The size of the asteroid Julietta
- Round
- Theory
- Points
- Not given
- Level
- Silver
- Taught in
- Section 6.5
Adapted from IOAA 2015 (Magelang), theory problem S3. We have restated it in our own words and kept the original data. The book gives only the weights 20, 25, 40 and 15 for the steps of its solution. The official solution takes the shadow's speed as Earth's speed minus Julietta's speed times the cosine of the angle at the Sun, 13.98 km/s, which gives 86.7 km. That leaves out the direction of the line of sight, and the speed across it is 9.79 km/s. Like the official solution, we ignore the rotation of Earth. The shadow moved west while Borobudur was carried east at 0.46 km/s, which would raise the size to about 64 km. The solution was last checked on 23 October 2026.
On 27 May 2015 the asteroid 1285 Julietta passed in front of the star δ Sagittarii, as seen from the Borobudur temple in Indonesia, which lay on the centre line of the asteroid's shadow. The star was hidden for 6.201 s. At the time Julietta was 3.076 au from the Sun and 2.156 au from Earth, near the far end of its orbit, whose semi-major axis is 2.9914 au. Take Earth's orbit as a circle and Julietta's orbit as lying in the plane of the ecliptic, with both bodies moving the same way round the Sun. Estimate Julietta's diameter.
- Earth moves at
- In the triangle of the Sun, Earth and Julietta, the law of cosines gives the angle at the Sun with
- Placing Earth at
- With Borobudur on the centre line, the chord is a diameter,
AnswerJulietta is about 61 km across.
This problem also appears among the olympiad problems of Section 6.5, Transits and occultations, which teaches the ideas it uses. Equation and figure numbers refer to the course.