Right ascension, declination and hour angle

Builds on: Sections 1.1 and 1.2
Reading time: 40 min
Exercises: 6
Olympiad problems: 2

Mintaka, the star at the western end of Orion's belt, rises almost exactly due east and sets almost exactly due west. This is true in Ankara and in Sydney, and anywhere else on Earth away from the poles, whenever it rises. The reason is that Mintaka lies almost exactly on the celestial equator, the great circle halfway between the celestial poles. The place of a star on the celestial sphere can be given by two angles measured from that circle, the declination and the right ascension. Unlike altitude and azimuth they are the same for every observer and change only very slowly over the years, so they are the coordinates listed in star catalogues and given in olympiad problems. In this section we define them, together with a third angle, the hour angle, which connects them with the sky that a particular observer sees at a particular moment.

Figure 1.7The eastern sky in right ascension and declinationinteractive

Seen from

Facing

Grid

Local time21:00

Star
Right ascension
Declination
Hour angle

The eastern sky over Ankara (39.9° N) at 21:00 on 15 November, drawn from a star catalogue, with circles of declination every 10° and an hour circle for every hour of right ascension. The dotted line is the celestial equator. Mintaka has just risen, almost exactly due east, where the celestial equator crosses the horizon. Move the time and the grid turns with the stars, so that each star keeps its right ascension and declination while its hour angle grows. Switch to Sydney, where Mintaka rises at almost the same local time and again due east, or switch the grid to the horizon system of Section 1.2 to compare. The coordinates shown are for 2026.

1.3.1The celestial equator and declination

Before you readMake a guess first

Seen from Ankara, Mintaka rises due east. Where does it rise as seen from Sydney, in the southern hemisphere?

In Section 1.2 the horizon was the base circle, and the zenith and the nadir were its poles. For coordinates that belong to the sky rather than to the observer, the base circle is the , and its poles are the north and south of Section 1.1. A great circle through the two celestial poles is called an hour circle, and a star's hour circle is the one that passes through the star. It meets the celestial equator at right angles.

The declination $δ$ of a star is the arc of its hour circle from the celestial equator to the star. It is positive for stars north of the celestial equator and negative for stars south of it, and it runs from +90° at the north celestial pole to −90° at the south celestial pole. Declination does on the sky what latitude does on Earth. As the sky turns, each star keeps the same angular distance from the poles while it goes round its daily circle, so its declination does not change during the night, and it is the same for every observer. The arc from the north celestial pole to the star along the same hour circle is the star's polar distance, the angle $p$ that appeared in (1.3).

ResultPolar distance(1.7)

p = 90 ° - δ

$p$	polar distance, the angle from the north celestial pole to the star
$δ$	declination, positive north of the celestial equator and negative south of it

The celestial equator lies directly above Earth's equator, in the same plane. On a round Earth the plumb line points straight away from Earth's centre, so the zenith lies along the line from the centre through the observer. The angle between this line and the plane of the equator is the observer's $φ$ , so the zenith is at declination $φ$ . A star passes through the zenith, at the moment it crosses the meridian, as seen from every place whose latitude equals the star's declination. Sirius ( $δ = - 16.7 °$ ) passes about a quarter of a degree from the zenith of La Paz in Bolivia (16.5° S).

NoteEarth is slightly flattened at the poles, so the plumb line misses Earth's centre by up to 0.19°, at latitude 45°. Latitudes on maps are measured from the plumb line; the difference matters only in precise work.

ResultDeclination of the zenith(1.8)

δ_{Z} = φ

$δ_{Z}$	declination of the observer's zenith
$φ$	latitude of the observer, positive north and negative south

Quick checkUse the result above

Vega has declination +38.8°. Over which of these cities does it pass almost exactly overhead?

The horizon and the celestial equator are both great circles, so they cross at two points opposite each other, and these turn out to be the east and west points. The east point is 90° from N, from S and from the zenith, so it is 90° from every point of the meridian through them, and the same goes for the west point. One of the points on the meridian is the celestial pole, and every point 90° from the pole lies on the celestial equator. So E and W are on the celestial equator as well as on the horizon, and they are the two points where the circles cross. A star on the celestial equator therefore rises due east and sets due west, as seen from anywhere on Earth except the poles, where the celestial equator lies along the horizon. A star north of the celestial equator goes round a daily circle parallel to the equator and north of it, so it rises north of east and sets north of west, and a star south of the equator rises south of east and sets south of west. The Sun is on the celestial equator at the equinoxes, around 20 March and 22 or 23 September, and on those days it rises almost due east and sets almost due west everywhere.

The celestial equator is tilted to the horizon. Let Q be the point where it crosses the meridian above the horizon, halfway between E and W. Q has declination 0 and the zenith has declination $φ$ , so Q is $| φ |$ from the zenith and stands at an altitude of $90 ° - | φ |$ . Q is the highest point of the celestial equator, 90° from both crossing points, so this is also the angle at which the celestial equator meets the horizon at E and W. A star on the celestial equator rises and sets along a path that slopes at $90 ° - | φ |$ to the horizon, steeply near Earth's equator and gently near the poles.

Quick checkUse the text above

At what angle to the horizon does a star on the celestial equator set, as seen from Ankara ( $φ = 39.9 °$ N)?

degrees

1.3.2The hour angle

Before you readMake a guess first

Through what angle does the sky carry a star around the pole in two hours?

Declination tells us how far a star is from the celestial equator, but not where it is on its daily journey round the pole. For that we measure from the meridian, which is fixed for each observer, starting at Q, where the meridian crosses the celestial equator above the horizon. Let F be the foot of the star's hour circle, the point of the celestial equator nearest to the star. The hour angle $H$ of the star is the arc QF, measured westward from Q, which is the way the sky turns. It is also the angle at the celestial pole between the meridian and the star's hour circle. Hour angle is normally given in hours rather than degrees, with 24 hours for a whole turn, because it then says how long ago the star crossed the meridian.

ResultHours and degrees(1.9)

24 h = 360 °, 1 h = 15 °

In the same way 1 minute of time is 15′ of angle and 1 second of time is 15″. Read the other way round, 1° corresponds to 4 minutes of time.

A star crosses the meridian at its highest point, its upper culmination (Section 1.4), when its hour angle is zero. From then on the hour angle grows steadily as the sky turns. A star on the celestial equator reaches the west point at an hour angle of 6h. Any star is at its lower culmination, below the pole, at 12h, and back on the meridian one sidereal day after it left, when the hour angle reaches 24h, the same as 0h. An hour angle of 21h can also be written −3h, which means that the star is 3 hours east of the meridian and will cross it in 3 hours. The hours of hour angle are sidereal hours, because the sky turns once in a sidereal day of 23 h 56 min 4 s. A sidereal hour is about 10 seconds shorter than an ordinary hour, which matters only over many hours.

Figure 1.8Hour angle and declinationinteractive

Latitude

φ

Hour angle

H

Declination

δ

Watch

Hour angle
In the sky

The celestial sphere for an observer O at latitude

φ

, seen from outside, with the horizon, the zenith Z and the meridian. The north celestial pole P stands at altitude

φ

above the north point, P′ is the south celestial pole, and the celestial equator (dotted) crosses the horizon at E and W and the meridian at Q. The star X lies on its hour circle through P and P′, which meets the equator at F. The heavy arcs are the hour angle

H

, measured westward from Q to F, and the declination

δ

, from F to X, and the thin circle parallel to the equator is the star's daily path. Press Turn the sky to watch

H

grow while

δ

stays the same. Past 12h the arc for

H

is drawn the short way, east of Q, as a negative hour angle. Try , , , and .

Example 1.7An hour angle in degrees

A star has an hour angle of 2h 37m 20s. What is this in degrees and arcminutes?

Solution

By (1.9), 2h is 30°, 37m is $37 \times 15^{'} = 555^{'} = 9 ° 15^{'}$ , and 20s is $20 \times 15^{″} = 300^{″} = 5^{'}$ . Adding these gives $30 ° + 9 ° 15^{'} + 5^{'} = 39 ° 20^{'}$ .

Answer39°20′, or 39.33°.

Quick checkUse the result above

Convert an hour angle of 4h 20m into degrees.

degrees

Quick checkUse the text above

A star has an hour angle of 21h. Where is it, and when does it next cross the meridian?

1.3.3Right ascension

Before you readMake a guess first

On an evening in January, Aldebaran (right ascension 4h 36m), Betelgeuse (5h 55m) and Sirius (6h 45m) all cross the meridian. Which crosses first?

The hour angle changes all the time, so it cannot go into a star catalogue. A catalogue needs a starting point on the celestial equator that turns with the stars, and the Sun provides one. Over a year the Sun moves eastward among the stars, by about a degree a day, along a great circle called the , which is tilted to the celestial equator by 23.4° (Section 1.6). The two circles cross at two points. The one where the Sun crosses the celestial equator going north, around 20 March each year, is the March equinox point. Olympiad problems usually call it the vernal equinox or the spring equinox, after the northern spring, and older books the first point of Aries. It is often marked with the sign of Aries, , because it lay in the constellation Aries when Greek astronomers named it, more than two thousand years ago.

The right ascension $α$ of a star is the arc of the celestial equator from the March equinox point eastward to F, the foot of the star's hour circle. Like hour angle it is given in hours, from 0h to 24h. Right ascension and declination together are the equatorial coordinates of the star. Both are measured from points and circles that turn with the sky, so they stay the same while the sky turns, and they are the same for every observer.

Right ascension is measured eastward, the opposite way to hour angle, so that stars cross the meridian in order of increasing right ascension. As the sky turns westward, a star further east along the celestial equator reaches the meridian later, and the difference between two right ascensions is the time between the two crossings, in sidereal hours.

NoteRight ascension and declination do change slowly, because Earth's axis precesses (Chapter 2) and the March equinox point drifts westward along the ecliptic by about 1.4° a century. Catalogues give coordinates for a stated year, usually 2000, and those in Figure 1.7 are for 2026.

Quick checkUse the text above

Seen from a certain place, Aldebaran ( $α$ = 4h 36m) crosses the meridian at 23:30. At what time by the clock does Sirius ( $α$ = 6h 45m) cross it?

hours

1.3.4Sidereal time

Before you readMake a guess first

A star rises at 22:00 tonight. At what time does it rise tomorrow night?

The March equinox point turns with the sky like any star, so at any moment it has an hour angle too. The hour angle of the March equinox point is called the local sidereal time, written LST. Along the celestial equator, the arc from Q westward to the equinox point is the local sidereal time, and the foot F of a star's hour circle divides it into the star's hour angle and its right ascension, as Figure 1.9 shows.

ResultHour angle and sidereal time(1.10)

H = LST - α

$H$	hour angle of the star, measured westward from the meridian
$LST$	local sidereal time, the hour angle of the March equinox point
$α$	right ascension of the star, measured eastward from the March equinox point

All three are in hours. If the result is negative, add 24h. An hour angle between 12h and 24h means that the star is east of the meridian. Some papers write the local sidereal time as

θ

Figure 1.9Sidereal time, hour angle and right ascensioninteractive

Latitude

φ

Sidereal timeRight ascension

α

Declination

δ

Watch

Hour angle
Sidereal time
In the sky

The sphere of Figure 1.8 with the March equinox point marked on the celestial equator by the sign of Aries. Its hour angle, the arc of the equator from Q westward to the sign, is the local sidereal time. When

α

is less than the sidereal time, the foot F of the star's hour circle splits this arc into the star's hour angle

H

, measured westward from Q, and its right ascension

α

, measured eastward from the equinox point. When

α

is more,

LST - α

comes out negative, as in Example 1.8, and adding 24h gives the same hour angle. Press Turn the sky and the sidereal time and the hour angle grow together, while

α

and

δ

stay the same.

Setting $H = 0$ in (1.10) shows that a star is at upper culmination, crossing the meridian on the same side of the pole as the zenith, when the local sidereal time equals its right ascension. Sidereal time runs through 24 sidereal hours in one sidereal day, gaining about 3 min 56 s on an ordinary clock every day, so each star crosses the meridian about 4 minutes earlier each night. It also depends on where we are. A place further east sees each star cross its meridian sooner, so its sidereal time is later, by one hour for every 15° of longitude. Working out the sidereal time from the date and the time on a clock is part of Chapter 2. In the observation round it is often quicker to read it off the sky, because a star crossing the meridian at upper culmination has a right ascension equal to the local sidereal time. A star crossing below the pole, at lower culmination, has a right ascension 12h different. That is how the panorama station at IOAA 2017, described in Section 1.2, could be answered.

Example 1.8How long until Regulus crosses the meridian

At a certain moment Betelgeuse ( $α$ = 5h 55m) is at upper culmination, crossing the meridian. What is the hour angle of Regulus ( $α$ = 10h 08m), and how long is it until Regulus crosses the meridian?

Solution

Betelgeuse is at upper culmination, so by (1.10) the local sidereal time equals its right ascension, 5h 55m. For Regulus

H = 5 h 55 m - 10 h 08 m = - 4 h 13 m,

which is the same as 19h 47m. Regulus is east of the meridian and reaches it after 4h 13m of sidereal time. A sidereal hour is about 10 seconds shorter than an ordinary hour, so this is 4 h 12 min by an ordinary clock.

AnswerThe hour angle is −4h 13m, or 19h 47m. Regulus crosses the meridian about 4 h 12 min later.

Example 1.9Vega on a November eveningfinish it

At 21:00 on 15 November 2026 the local sidereal time in Ankara is 23h 51m, and the right ascension of Vega, for that year, is 18h 38m. What is the hour angle of Vega, and is it east or west of the meridian?

The start

By (1.10), $H = LST - α = 23 h 51 m - 18 h 38 m$ .

Your stepFinish it: find

H

in hours.

hours

Common errorsSection 1.3

Measuring hour angle eastward, or right ascension westward. Hour angle runs westward from the meridian, the way the sky turns, and right ascension runs eastward from the March equinox point.
Putting hours into a trigonometric function. Convert them to degrees first, at 15° to the hour, so an hour angle of 3h is 45°.
Misreading a negative hour angle. −3h and 21h are the same hour angle, and both mean that the star is 3 hours east of the meridian. Add or take away 24h to move between the two forms.
Treating sidereal hours as ordinary hours over a long time. They differ by about 10 seconds an hour, which adds up to about 4 minutes a day.

ExercisesSection 1.36 questions

1.3.1
The right ascension of Spica is 13h 25m. What is this in degrees?
degrees
1.3.2
Which pair of coordinates of a star stays the same through a whole night, as seen from one place?
1.3.3
The local sidereal time is 2h 10m. Rigel has right ascension 5h 15m. How many sidereal hours is it until Rigel crosses the meridian?
hours
1.3.4
Seen from Hanoi, Vega ( $α$ = 18h 37m) sets when its hour angle is 7h 12m, as Section 1.4 shows. What is the local sidereal time at that moment?
hours
1.3.5
A star rises at 22:00 tonight. At about what time does it rise one month later?
1.3.6
In Figure 1.7, set the time to 23:30 in Ankara and face south. Tap Hamal, the brightest star of Aries, which is just past the meridian. Use its right ascension and hour angle to find the local sidereal time.
hours

Olympiad problemsPast IOAA problems, restated2 problems

2007Solar days and sidereal days2 points

Adapted from IOAA 2007 (Chiang Mai), theory problem 1.3. Restated in our own words; the data are unchanged. Solution checked 15 October 2026.

From noon on 1 July to noon on 31 December is 183 solar days. How long is this interval in sidereal days?

I solved it without the solution

Answer183.50 sidereal days, half a sidereal day more than the number of solar days.

2018Height of a hill10 points

Adapted from IOAA 2018 (Beijing), theory problem 4. Restated in our own words; the data are unchanged, with Earth's radius added. Solution checked 15 October 2026.

Two friends want to measure the height of a hill next to their village, at latitude 40°. One of them climbs to the top of the hill and agrees to send a light signal to the other, in the village, the moment she sees the Sun set. On 21 March, when they do the experiment, the friend in the village receives the signal 4.1 minutes after the Sun has set as seen from the village. Estimate the height of the hill, and the distance to the horizon for the friend on the hilltop. Ignore atmospheric refraction, and take Earth's radius to be 6378 km.

I solved it without the solution

AnswerAbout 600 m, with the horizon about 87 km away. The official solution lets the Sun move at the stars' rate, 15.04′ a minute, and gets 602 m and 87.6 km; the method is the same.