Distance to a near-Earth asteroid
- Round
- Theory
- Points
- 25
- Level
- Silver
- Taught in
- Section 1.7
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
and
-
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%.
This problem also appears among the olympiad problems of Section 1.7, Problems for Chapter 1, which teaches the ideas it uses. Equation and figure numbers refer to the course.