AstroSpeedrunSearch
2010

A transit of Venus seen from the North Pole

Round
Theory
Points
10
Level
Silver
Taught in
Section 6.5

Adapted from IOAA 2010 (Beijing), theory problem S14. We have restated it in our own words and kept the original data, describing in words the path that the original shows in a figure. The official solution turns the chord into an angle round the Sun and uses the planets' angular speeds round the Sun, which is the same method, and gets 14h 57m. Its factor cos ε, never defined, is cos 3.4° for Venus's inclination. It corrects for the tilt of the path, but seen from Earth the path is tilted 8.8° to the ecliptic, not 3.4°. The official solution takes the Sun's angular diameter as 32.0′ and Earth's radius as 6378 km, which we state. The solution was last checked on 23 October 2026.

An observer near Earth's North Pole watches a transit of Venus. The path of Venus's centre across the Sun is a straight line. It crosses the Sun's limb at the points A and B, which are 90° apart as seen from the Sun's centre O. The first contact, when Venus's disc first touches the Sun's from outside, is at 9:00 UT. Find the time of the last contact. Take Venus's sidereal period as 224.70 days, Earth's as 365.25 days, the radius of Venus's orbit as 0.723 au, Venus's radius as 0.949 times Earth's radius of 6378 km, and the Sun's angular diameter as 32.0′.

AnswerWith the data given, the last contact comes about 5.97 hours after the first, at about 14:58 UT. Allowing for the tilt of Venus's path brings it to about 14:54 UT.

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.