What you can see from where you stand

Builds on: Sections 1.1 to 1.3
Reading time: 35 min
Exercises: 4
Olympiad problems: 2

From Hanoi on a clear February evening Canopus, the second brightest star in the night sky, can be seen about sixteen degrees above the southern horizon. From Beijing it never rises at all. Polaris is visible from both cities and barely moves during the night, while the stars of Orion rise in the east and set in the west in the same way as the Sun. All of these differences come from the way the sky is tilted for an observer at a particular latitude. In this section we work out that tilt and use it to predict when, and how high, a given star can be seen from a given place on Earth.

Figure 1.4The evening sky of 15 Februaryinteractive

Seen from

Local time20:40

Watch

Polaris
Canopus
Betelgeuse, in Orion

The whole sky from horizon to zenith, drawn from a star catalogue: every star down to magnitude 4, with the lines of a few constellations used in this section. North is at the top and east on the left, as on a star chart held overhead. Circles mark altitudes of 30° and 60°, and the dotted curve is the celestial equator. Move the time, or play the night, and watch Polaris stay put while Orion swings across the south. Then switch to Beijing, where Canopus never comes above the horizon.

1.4.1The altitude of the pole

Before you readMake a guess first

Seen from the North Pole, how high above the horizon is Polaris?

Consider first an observer standing at the North Pole. Earth's axis of rotation passes vertically through the observer, so the is at the , at an of 90°. For an observer on the equator the axis is horizontal and points towards the north point of the , so the pole has an altitude of 0°. Between these two cases the altitude of the pole changes in step with the of the observer.

NotePolaris is now within 0.7° of the north celestial pole, so its altitude gives the latitude to better than a degree. Navigators in the northern hemisphere used it in this way for centuries.

Figure 1.5 shows why the relation is exact. The zenith of an observer lies along the line from Earth's centre through the observer, and the angle this line makes with the plane of the equator is, by definition, the latitude $φ$ . The celestial pole is so distant that the direction to it is the same from every point on Earth, namely parallel to Earth's axis. The horizon plane is perpendicular to the zenith direction, and the axis is perpendicular to the equatorial plane, so the angle between the horizon and the direction of the pole equals the angle between the zenith and the equator.

ResultAltitude of the celestial pole(1.4)

h_{pole} = φ

$h_{pole}$	altitude of the celestial pole above the horizon
$φ$	latitude of the observer

In the southern hemisphere it is the south celestial pole that can be seen, at an altitude

| φ |

above the south point of the horizon.

Figure 1.5Why the pole stands at your latitude

A section through Earth along the observer's meridian. The observer O is at latitude

φ

. The line from the centre C through O points to the zenith Z, and the observer's horizon is perpendicular to it. Because the celestial pole is effectively at infinity, the direction to it from O is parallel to Earth's axis. The angle between the north point of the horizon and the pole is therefore also

φ

1.4.2The sky seen along the meridian

Each star moves once a day around a circle centred on the celestial pole. To find how much of this circle lies above the horizon it is convenient to look at the celestial sphere from the side, so that the plane of the is the plane of the page, as in Figure 1.6. In this view the horizon, the and the daily circle of every star are all seen edge-on and appear as straight lines. The daily circles are parallel to the equator and are displaced from it towards the pole by an amount that depends on the . The part of a star's line that lies above the horizon corresponds to the part of the day during which the star is up.

Figure 1.6The meridian diagraminteractive

Latitude

φ

Declination

δ

The star
Upper culmination
Lower culmination
Time above the horizon

The celestial sphere seen from the west and cut along the meridian. N and S are the north and south points of the horizon, Z is the zenith, P and P′ are the north and south celestial poles, and QQ′ is the celestial equator, drawn dotted. The heavy line is the daily path of a star of declination

δ

, with X and X′ marking upper and lower culmination. The region below the horizon is hatched. Places: , , , , . Stars: , , , , , .

The figure makes the two limiting cases easy to see. At latitude 0° both poles lie on the horizon, which then cuts every daily circle into two equal halves, so every star is above the horizon for twelve hours. At the North Pole the daily circles are parallel to the horizon and no star rises or sets. Canopus is a useful intermediate case, since its path moves from partly above the horizon at Hanoi to entirely below it at Beijing.

1.4.3Culmination

Before you readMake a guess first

From Hanoi ( $φ = 21 °$ N), how high does a star on the celestial equator climb?

A star reaches its greatest altitude when it crosses the meridian on the same side of the pole as the zenith. This is called upper culmination, or transit. At that moment the star and the zenith lie on the same great circle, so their angular separation is the difference between the latitude and the declination.

ResultAltitude at upper culmination(1.5)

h_{max} = 90 ° - | φ - δ |

$h_{max}$	greatest altitude of the star
$φ$	latitude, positive north and negative south
$δ$	declination, positive north and negative south

A star with

δ < φ

culminates south of the zenith, and one with

δ > φ

culminates north of it.

Quick checkUse the result above

How high does Sirius ( $δ = - 16.7 °$ ) climb above the southern horizon of Beijing ( $φ = 39.9 °$ N)?

degrees

Half a sidereal day later the star crosses the meridian again, on the far side of the pole. This is lower culmination.

ResultAltitude at lower culmination(1.6)

h_{min} = | φ + δ | - 90 °

A negative value means that lower culmination takes place below the horizon.

1.4.4Circumpolar stars

If a star is still above the horizon at lower culmination, it never sets, and it is called circumpolar. For an observer in the northern hemisphere, equation (1.6) gives $h_{min} \geq 0$ when $δ \geq 90 ° - φ$ . The same argument applied to the southern part of the sky shows that a star with $δ \leq - (90 ° - φ)$ never rises.

ResultCircumpolar and invisible stars, northern observer(1.7)

\begin{aligned} never sets: & δ \geq 90 ° - φ \\ never rises: & δ \leq - (90 ° - φ) \end{aligned}

In the southern hemisphere the two conditions exchange roles, with

| φ |

in place of

φ

From Hanoi ( $φ = 21.0 °$ ) the circumpolar stars are those north of declination $+ 69.0 °$ , and stars south of $- 69.0 °$ are never seen. For Beijing ( $φ = 39.9 °$ ) the corresponding limits are $\pm 50.1 °$ .

Quick checkUse the limits above

Which of these stars never sets as seen from Beijing?

Example 1.4Canopus from Hanoi and Beijing

Canopus has declination $δ = - 52.7 °$ . Decide whether it rises at Hanoi ( $φ = 21.0 °$ N) and at Beijing ( $φ = 39.9 °$ N), find its greatest altitude where it does rise, and find the most northerly latitude from which it can be seen.

Solution

At Beijing the lower limit in (1.7) is $- 50.1 °$ . Canopus lies further south than this, so it never rises there. At Hanoi the limit is $- 69.0 °$ , so Canopus does rise, and by (1.5) it culminates at

\begin{aligned} h_{max} & = 90 ° - | 21.0 ° - (- 52.7 °) | \\ = 16.3 °, \end{aligned}

south of the zenith because $δ < φ$ . The star can only just be seen where its upper culmination reaches the horizon, that is, where $90 ° - (φ + 52.7 °) = 0$ . This gives $φ = 37.3 °$ N. Atmospheric refraction raises objects near the horizon by a little over half a degree, which moves the limit to about 37.9° N.

AnswerRises at Hanoi, reaching 16.3° in the south. Never rises at Beijing. Visible south of about 37.3° N (37.9° N with refraction).

1.4.5Rising and setting

Before you readMake a guess first

A star on the celestial equator, seen from Beijing: how long is it above the horizon?

The time a star spends above the horizon is twice the $H_{0}$ at which it sets. In Section 1.5 we derive this hour angle from the astronomical triangle; here we simply quote the result.

ResultHour angle of rising and setting(1.8)

\cos H_{0} = - \tan φ \tan δ

$H_{0}$	hour angle at setting; the star rises at $- H_{0}$
$2 H_{0}$	time above the horizon, with 15° of hour angle equal to one sidereal hour

NoteA sidereal hour is shorter than an ordinary hour by about 0.3 per cent, or ten seconds. For rising and setting times this is usually smaller than other uncertainties, but it matters when times are needed to the minute over many hours (Chapter 2).

Equation (1.8) agrees with Figure 1.6. When $δ = 0$ it gives $H_{0} = 90 °$ at every latitude, so a star on the celestial equator is above the horizon for twelve hours. When $φ$ and $δ$ have the same sign, $\cos H_{0}$ is negative and $H_{0}$ is greater than 90°, so the star is up for more than half a day. If the right-hand side of (1.8) is less than $- 1$ the equation has no solution because the star never reaches the horizon and is circumpolar. If it is greater than $+ 1$ the star never rises.

Example 1.5Vega from Hanoi

For how long is Vega ( $δ = + 38.8 °$ ) above the horizon at Hanoi ( $φ = 21.0 °$ N)?

Solution

From (1.8),

\begin{aligned} \cos H_{0} & = - \tan 21.0 ° \tan 38.8 ° \\ = - (0.3839) (0.8040) = - 0.309, \end{aligned}

so $H_{0} = 108.0 ° = 7.20$ h, and Vega is above the horizon for $2 H_{0} = 14.4$ hours. As expected for a northern star seen from the northern hemisphere, this is more than half a day.

Answer14.4 hours, or 14 h 24 min.

Quick checkUse the result above

For how many hours is Sirius ( $δ = - 16.7 °$ ) above the horizon at Beijing ( $φ = 39.9 °$ N)?

hours

1.4.6The polar circles and the white nights

The Sun can be treated as a star whose declination changes slowly through the year, from $+ 23.44 °$ at the June solstice to $- 23.44 °$ at the December solstice. At the June solstice (1.7) shows that the Sun is circumpolar wherever $23.44 ° \geq 90 ° - φ$ , which is to say for $φ \geq 66.56 °$ . This is the latitude of the Arctic Circle, and it is in fact how the Arctic Circle is defined.

A related effect reaches much further south. Astronomical twilight ends when the Sun is 18° below the horizon. At the June solstice the altitude of the Sun at lower culmination is, by (1.6), $φ + 23.44 ° - 90 °$ , and this stays above $- 18 °$ at every latitude north of 48.56°. In Paris (48.9° N) and London (51.5° N), the sky around midsummer never becomes fully dark, even at midnight.

Common errorsSection 1.4

In Figure 1.6 equal lengths along a star's path do not correspond to equal times, because the daily circle is seen edge-on. The figure shows whether a star is above the horizon at some point in its circle. The time it spends there must be found from (1.8).
Sunrise and sunset are timed by the upper edge of the Sun, about 16′ above its centre, and refraction raises the Sun near the horizon by about 34′. At the moment of sunrise the centre of the Sun is therefore 34′ + 16′ = 50′ below the horizon, at an altitude of about $- 0.83 °$ , and the general formula of Section 1.5 must be used instead of (1.8). At latitudes between about 20° and 40° the correction makes the day between 7 and 10 minutes longer, and further from the equator it adds more.
For places in the southern hemisphere $φ$ is negative. Every equation in this section remains valid as long as the sign is kept, and most errors with southern observers come from dropping it.

ExercisesSection 1.44 questions

1.4.1
From Sydney ( $φ = - 33.9 °$ ), which of these stars never sets?
1.4.2
At which latitude is every star above the horizon for exactly twelve hours?
1.4.3
Antares has $δ = - 26.4 °$ . For how many hours is it above the horizon at Hanoi ( $φ = 21.0 °$ N)?
hours
1.4.4
What is the greatest altitude of Antares as seen from Hanoi?
degrees

Olympiad problemsPast IOAA problems, restated2 problems

2022Georgia to Georgia20 points

Adapted from IOAA 2022 (Kutaisi), theory problem 7. Restated in our own words; the data are unchanged, with Earth's radius added. Solution checked 14 October 2026.

An astronomer flies west overnight from Tbilisi, the capital of Georgia (41°43′ N, 41°48′ E), to Atlanta, the capital of the US state of Georgia, along the shortest possible route. The flight lasts 11 h 25 min at an average speed of 875 km/h. Throughout the flight she can see the star Furud ( $δ = - 30 ° 04^{'}$ ) from a window. At one moment it just touches the horizon, and at that moment it is exactly due south.

Find the latitude and longitude of Atlanta. Treat Earth as a sphere of radius 6378 km, and ignore Earth's rotation during the flight, the height of the aircraft, refraction and wind.

I solved it without the solution

Answer33°48′ N, 84°20′ W. The real Atlanta is at 33°45′ N, 84°23′ W.

2024Castaway20 points

Adapted from IOAA 2024 (Rio de Janeiro), theory problem 7. Restated in our own words; the data are unchanged. Solution checked 14 October 2026.

A sailor is shipwrecked on a small island in the southern hemisphere. His eyesight is poor, so he cannot use the stars, and he does not know the date. He does notice that the days are longer than the nights.

(a) On his first day the angle between the points on the horizon where the Sun rises and where it sets is 120°. Find the range of possible latitudes for the island. Ignore the change in the Sun's declination during a single day. [7 points]

(b) The angle grows a little every day, and 40 days later it is 163°. Estimate the latitude of the island, ignoring the eccentricity of Earth's orbit. [13 points]

I solved it without the solution

Answer(a) Anywhere south of 37.3° S. (b) About 50.5° S. The official solution takes $ε = 23.5 °$ and so gets 37.1° S and 50.4° S; the method is the same.