset in Mumbai in January as 18:30, give the date of the festival in each year from 2006 to 2013. [4 points]

I solved it without the solution

Answer(a) Around AD 331. (b) At 06:54:44 on 14 January 2013. (c) The festival falls on 14 January in 2006, 2007, 2009, 2010 and 2013, and on 15 January in 2008, 2011 and 2012.