Problems for Chapter 1
- Builds on
- Sections 1.1 to 1.6
- Reading time
- 1 h
- Exercises
- 0
- Olympiad problems
- 8
The problems in this section come from past olympiads and use the ideas of the whole chapter, from angles on the sky to the astronomical triangle. They are not labelled with the sections they need, because recognising which idea a problem calls for is a large part of the skill that the exam tests. They run roughly from shorter to longer. Each one opens in place, with a hint and a full solution that can be revealed when needed. A real attempt on paper before opening either teaches far more than reading the solution. Two of the problems, on the Large Magellanic Cloud and on a shipwreck, need the sidereal time at a given date and clock time, which Chapter 2 covers in full. Their solutions explain each step, and they can also be left until after Chapter 2. The first problem, Shadows, is worked through in detail below, to show how a problem that looks as if it is missing information can be cracked with the ideas of this chapter.
1.7.1From textbook to olympiad
Shadows, from IOAA 2016, gives the shortest and the longest shadow of a stick on one day and asks for the latitude and the Sun's declination. Read like a textbook exercise, it seems to contain a mistake. A shadow is longest at sunrise and sunset, when the Sun is on the horizon and the shadow is infinitely long, so how can the longest shadow be 5.671 m? The way through is the method of Section S2, taken one step at a time.
- We start by drawing it. The stick, its shadow and a ray of sunlight make a right-angled triangle, and the Sun's altitude
- Next we use every piece of information. The longest shadow has a finite length, so the Sun's least altitude that day is 10.0°, above the horizon. The Sun never sets. It is circumpolar (Section 1.4), and its least altitude is reached at lower culmination, below the pole, at midnight.
- Then we name the physics. A circumpolar Sun has two culminations, and Section 1.4 gives the altitude of each,
- Now we work back from what is asked. There are two unknowns and two equations,
- Last, we check the answer. At latitude 80° N a body is circumpolar if its declination is more than 10°, and the Sun at +20° is. This happens in late May or late July, in the far north, where the midnight Sun lasts for months.
The step that unlocks the problem is the second one. Every number in an olympiad problem is there for a reason, and when a fact seems impossible, it usually means that the situation is not the ordinary one. Figure 1.17 shows the Sun's path for this observer.
| The star | |
|---|---|
| Upper culmination | |
| Lower culmination | |
| Time above the horizon |
2016Shadows10 points
Adapted from IOAA 2016 (Bhubaneswar), theory problem 4. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.
An observer in the northern hemisphere finds that on a certain day the shortest shadow of a vertical stick 1.000 m tall is 1.732 m long, and the longest shadow of the same stick is 5.671 m long. Find the observer's latitude and the Sun's declination on that day. Treat the Sun as a point and ignore refraction.
- The Sun's altitude follows from
- The longest shadow is finite, so the Sun never sets. It is circumpolar, and its least altitude is at lower culmination. By (1.12) and (1.13),
- With
AnswerThe observer is at latitude 80° N, and the Sun's declination is +20°.
2008The Moon from two sides of Earth20 points
Adapted from IOAA 2008 (Bandung), theory problem 1. We have restated it in our own words and kept the original data, adding values for Earth's radius and the Moon's distance. The solution was last checked on 15 October 2026.
Two observers on Earth's equator, nearly 180° apart in longitude, observe the position of the Moon against the background stars at the same moment. The Moon's declination is zero. Sketch the situation, and find the difference between the right ascensions of the Moon that the two observers measure. Take Earth's radius to be 6371 km and the Moon's distance to be 384 400 km.
- The two observers are at opposite ends of a diameter of the equator, and the Moon is in the plane of the equator. Both can see it at once only if it is on both horizons, rising for one observer and setting for the other.
- Each observer is then 6371 km to one side of Earth's centre, at right angles to the line to the Moon, so each sees the Moon displaced against the stars by the angle
- The two directions differ by twice this angle, 1.899°, or 113.96′. Along the celestial equator 15′ of arc is 1 minute of right ascension, so the difference is
AnswerThe two right ascensions differ by about 7.6 minutes of time, which is 1.90° on the sky.
2017The Large Magellanic Cloud in Phuket10 points
Adapted from IOAA 2017 (Phuket), theory problem 1. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.
The Large Magellanic Cloud has right ascension 5h 24min and declination −70°00′. Phuket is at latitude 7°53′ N and longitude 98°24′ E, in the time zone UT+7. On what date of the year does the Large Magellanic Cloud culminate at 9 p.m. as seen from Phuket? The Greenwich sidereal time at 0h UT on 1 January is about 6h 43min.
- At upper culmination the local sidereal time equals the right ascension, 5h 24m. The cloud does rise at Phuket, since it culminates
- Phuket is 98°24′ east, which is 6h 33.6m, so at that moment the Greenwich sidereal time is
- 9 p.m. in Phuket is 14:00 UT. During 14 hours of ordinary time the sidereal time advances by
- Setting this equal to 22h 50.4m gives
- 31 days after 1 January is 1 February, when the cloud culminates about 3 minutes after 9 p.m., and on 2 February it culminates about 1 minute before.
AnswerThe Large Magellanic Cloud culminates at 9 p.m. in Phuket on about 2 February.
2012A sundial in Rio de Janeiro
Adapted from IOAA 2012 (Rio de Janeiro), theory problem 1. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.
Above the door of the dome of the 32 cm telescope at Brazil's National Observatory, in Rio de Janeiro (22°54′ S, 43°12′ W), there is a sundial facing north. Its dial lies in the vertical plane through the east point, the zenith and the west point, and its rod is parallel to Earth's axis. For which declinations of the Sun, and during which part of the year (months and seasons), does the sundial (i) fail to work for at least part of the day, and (ii) not work at all during the day?
- The dial faces north, so the Sun lights it, and the rod can cast a shadow on it, only while the Sun is north of the vertical plane through E, Z and W, the prime vertical (Section 1.2).
- The Sun rises north of east and sets north of west when its declination is positive, and south of them when it is negative (Section 1.3). So when
- The Sun crosses the meridian north of the zenith only if its declination is more than the declination of the zenith, which is the latitude, −22°54′ ((1.8)). If
Answer(i) The sundial fails near sunrise and sunset whenever the Sun's declination is negative, from the September equinox to the March equinox, in the southern spring and summer. (ii) It does not work at all when the declination is −22°54′ or less, for about three weeks around the December solstice, which is the southern summer solstice.
2017A synchronous satellite10 points
Adapted from IOAA 2017 (Phuket), theory problem 5. We have restated it in our own words and kept the original data, adding a value for Earth's radius. The solution was last checked on 15 October 2026.
A synchronous satellite goes round Earth in exactly the time Earth takes to turn once, at a height of 35 786 km above the surface. One such satellite is placed in an orbit tilted at 6.69° to the plane of the equator. Find the greatest altitude the satellite can have for an observer at latitude 51.49°. Ignore refraction, and take Earth's radius to be 6378 km.
- Seen from Earth's centre, the satellite stays within 6.69° of the plane of the equator. It is highest for the observer when it crosses the observer's meridian at its furthest north, 6.69° north of the equatorial plane.
- In the plane of that meridian, the angle at Earth's centre C between the observer O and the satellite S is
- Dropping a perpendicular from O onto the line CS splits the triangle COS into two right-angled triangles. The perpendicular has length
- The zenith distance of the satellite is the outside angle of the triangle at O, which equals the sum of the angles at C and S,
AnswerThe greatest altitude is 38.4°. Treating the satellite as if it were infinitely far away would give
2019South, east and north20 points
Adapted from IOAA 2019 (Keszthely), theory problem 10. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.
Treat Earth as a rigid sphere of radius 6378 km. From some points on its surface we can travel 6378 km due south, then 6378 km due east, then 6378 km due north, and arrive back where we started. Find all such points and the paths, give the geographic coordinates of the points where the path turns, and draw the paths. Paths that differ only by a turn about Earth's axis count as the same.
- A distance of 6378 km, one Earth radius, is an arc of 1 radian, or 57.296°, along a great circle.
- The southward and northward legs run along meridians and cancel in latitude, so the path ends at the starting latitude. It closes only if the eastward leg ends on the meridian where it began, or if the start is the North Pole, where all meridians meet.
- From the North Pole, the first leg reaches latitude
- Otherwise the eastward leg must go round its parallel a whole number of times
- Each larger
AnswerOne path starts at the North Pole and turns at 32°42′ N. Every other path starts in the southern hemisphere and runs south to a parallel near the South Pole, goes round that parallel a whole number of times and comes back north. Going round once, it starts at 23°33′ S and turns at 80°51′ S. Going round twice, it starts at 28°08′ S and turns at 85°26′ S, and going round three times, it starts at 29°40′ S and turns at 86°58′ S. The more times the path goes round, the closer its start comes to 32°42′ S.
2019Distance to a near-Earth asteroid25 points
Adapted from IOAA 2019 (Keszthely), theory problem 12. We have restated it in our own words and kept the original data, adding values for Earth's radius and the Moon's distance. The solution was last checked on 15 October 2026.
A near-Earth asteroid is observed at the same instant by two astronomers whose cities both lie at longitude 17° E, as it crosses their meridian. In Nagykanizsa, Hungary (46°27′ N), it is 25° south of the zenith, and in Windhoek, Namibia (22°34′ S), it is 45° north of the zenith. Both observers are at sea level.
(a) Draw a diagram of the situation. [5 points]
(b) Find the distance of the asteroid from Earth's centre, in Earth radii and in units of the Moon's mean distance. Use all the information given. Ignore refraction, and take Earth's radius to be 6378 km and the Moon's distance 384 400 km. [20 points]
- (a) The diagram is a section through Earth along the 17° E meridian. It shows Earth's centre C, the two observers N and W on its circle, the zenith direction at each, which continues the line from C, and the two lines of sight, 25° south of the zenith at N and 45° north of the zenith at W, which meet at the asteroid A far beyond the circle.
- (b) Seen from C, the observers are
- At N, the line towards W makes 55.49° with the downward vertical NC, so it makes
- By the sine rule for the flat triangle NWA,
- In the triangle CNA the angle at N is
- In units of the Moon's distance this is
AnswerThe asteroid is about 65.8 Earth radii from Earth's centre, about 1.09 times as far as the Moon. The angle at the asteroid is less than a degree, so an error of 0.1° in either zenith distance would change the answer by about 10 per cent.
2017Shipwreck40 points
Adapted from IOAA 2017 (Phuket), theory problem 12. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.
You are shipwrecked on an island. You still have a watch set to Bangkok time (UT+7), a compass, an atlas and a calculator. You wake to find that it has recently become dark, but the sky is cloudy. An hour or so later you see Orion through a gap in the clouds. You estimate that Rigel is about 52.5° above the horizon, and with the compass you find its azimuth, measured from north through east, to be 109°. Your watch reads 01:00 on 21 November 2017. You remember that the Greenwich sidereal time at 0h UT on 1 January 2017 was about 6h 43min, and that Rigel has right ascension 5h 15min and declination −8°11′.
(a) Find the local hour angle of Rigel. [10 points]
(b) Find the Greenwich sidereal time at that moment. [10 points]
(c) Find the longitude of the island. [5 points]
(d) Find the latitude of the island, to the nearest arcminute. [15 points]
- (a) The zenith distance is
- The azimuth of 109° puts Rigel east of the meridian, so its local hour angle is
- (b) 01:00 on 21 November in Bangkok is 18:00 UT on 20 November, 323 days after 1 January. In 323 days the sidereal time at 0h UT gains
- In the next 18 hours of ordinary time it advances by
- (c) The local sidereal time, after taking away 24h, is
- (d) The cosine rule for the side PX, as in (1.20), gives
- An expression
- As a check, (1.19) rearranged with
Answer(a) The local hour angle is 21h 37.8m, Rigel being 2h 22.2m east of the meridian. (b) The Greenwich sidereal time is 21h 59.4m. (c) The longitude is 73.3° E. (d) The latitude is 4°00′ N, which puts the island in the Maldives. The main official solution gives 21h 58m and 73.75° E because of a slip in adding up the sidereal time, but its alternative method gives 21h 59m 24.8s, as here.