Ecliptic and galactic coordinates

Builds on: Sections 1.3 and 1.5
Reading time: 25 min
Exercises: 6
Olympiad problems: 3

On some evenings the Moon and two or three bright planets stand in a slanting line across the sky. A month later the planets have moved, but they still lie along the same line among the stars. The Moon and every planet keep close to one great circle on the celestial sphere, the path along which the Sun moves among the stars in the course of a year. It is called the , because eclipses of the Sun and the Moon can happen only when the Moon is near it. A second great circle can be seen on a dark night as the faint band of the Milky Way, which is the disc of our own Galaxy seen from inside. Each of these circles is the base of a coordinate system. Ecliptic coordinates suit the Sun, the Moon and the planets, and galactic coordinates suit the stars and the gas of the Milky Way. In this section we set up both systems and learn to convert between them and the right ascension and declination of Section 1.3.

1.6.1The ecliptic

Before you readMake a guess first

Over a year the Sun moves once round the sky against the background of stars. Roughly how far does it move in a day?

Earth goes round the Sun once a year, so from Earth the Sun appears to go round the sky once a year against the background stars, moving eastward by about a degree a day. Its path among the stars is the ecliptic. The plane of Earth's orbit passes through the Sun and through Earth, and the ecliptic is that plane extended until it meets the celestial sphere, so the ecliptic is a great circle. The celestial equator is the plane of Earth's equator extended in the same way. Earth's axis is tilted by 23.44° from the perpendicular to its orbit, so the two planes, and the two circles, cross at that angle. It is called the obliquity of the ecliptic, $ε = 23.44 °$ . The ecliptic has its own poles, 90° from every point on it. The north ecliptic pole K lies in Draco, 23.44° from the north celestial pole, at right ascension 18h and declination +66.56°.

The ecliptic and the celestial equator cross at two points opposite each other. The Sun passes the March equinox point, going north, around 20 March, and the September equinox point, going south, around 23 September. Halfway between them it is furthest north of the equator, at declination +23.44°, around 21 June, and furthest south, at −23.44°, around 21 December. These two points are the June and December solstices. When the Sun is north of the celestial equator, days in the northern hemisphere are longer than nights and the noon Sun stands higher, as Example 1.14 found for Ankara, and this is what makes the seasons.

The position of the Sun along the ecliptic is given by its ecliptic longitude $λ_{⊙}$ , the arc of the ecliptic from the March equinox point eastward to the Sun, from 0 to 360°. The Sun, the March equinox point and the foot of the Sun's hour circle on the celestial equator make a spherical triangle with a right angle at the foot. Its hypotenuse is $λ_{⊙}$ , its angle at the equinox point is $ε$ , and its other two sides are the Sun's declination and right ascension, so the results for right-angled triangles in (1.18) give both.

ResultThe Sun's declination and right ascension(1.24)

\begin{aligned} \sin δ_{⊙} & = \sin ε \sin λ_{⊙} \\ \tan α_{⊙} & = \cos ε \tan λ_{⊙} \end{aligned}

$λ_{⊙}$	the Sun's ecliptic longitude, measured from the March equinox point
$δ_{⊙}, α_{⊙}$	the Sun's declination and right ascension
$ε$	the obliquity of the ecliptic, 23.44°

The right ascension lies in the same quarter of the circle as the ecliptic longitude. Both formulas assume that the Sun is exactly on the ecliptic, which is true to within a few arcseconds.

Figure 1.15The ecliptic on the celestial sphereinteractive

Date in 2026

Watch

Ecliptic longitude
Equatorial
In Ankara

The figure shows the celestial sphere seen from outside, with the celestial equator (dotted) and the ecliptic (thin line), which cross at the March equinox point, marked with the sign of Aries, at the angle

ε

. P is the north celestial pole and K is the north pole of the ecliptic, 23.44° from P. The slider moves the Sun through the days of 2026 using its true position, and the heavy arcs measure its ecliptic longitude

λ

along the ecliptic and its declination

δ

from the equator. The table gives its coordinates and the length of the day in Ankara. Running through the year shows the Sun climbing to +23.44° at the June solstice and sinking to −23.44° at the December solstice. The figure can be set to , , or .

Example 1.17The Sun in late April

Around 20 April the Sun has moved 30° along the ecliptic from the March equinox point. Find its declination and right ascension.

Solution

By (1.24),

\begin{aligned} \sin δ_{⊙} & = \sin 23.44 ° \sin 30 ° \\ = 0.3978 \times 0.5 = 0.1989, \end{aligned}

so $δ_{⊙} = + 11.47 °$ , and

\begin{aligned} \tan α_{⊙} & = \cos 23.44 ° \tan 30 ° \\ = 0.9175 \times 0.5774 = 0.5297, \end{aligned}

so $α_{⊙} = 27.91 °$ , which at 15° to the hour is 1h 51.6m.

AnswerThe Sun's declination is +11.47° and its right ascension is 1h 51.6m, or 27.91°.

The right ascension in this example is less than the ecliptic longitude. Near the equinoxes the ecliptic slants across the hour circles, so the Sun gains less in right ascension than it moves along the ecliptic, and near the solstices it gains more. This is one of the two reasons why sundials and clocks drift apart during the year, which Chapter 2 explains.

1.6.2Ecliptic coordinates

Before you readMake a guess first

The Moon's path is tilted a little to the ecliptic. How far from the ecliptic can the Moon be?

The ecliptic coordinates of any object are measured in the same way as the Sun's longitude. The ecliptic longitude $λ$ is measured eastward along the ecliptic from the March equinox point, from 0 to 360° and in degrees rather than hours. The ecliptic latitude $β$ is the angle north or south of the ecliptic, positive towards K. The Sun's latitude is always zero and the Moon's stays within about 5.3° of zero. The planets stay close to the ecliptic too, because their orbits are tilted by less than 4° to Earth's, except Mercury's, which is tilted by 7°. Seen from Earth a planet can appear further from the ecliptic than the tilt of its orbit when it passes close to us, Venus up to about 9° and Mars about 7°.

The equatorial and ecliptic systems share the direction of the March equinox point and differ only by a turn of $ε$ about it. Converting between them is a matter of turning a unit vector through that angle.

ResultFrom ecliptic to equatorial coordinates(1.25)

\begin{aligned} \sin δ & = \sin β \cos ε + \cos β \sin ε \sin λ \\ \cos δ \cos α & = \cos β \cos λ \\ \cos δ \sin α & = \cos β \cos ε \sin λ - \sin β \sin ε \end{aligned}

$λ, β$	ecliptic longitude and latitude
$α, δ$	right ascension and declination

The first line gives the declination. The other two give

\cos α

and

\sin α

, each multiplied by the same positive number

\cos δ

. Dividing the third line by the second gives

\tan α

, and the signs of the two lines decide which quarter of the circle

α

is in.

ResultFrom equatorial to ecliptic coordinates(1.26)

\begin{aligned} \sin β & = \sin δ \cos ε - \cos δ \sin ε \sin α \\ \cos β \cos λ & = \cos δ \cos α \\ \cos β \sin λ & = \cos δ \cos ε \sin α + \sin δ \sin ε \end{aligned}

Show the derivation

Write the direction of the star as a unit vector. In the equatorial frame, with the $x$ axis towards the March equinox point, the $y$ axis towards right ascension 6h on the celestial equator and the $z$ axis towards P, its components are $(\cos δ \cos α, \cos δ \sin α, \sin δ)$ . In the ecliptic frame, which has the same $x$ axis but its $z$ axis towards K, they are $(\cos β \cos λ, \cos β \sin λ, \sin β)$ . The ecliptic frame is the equatorial frame turned through $ε$ about the $x$ axis, so the $x$ components are equal, and the other two are related by a rotation through $ε$ ,

\begin{aligned} y_{equ} & = y_{ecl} \cos ε - z_{ecl} \sin ε, \\ z_{equ} & = y_{ecl} \sin ε + z_{ecl} \cos ε . \end{aligned}

Putting in the components gives (1.25). Turning back through $- ε$ gives (1.26). For the Sun, with $β = 0$ , the first line of (1.25) is the declination in (1.24).

Example 1.18Regulus and the ecliptic

Regulus, the brightest star in Leo, has right ascension 10h 08.4m and declination +11.97° for the year 2000. Find its ecliptic latitude and longitude.

Solution

The right ascension is $α = 152.10 °$ . By the first line of (1.26),

\begin{aligned} \sin β & = \sin 11.97 ° \cos 23.44 ° \\ - \cos 11.97 ° \sin 23.44 ° \sin 152.10 ° \\ = 0.1903 - 0.1821 = 0.0082, \end{aligned}

so $β = + 0.47 °$ . The other two lines give $\cos β \cos λ = \cos δ \cos α = - 0.8645$ and $\cos β \sin λ = 0.5025$ . The cosine is negative and the sine positive, so $λ$ is between 90° and 180°, and $\tan λ = 0.5025 / (- 0.8645)$ gives $λ = 149.83 °$ .

AnswerRegulus is at ecliptic longitude 149.83°, less than half a degree north of the ecliptic. The Sun passes almost in front of it each year around 23 August, when its own longitude reaches 150°.

Regulus is one of four bright stars close enough to the ecliptic for the Moon to pass in front of them from time to time. The others are Aldebaran, Spica and Antares, at ecliptic latitudes of −5.47°, −2.05° and −4.57°. Aldebaran lies a little beyond 5.3°, but the Moon's disc is about half a degree wide, and seen from Earth's surface rather than its centre the Moon can be shifted by up to about 1°, so it can still cover the star. Pollux, at +6.68°, is just too far north even so.

Example 1.19The Moon at its furthest northfinish it

With its orbit at its average tilt of 5.15°, the furthest north the Moon gets is ecliptic latitude +5.15°. Find its declination when this happens at ecliptic longitude 90°.

The start

By the first line of (1.25),

\begin{aligned} \sin δ & = \sin 5.15 ° \cos 23.44 ° \\ + \cos 5.15 ° \sin 23.44 ° \sin 90 ° . \end{aligned}

Your stepNow find

δ

degrees

1.6.3Galactic coordinates

Before you readMake a guess first

The band of the Milky Way crosses the celestial equator. At roughly what angle?

The Milky Way is the combined light of the many distant stars in the disc of our Galaxy, seen from inside the disc. The great circle along the middle of the band is the galactic equator. Its poles are the north galactic pole, at right ascension 12h 51.4m and declination +27.13° in Coma Berenices, and the south galactic pole opposite it. The centre of the Galaxy lies on the galactic equator at right ascension 17h 45.6m and declination −28.94°, in Sagittarius, about 26 000 light years away and hidden from view by dust.

The galactic latitude $b$ is the angle of an object north or south of the galactic equator. The galactic longitude $l$ is measured along the galactic equator from the direction of the galactic centre, from 0 to 360°, through Aquila and Cygnus at longitudes between about 30° and 90°, then Cassiopeia and Perseus, to the point opposite the centre at 180° on the border of Taurus and Auriga, and on through Carina and Crux back to Sagittarius. The conversion uses the position of the north galactic pole given above and the galactic longitude of the north celestial pole, 122.93°, both for the year 2000.

ResultGalactic latitude and longitude(1.27)

\begin{aligned} \sin b & = \sin δ \sin δ_{G} \\ + \cos δ \cos δ_{G} \cos (α - α_{G}) \\ \cos b \sin (l_{P} - l) & = \cos δ \sin (α - α_{G}) \\ \cos b \cos (l_{P} - l) & = \sin δ \cos δ_{G} \\ - \cos δ \sin δ_{G} \cos (α - α_{G}) \end{aligned}

$b, l$	galactic latitude and longitude
$α_{G}, δ_{G}$	right ascension and declination of the north galactic pole, 192.86° and +27.13° for the year 2000
$l_{P}$	galactic longitude of the north celestial pole, 122.93°

The first line is (1.23) applied to the star and the north galactic pole, since the star's distance from the pole is

90 ° - b

. The other two lines come from the triangle formed by the star and the two poles. Its angle at the celestial pole is

α - α_{G}

, and its angle at the galactic pole is

l_{P} - l

, the difference between the galactic longitudes of the celestial pole and the star. Together the two lines fix

l

in the same way as (1.25) fixes

α

Figure 1.16The ecliptic and the Milky Way among the starsinteractive

Star
Equatorial
Ecliptic
Galactic

The figure shows the sky between declinations −70° and +70°, drawn from a star catalogue for the year 2000, with stars down to magnitude 4 and the lines of the thirteen constellations that the ecliptic passes through. Right ascension grows to the left, as on a star chart held up against the sky. The straight line of fine dots is the celestial equator, the curve of heavier dots is the ecliptic, with the sign of Aries where it crosses the equator going north, and the line of larger, wider dots is the galactic equator, with the galactic centre and the north galactic pole marked. The small ticks on the ecliptic show where the Sun is on the 21st of each month, and the four that are labelled mark the equinoxes and the solstices. The map can be dragged sideways, and tapping a star, or choosing one from the list, gives its coordinates in all three systems.

Example 1.20Deneb and Vega

Deneb (α = 20h 41.4m, δ = +45.28°) and Vega (α = 18h 36.9m, δ = +38.78°) are two corners of the Summer Triangle. Find the galactic latitude of each and the galactic longitude of Deneb, and say which star lies in the band of the Milky Way.

Solution

For Deneb, $α = 310.35 °$ and $α - α_{G} = 117.49 °$ . By the first line of (1.27),

\begin{aligned} \sin b & = \sin 45.28 ° \sin 27.13 ° \\ + \cos 45.28 ° \cos 27.13 ° \cos 117.49 ° \\ = 0.3240 - 0.2891 = 0.0350, \end{aligned}

so $b = + 2.0 °$ . The other two lines give $\cos b \sin (l_{P} - l) = 0.6242$ and $\cos b \cos (l_{P} - l) = 0.7805$ . Both are positive, so $l_{P} - l = 38.65 °$ and Deneb's longitude is $l = 122.93 ° - 38.65 ° = 84.28 °$ , in the direction of Cygnus. For Vega, $α = 279.225 °$ and $α - α_{G} = 86.365 °$ , and

\sin b = 0.2856 + 0.0440 = 0.3296,

so $b = + 19.2 °$ .

AnswerDeneb is only 2.0° from the galactic equator, in the middle of the Milky Way, while Vega is 19.2° north of it, to one side of the band. Deneb's galactic longitude is 84.28°, and the same steps give 67.44° for Vega.

Common errorsSection 1.6

Ecliptic longitude is measured in degrees along the ecliptic, and right ascension in hours along the celestial equator. For a point on the ecliptic the two agree only at the equinox and solstice points, as Example 1.17 shows.
A tangent alone cannot tell the quarter of the circle. When finding $α$ , $λ$ or $l$ , the signs of the sine and cosine lines of the conversion decide it.
The Sun's ecliptic latitude is zero all year, while its declination swings between −23.44° and +23.44°, so a latitude in one system says little about the latitude in another.
Galactic longitude starts from the direction of the galactic centre, and ecliptic longitude and right ascension start from the March equinox point.

ExercisesSection 1.66 questions

1.6.1
In which constellation does the centre of our Galaxy lie?
1.6.2
Around 21 May the Sun's ecliptic longitude is 60°. What is its declination?
degrees
1.6.3
Spica has right ascension 13h 25.2m and declination −11.16°. Find its ecliptic latitude.
degrees
1.6.4
What is the galactic latitude of the north celestial pole?
degrees
1.6.5
On about which date is the Sun's ecliptic longitude 270°?
1.6.6
What is the greatest altitude that the Sun reaches in Ankara ( $φ = 39.9 °$ N) during the year?
degrees

Olympiad problemsPast IOAA problems, restated3 problems

2014Stars with Romanian names

Adapted from IOAA 2014 (Suceava), theory problem 13. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.

In September 2013 the Romanian astronomers Ovidiu Tercu and Alex Dumitriu, working at the observatory of the Museum Complex of Natural Sciences in Galați, discovered two variable stars, now named Galati V1 and Galati V2. Both are circumpolar as seen from Romania, one in Cassiopeia and the other in Andromeda. Their galactic coordinates are $l = 114.371 °$ , $b = - 11.35 °$ for Galati V1 and $l = 113.266 °$ , $b = - 16.177 °$ for Galati V2. Another star with a Romanian name, Sanduleak −69°202, which exploded as the supernova SN 1987A in the Large Magellanic Cloud, is at α = 5h 35m 28.03s, δ = −69°16′11.79″, $l = 279.7 °$ , $b = - 31.9 °$ . Estimate the angular distance between Galati V1 and Galati V2.

I solved it without the solution

AnswerGalati V1 and Galati V2 are about 4.94° apart in the sky.

2021The Moon over Bogotá6 points

Adapted from IOAA 2021 (Bogotá, held online), theory problem 8, part 3. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.

At 8 p.m. local time on 21 September 2021, the Moon was observed from the National Astronomical Observatory of Colombia in Bogotá (4°35′53″ N, 74°04′37″ W), towards the hills east of the city. Its azimuth, measured from north through east, was 90°42′59″, its altitude 19°01′42″, its ecliptic longitude 12°20′16″ and its ecliptic latitude −4°24′14″. Find the right ascension and declination of the Moon at that moment.

I solved it without the solution

AnswerThe Moon's right ascension was 0h 52.3m, or 13.06°, and its declination +0.82°. The official solution gives 0h 50m 52s and +1°34′. The check in step 4 shows that its declination is a slip, and its right ascension is close to the angle whose tangent is 0.2260, as if the division by 0.9740 in step 3 had been left out.

2013The highest full Moon10 points

Adapted from IOAA 2013 (Volos), theory problem 11. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.

What is the greatest altitude at which the full Moon can be seen from Thessaloniki, at latitude 40°37′ N? Take into account as many effects as you can.

I solved it without the solution

AnswerWith the average tilt of the Moon's orbit, which is what the problem expects, the full Moon can reach an altitude of about 77.8° over Thessaloniki. The tilt itself varies between about 5.0° and 5.3°, and in this arrangement, with the Sun 90° from the points where the Moon's path crosses the ecliptic, it is close to its smallest. A real full Moon therefore reaches about 0.15° lower, about 77.6°. The official solution prints 79.8° before the correction for the Moon's distance, but its own sum, $90 ° - 40.6 ° + 23.5 ° + 5.3 °$ , is 78.2°.