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′.
- Since A and B are 90° apart round the limb, the path passes
from the Sun's centre. - Venus's angular radius is
seen from au, which is 0.50′. At the contacts Venus's centre is from O, so half the chord between them is , and the whole chord is 24.0′. - Against the Sun Venus moves at
, which is 4.02′ an hour. - The transit lasts
hours, so the last contact is at about 14:58 UT. - Venus also drifts north or south as it crosses, so its path is tilted 8.8° to the ecliptic and its speed along it is 4.07′ an hour, as in the worked example. With that speed the transit lasts 5.90 hours and ends at about 14:54 UT. The problem gives no inclination, so 14:58 UT is the answer the data allow.
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.