The highest full Moon
- Round
- Theory
- Points
- 10
- Level
- Silver
- Taught in
- Section 1.6
Adapted from IOAA 2013 (Volos), theory problem 11. We have restated it in our own words and kept the original data. The solution was last checked on 15 October 2026.
What is the greatest altitude at which the full Moon can be seen from Thessaloniki, at latitude 40°37′ N? Take into account as many effects as you can.
- A full Moon is opposite the Sun, so it is furthest north when the Sun is furthest south, near the December solstice. The full Moon is then close to ecliptic longitude 90°, where the ecliptic is 23.44° north of the equator.
- The Moon's orbit is tilted by 5.15° to the ecliptic on average. When its most northerly point falls at longitude 90°, which happens once every 18.6 years, the Moon's declination reaches
- By (1.12), seen from Earth's centre the full Moon would then culminate at
- The Moon is only about 60 Earth radii away, so an observer on the surface sees it lower than an observer at Earth's centre would. The shift is about
- Refraction lifts an object at an altitude of 78° by only about 0.2′, which is negligible.
AnswerWith the average tilt of the Moon's orbit, which is what the problem expects, the full Moon can reach an altitude of about 77.8° over Thessaloniki. The tilt itself varies between about 5.0° and 5.3°, and in this arrangement, with the Sun 90° from the points where the Moon's path crosses the ecliptic, it is close to its smallest. A real full Moon therefore reaches about 0.15° lower, about 77.6°. The official solution prints 79.8° before the correction for the Moon's distance, but its own sum,
This problem also appears among the olympiad problems of Section 1.6, Ecliptic and galactic coordinates, which teaches the ideas it uses. Equation and figure numbers refer to the course.