The celestial sphere

Builds on: Angles and radians (Section S3)
Reading time: 25 min
Exercises: 5
Olympiad problems: 2

Leave a camera on a tripod facing north on a clear, dark night, open its shutter for two hours, and every star draws an arc of light. All the arcs are parts of circles with a common centre, a point less than a degree from Polaris. Stars close to that point draw short arcs and stars far from it draw long ones, yet every arc spans the same angle around the centre. Nothing in the picture says how far away any of the stars is. It records only the direction to each star and how the directions change as Earth turns. Astronomers describe the sky in the same way, as directions drawn on a sphere around the observer, and in this section we build that sphere and see how it turns.

Figure 1.1Star trails over Ankarainteractive

Camera faces

Exposure

Watch

Shutter open
The sky has turned

The sky over Ankara (39.9° N) on the evening of 15 January from 19:00, drawn from a star catalogue. Every star down to 4 leaves a trail while the shutter is open, and P marks the north celestial pole, the point about which every trail turns. Facing north, the trails are arcs of circles around P. Facing east, the stars rise along slanted lines, and facing south their trails curve the other way, around the south celestial pole far below the horizon. However long the exposure, every star turns through the same angle about the pole, about 15° for each hour.

1.1.1A sphere of directions

Before you readMake a guess first

Two stars look close together in the sky. Are they close together in space?

When we look at the night sky, nothing tells us how far away the stars are. Each star is a point of light in a particular direction, and the sky looks like the inside of a dome resting on the . Astronomers keep this picture and complete it. Below the horizon the dome carries on under our feet, so the whole sky forms a sphere with the observer at its centre, called the celestial sphere. Each star, planet or galaxy is drawn at the point where the line from the observer towards it meets the sphere.

NoteA light year is the distance light travels in one year, about 9.46 × 10¹² km.

The size of the celestial sphere makes no difference, because only directions are drawn on it. Two stars at very different distances along nearly the same line of sight appear next to each other, as Dubhe and Merak do. For the same reason, separations on the sphere are never given in kilometres. The separation of two objects is the angle between the directions to them, measured at the observer, and we call it their angular distance.

Figure 1.2Directions on the celestial sphere

The observer O is at the centre of the celestial sphere. The stars X and Y are at different distances, but each is drawn where the line from O towards it meets the sphere, at X′ and Y′. Their angular distance

θ

is the angle X′OY′, which is also the arc X′Y′ (heavy line) of the great circle through the two points. A small circle is drawn for comparison; hidden parts of both circles are dashed.

Any plane through the centre of a sphere cuts it along a great circle, the largest circle that can be drawn on the sphere. Every other circle on the sphere is a small circle. Through any two points that are not exactly opposite each other there is just one great circle, and the shorter of its two arcs between the points is the shortest path between them on the sphere. The angular distance between two stars is that arc, measured as an angle. The same geometry explains why long flights follow great circles on Earth, as in the problem “Georgia to Georgia” in Section 1.4.

1.1.2Angles on the sky

Before you readMake a guess first

Hold your hand out at arm's length. How much of the full Moon does the tip of your little finger cover?

Angles on the sky are measured in degrees and, for small angles, in fractions of a degree. One arcminute, written 1′, is a sixtieth of a degree, and one arcsecond, written 1″, is a sixtieth of an arcminute. The full Moon is about 31′ across.

Your hand makes a rough protractor. Held at arm's length, the tip of the little finger covers about 1°, a clenched fist about 10°, and the spread hand from thumb to little finger about 20°. These are only rough guides, but they are good enough for finding your way around the sky and for the estimates of angles asked for in the observation round.

For calculations it is easier to measure angles in radians. An angle of one radian at the centre of a circle cuts off an arc as long as the radius, so a full turn of 360° is $2 π$ radians and one radian is 57.30°. If an object of size $D$ at a distance $d$ covers a small angle $θ$ , an arc of radius $d$ drawn across the object is almost exactly as long as the object itself. This gives the small-angle relation.

ResultSmall-angle relation(1.1)

θ = \frac{D}{d}

$θ$	angle covered by the object, in radians
$D$	size of the object, or the separation of two points
$d$	distance to the object, in the same unit as $D$

Valid when the angle is small. For angles of up to 5° the error is less than 0.1 per cent.

Show the derivation

Draw a circle of radius $d$ centred on the observer. By the definition of the radian, an arc of length $s$ on this circle covers an angle $s / d$ at the centre, and a small object is very nearly a short arc of this circle. To find the angle exactly, suppose the object stands at right angles to the line of sight, with its midpoint at distance $d$ . Each half of the object, of length $D / 2$ , makes a right-angled triangle with the line of sight, so $\tan (θ / 2) = D / (2 d)$ and

θ = 2 \arctan \frac{D}{2 d} .

For small $x$ , $\arctan x$ is very nearly $x$ , which gives $θ = D / d$ .

NoteUseful conversions: 1 rad = 57.30° = 3438′ = 206 265″.

Quick checkUse the result above

The Moon is 3474 km across and on average 384 400 km away. What is its angular diameter, in arcminutes?

arcminutes

1.1.3Near and far

The picture of a sphere centred on the observer has a weakness. People in different places are at different centres, and Earth carries all of us around the Sun once a year. If the directions to the stars changed noticeably as we moved, the constellations would change shape from place to place and through the year. In practice they hardly change at all, because the stars are so far away. The change in the direction to an object when the observer moves is called parallax. By (1.1) it is at most about the distance the observer moves divided by the distance to the object, and it reaches this largest value when the observer moves at right angles to the line of sight.

Example 1.1The nearest star from two sides of Earth's orbit

Proxima Centauri, the nearest star to the Sun, is $4.00 \times 10^{13}$ km away. In six months Earth moves to the far side of its orbit, $2.992 \times 10^{8}$ km from where it started. By how much, at most, does the direction to Proxima Centauri change between two observations six months apart?

Solution

The change is largest when the line joining the two positions of Earth is at right angles to the direction of the star. Then by (1.1)

θ = \frac{2.992 \times 10^{8}}{4.00 \times 10^{13}} = 7.48 \times 10^{- 6} rad,

and multiplying by 206 265 gives $θ = {1.54}^{″}$ .

AnswerAt most 1.54″, less than a thousandth of the width of the full Moon.

When astronomers give the parallax of a star, they mean half of this six-month change. That is the change for a baseline of 1 au, the astronomical unit, which is the average distance between Earth and the Sun, 1.496 × 10⁸ km. The parallax of Proxima Centauri is therefore 0.77″.

Every other star is further away, so its direction changes even less. Measuring these tiny shifts is how the distances of the nearest stars are found, as we will see later in the course. For describing where the stars appear, though, the shifts can be ignored, and the celestial sphere is the same for every observer. The Moon is the exception. Two observers near opposite sides of Earth, almost 12 700 km apart, can both see the Moon low in the sky, and they see it against the stars in directions that differ by almost 2°, more than three and a half times its own width.

1.1.4The turning sky

Before you readMake a guess first

Through what angle does the sky turn in one hour?

Earth turns on its axis from west to east once a day. Standing on Earth we do not feel this, and instead the whole sky seems to turn the other way, from east to west, carrying the Sun, the Moon and the stars with it. This daily turning is called diurnal motion. The sky turns about the line through Earth's axis, and that line meets the celestial sphere at two points, the north and south celestial poles. Midway between them lies the celestial equator, the great circle directly above Earth's equator.

Figure 1.3The celestial poles and the celestial equator

Earth at the centre of the celestial sphere. Its axis, extended, meets the sphere at the north and south celestial poles P and P′, and its equator, extended, meets the sphere along the celestial equator (dotted). As the sky turns, a star X moves around its daily circle (heavy line, hidden part dashed), which is parallel to the celestial equator. Earth is drawn far too large, since on the scale of the celestial sphere it would be smaller than a point.

Each star keeps a fixed angular distance from the poles, so as the sky turns it moves around a small circle centred on the pole, its daily circle. Stars on the celestial equator are the exception, since their daily circle is the celestial equator itself. Polaris is less than a degree from the north celestial pole, so it barely moves, while the trails of the other stars in Figure 1.1 circle around the pole close to it. There is no bright star near the south celestial pole. The nearest one visible to the naked eye, σ Octantis, is fainter than magnitude 5 and hard to see from a city.

NoteStar positions in this course are catalogue values for the year 2000. Earth's axis slowly changes direction, an effect called precession (Chapter 2), and the celestial poles drift among the stars by about a degree every 180 years. Polaris was 0.74° from the north celestial pole in 2000 and is 0.63° from it in 2026.

Earth turns once relative to the stars in one , 23 h 56 min 4 s, so each star comes back to the same place in our sky after that time. This is a little less than the 24 hours of an ordinary day, for a reason explained in Chapter 2. Dividing 360° by this time gives the rate at which the sky turns.

ResultRate at which the sky turns(1.2)

ω = \frac{360 °}{23.93 h} = 15.04 ° per hour

$ω$	angle through which the sky turns about the pole in one hour

The same number gives the rate in smaller units: 15.04′ per minute of time, and 15.04″ per second.

Quick checkUse the result above

In a photograph of the northern sky, the trails close to Polaris are arcs of 45° around the pole. For how many hours was the shutter open?

hours

Every star turns through the same angle about the pole, but the trails are not all the same length. A star close to the pole goes round a small circle, while a star on the celestial equator goes round the largest circle the sky has. If a star is at angular distance $p$ from the pole, its daily circle has a radius of $\sin p$ , in units of the radius of the celestial sphere, and the length of its trail follows from the length of an arc.

ResultLength of a star trail(1.3)

L = ω t \sin p

$L$	angular length of the trail
$ω t$	angle through which the sky turns in a time $t$ , from (1.2)
$p$	angular distance of the star from the celestial pole

On the celestial equator

p = 90 °

and the trail is as long as the angle turned. At the pole

p = 0

and the star does not move.

Show the derivation

Give the celestial sphere a radius of 1. A star at angular distance $p$ from the pole lies on a circle in a plane at right angles to the axis, and the radius of that circle is $\sin p$ . When the sky turns through an angle $ω t$ , the star moves along an arc of this circle of length $ω t \sin p$ . On a sphere of radius 1, a short length along the surface is equal to the angle it covers at the centre, so this is also the angular length of the trail.

Example 1.2Two trails in one exposure

A camera's shutter is open for two hours. Kochab, in the Little Bear, is 15.8° from the north celestial pole, and Mintaka, in Orion's belt, is almost exactly on the celestial equator. Through what angle does each star turn about the pole, and how long is each trail?

Solution

By (1.2) the sky turns through $2 \times 15.04 ° = 30.08 °$ in two hours, and both stars turn through this angle about the pole. By (1.3) the trail of Kochab is

L = 30.08 ° \times \sin 15.8 ° = 30.08 ° \times 0.272 = 8.2 °

long. Mintaka is 90.3° from the pole, so $\sin p$ is almost exactly 1 and its trail is 30.1° long, nearly four times as long as Kochab's.

AnswerBoth turn through 30.1° about the pole. The trails are 8.2° long for Kochab and 30.1° long for Mintaka.

Example 1.3A coin seen from far awayfinish it

In Example 1.1 the direction to Proxima Centauri changes by at most 1.54″ in six months. How far away would a coin 2.0 cm across have to be to cover the same angle?

The start

Example 1.1 found the angle in radians, $θ = 7.48 \times 10^{- 6}$ rad, before turning it into arcseconds. Rearranging (1.1) gives $d = D / θ$ .

Your stepFinish it: find

d

in kilometres.

Common errorsSection 1.1

Using degrees in (1.1). The relation gives $θ$ in radians. To get degrees multiply by $180 / π = 57.30$ , and to get arcseconds multiply by 206 265.
Confusing the angle a star turns about the pole with the length of its trail. In the same exposure every star turns through the same angle, but the trail is shorter by the factor $\sin p$ in (1.3).
Treating the celestial sphere as a real surface at some distance. It is a sphere of directions only, and stars that look close together on it can be at very different distances, as Dubhe and Merak show.

ExercisesSection 1.15 questions

1.1.1
Which of these is a great circle on the celestial sphere?
1.1.2
Seen from Earth, the Sun is 32′ across, and it is $1.496 \times 10^{8}$ km away. Use (1.1) to find the diameter of the Sun, in kilometres.
km
1.1.3
A clear night runs from 21:00 to 02:30. Through what angle does the sky turn about the pole in that time?
degrees
1.1.4
In a long exposure facing north from Ankara, the star trails are circles around a point. Where is that point?
1.1.5
How far away does a metre stick have to be to cover one arcsecond?
km

Olympiad problemsPast IOAA problems, restated2 problems

2020Astrophotography10 points

Adapted from the 2020 Global e-Competition on Astronomy and Astrophysics (Estonia, online), held in place of the IOAA that year, theory problem 1. Restated in our own words; the data are unchanged. Solution checked 14 October 2026.

An astrophotographer on the equator uses a digital camera on a fixed tripod, with no tracking. The lens has a focal length of 150 mm and an aperture of 84 mm, and the light-collecting width of the sensor is 22.5 mm. The camera points straight up, at a star field at the zenith.

(a) Find the field of view, the angular width of the sky that falls on the sensor. [2 points]

(b) The pixels on the sensor are 3.23 µm apart. What is the longest exposure for which no star trails appear in a single frame? [5 points]

(c) For a better image the photographer takes many frames, each with the exposure from (b), and stacks them. The frames together span 30 s. What fraction of the field of view appears in every frame, and so gets the full benefit of stacking? [3 points]

I solved it without the solution

Answer(a) 8.58°. (b) About 0.3 s. (c) 98.5% of the field of view.

2007The smallest shift2 points

Adapted from IOAA 2007 (Chiang Mai), theory problem 1.5. Restated in our own words; the data are unchanged, with values for the parsec and the astronomical unit added. Solution checked 14 October 2026.

An observer manages to measure the change in the direction to a star 100 parsecs away that is caused by Earth's motion around the Sun. What is the smallest angle, in arcseconds, that the observer must be able to measure? Take 1 parsec $= 3.086 \times 10^{16}$ m and 1 au $= 1.496 \times 10^{11}$ m.

I solved it without the solution

AnswerAbout 0.02″.