Problems for Chapter 2

Builds on: Sections 2.1 to 2.7
Reading time: 1 h
Exercises: 0
Olympiad problems: 4

The problems in this section come from past olympiads and use the ideas of the whole chapter, from sidereal time to calendars, together with the coordinates of Chapter 1. As in Section 1.7, they are not labelled with the sections they need, because recognising which idea a problem calls for is part of the skill that the exam tests. One of them comes from the data-analysis round, where the method and an honest estimate of the uncertainty earn marks as well as the answer. Each problem opens in place, with a hint and a full solution that can be revealed when needed. The first problem, on a full Moon over Bosscha Observatory in Indonesia, is worked through in detail below, because the number it seems to leave out is hidden in its wording.

2.8.1From textbook to olympiad

The full Moon over Bosscha, from IOAA 2008, gives the date and time of a full Moon and the position of the observatory, and asks for the shortest and the longest time the Moon could stay above the horizon that night. Read like a textbook exercise, it looks like a single use of the formula for rising and setting, but no declination is given, and the question asks for two answers instead of one. The method of Section S2 takes it apart, with its steps taken in the order this problem needs them and a check at the end.

We start by drawing it. Bosscha is 6°49′ south of the equator, so the south celestial pole stands 6.8° above its southern horizon (Section 1.4). The Moon, like a star, goes round that pole once a day, and the time it spends above the horizon depends on how much of its daily circle lies above the horizon. Figure 2.9 shows the circle.
Next we use every piece of information. The words full Moon carry the missing number. At full Moon the Moon is opposite the Sun in the sky, and on 19 June the Sun is close to the June solstice, at declination +23.4° and right ascension 6h. The Moon is therefore near right ascension 18h, and if it moved along the ecliptic it would be at declination −23.4°. Its orbit is tilted by 5.15° to the ecliptic (Section 1.6), so it can be up to 5.15° north or south of that point, and its declination can be anything from −18.3° to −28.6°. This range is the reason the problem asks for a shortest and a longest time. The time of full Moon, 00:30 West Indonesian Time, is 00:40 by the local mean time of the observatory, which is 2°35′ east of the 105° E meridian, so the Moon crosses the meridian close to midnight and the whole of its time above the horizon falls in one night. The elevation of 1300 m would lower the horizon by about 1.2° if the ground around were at sea level, but the observatory stands in the highlands north of Bandung, where the land around is high too, so we use the ordinary horizon, as the official solution does.
Then we name the physics. This is rising and setting. Ignoring refraction and the size of the Moon's disc, the Moon sets at the hour angle $H_{0}$ given by (1.21) with $h_{0} = 0$ , which is $\cos H_{0} = - \tan φ \tan δ$ . It rises at hour angle $- H_{0}$ , so it stays above the horizon while its hour angle grows by $2 H_{0}$ .
Now we work back from what is asked. With $φ = - 6.817 °$ and $δ = - 18.29 °$ , the formula gives $\cos H_{0} = - 0.0395$ and $H_{0} = 92.26 °$ . With $δ = - 28.59 °$ it gives $\cos H_{0} = - 0.0651$ and $H_{0} = 93.74 °$ . At 15° to the hour, $2 H_{0}$ is 12h 18m in the first case and 12h 30m in the second. These are the answers of the official solution.
Last, we check the answer. Both times are a little more than 12 hours, as they should be. From the southern hemisphere a body south of the celestial equator spends more than half of each turn above the horizon, and Bosscha is so close to the equator that the excess is small. A second check shows that the answer is not finished. A star's hour angle grows by 15.04° in an hour of ordinary time (Section 2.1), but the Moon moves eastward among the stars by 13.2° a day on average, one turn in 27.32 days. That takes about 0.55° an hour off, so the Moon's hour angle grows by only about 14.49° an hour. The official answers, which divide by 15° an hour, are too short, and the Moon is above the horizon for roughly 12h 45m to 13h.

The step that unlocks the problem is the second one, because the missing declination is hidden in the words full Moon and in the date. The last step matters as well. The marking scheme gave full marks for 12h 18m and 12h 30m, which treat the Moon as a star, but the Moon's own motion lengthens both times by nearly half an hour, more than the whole range the problem asks about. In the exam a sentence on what a model leaves out takes a minute to write, and it shows that we understand the model.

Figure 2.9The full Moon over Bosschainteractive

Latitude

φ

Declination

δ

The star
Upper culmination
Lower culmination
Time above the horizon

The figure shows the meridian diagram of Section 1.4 set for Bosscha Observatory, at latitude 6.8° S, with the Moon at declination −23.4°, where it would be if it moved along the ecliptic. A little more than half of the Moon's daily path, the heavy line, is above the horizon. The diagram can be set to the Moon or as the tilt of its orbit allows, or to , which is above the horizon for less than 12 hours. The time in the table counts 15° of hour angle to the hour, as the official solution does.

Olympiad problemsPast IOAA problems, restated4 problems

2008The full Moon over Bosscha20 points

Adapted from IOAA 2008 (Bandung), theory problem 3. We have restated it in our own words and kept the original data. The solution was last checked on 17 October 2026.

A full Moon occurred on 19 June 2008 at 00h 30m West Indonesian Time, the civil time of western Indonesia, which is UT+7 and is the mean time of the 105° E meridian. Find the shortest and the longest possible time that the Moon spends above the horizon on that night for an observer at Bosscha Observatory, at longitude 107°35′ E, latitude 6°49′ S and elevation 1300 m.

I solved it without the solution

AnswerThe Moon is above the horizon for at least about 12h 45m and at most about 13h. The official solution gives 12h 18m and 12h 30m, which leave out the Moon's motion among the stars. On that night the Moon was 4.0° south of the ecliptic, at declination −27.4°, and by its computed position it rose at 17:23 on 18 June and set at 06:17 on 19 June, West Indonesian Time, so it was above the horizon for 12h 54m.

2024A sundial10 points

Adapted from IOAA 2024 (Rio de Janeiro), theory problem 1. We have restated it in our own words, drawn our own sundial from a calculation and reworded statement (e), as the solution explains. The solution was last checked on 17 October 2026.

Figure 2.10 shows a horizontal sundial from above. Its gnomon is a flat triangle standing upright on the heavy line, with its tip directly above the point F, and the shadow of the tip falls on the dial. The sundial is either between the Tropic of Cancer and the Arctic Circle or between the Tropic of Capricorn and the Antarctic Circle. Each hour has a solid line and a dashed line, and the Sun's hour angles on the two lines for the same hour differ by the difference in longitude between the sundial and the central meridian of its time zone. Through the year the shadow of the tip of the gnomon always falls between curves A and C. Is each of the following statements true or false?

(a) The sundial works properly only in the southern hemisphere.

(b) Curve A is the path of the shadow of the tip of the gnomon on the day of the winter solstice, in the sundial's own hemisphere.

(d) The solid lines give local mean solar time.

(e) The figure of eight around the dashed line for 12 shows where the shadow of the tip falls at 12:00 zone time through the year.

Figure 2.10The sundial

The figure shows the dial seen from above. The gnomon stands on the heavy line, with its tip directly above F. The hour numbers of the dashed lines are smaller and grey.

I solved it without the solution

AnswerStatements (a), (c) and (e) are true, and (b) and (d) are false, as in the official answers. The original statement (e) described the figure of eight as the shadow at true solar noon on the central meridian, but at that moment the shadow lies on the dashed line for 12 itself. The figure of eight shows it at mean noon on the central meridian, which is 12:00 zone time, so we have reworded the statement, and its answer stays true. We drew the sundial for Christchurch in New Zealand (43.53° S, 172.64° E), whose standard time is the mean time of 180° E, so each dashed line is 7.36° of hour angle, or 29 minutes, from the solid line for the same hour.

2007The Moon on Einstein's birthday8 points

Adapted from IOAA 2007 (Chiang Mai), data analysis problem 2. We have restated it in our own words. The original gave photographs of the Moon taken at the same hour each night, and we have replaced them with drawings worked out from the Moon's computed position. The solution was last checked on 17 October 2026.

Thailand, on UT+7, celebrated the 60th anniversary of King Bhumibol Adulyadej's accession to the throne from 8 to 13 June 2006. Figure 2.11 shows the Moon at 23:00 Thai time on each of those nights, with the fraction of its disc that was lit. Take Albert Einstein to have been born at noon on 14 March 1879 in Germany, on UT+1. Use the figure to find the Moon's age, the number of days since new Moon, at his birth. Show the method in detail and estimate the uncertainty of the result.

Figure 2.11The Moon from 8 to 13 June 2006

The figure shows the Moon at 23:00 Thai time (16:00 UT) on each night from 8 to 13 June 2006, worked out from its position, with the lit part shaded and the lit fraction of the disc written under the date. West is on the right, as for an observer in Thailand facing south.

I solved it without the solution

AnswerThe Moon was about 20.7 days old, give or take a day, so it was a waning Moon a little before last quarter. The official solution, which takes full Moon at the start of 11 June, gives 20.9 days. The Moon's computed position puts the new Moon 21.3 days before the birth, inside our range.

2019The chimney at Tiszaújváros19 points

Adapted from IOAA 2019 (Keszthely), theory problem 6. We have restated it in our own words and kept the original data. We have left out part (b) of the original, which needs the satellite images, so parts (b) and (c) here are parts (c) and (d) of the original. The solution was last checked on 17 October 2026.

The third tallest structure in Hungary is the chimney of a power station near the town of Tiszaújváros. The two Sentinel-2 satellites photograph each area every few days, always at the same local time to within a few minutes, with a resolution of 10 m. One image of the chimney was taken on 29 June and another on 16 December, close to the two solstices. On the two images its shadow is 125 m and 780 m long.

(a) On which of the two dates do we expect the shadow to be longer? [1 point]

(b) Estimate the height of the chimney from the two lengths of its shadow. For this part only, assume that both images were taken at local noon. [16 points]

(c) Which of the following could affect the accuracy of the height found in (b)? More than one may be correct. (A) The flattened shape of Earth. (B) The limited resolution of the images and the blurred edge of the shadow. (C) The height of the base of the chimney above sea level. (D) Changes in the tilt of Earth's axis through the year. (E) Refraction in the atmosphere. [2 points]

I solved it without the solution

Answer(a) The shadow is longer on 16 December. (b) The chimney is about 230 m tall. (c) The answers are (B) and (E). The chimney is in fact 250 m tall. The result is very sensitive to the angle used, since a change of 1° in $α_{1} - α_{2}$ changes $h$ by about 26 m. With the full 46.88° the equation has no solution at all, because these two shadows can give a difference in altitude of at most 46.37°, when $h = \sqrt{x_{1} x_{2}} = 312$ m. The images were taken in the late morning, at about 09:45 UT, when by our calculation the Sun's altitudes were 63.1° and 18.0°. These differ by 45.0°, so the rounded value happens to suit the data.