The times of minima of V1107 Cassiopeiae
- Round
- Data analysis
- Points
- 25
- Level
- Silver
- Taught in
- Section D8
Adapted from IOAA 2011 (Katowice), data-analysis problem 1. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.
V1107 Cas is an eclipsing binary of the W Ursae Majoris type, whose two stars touch, so that its light changes all the time and it shows two nearly equal minima in each orbit. The table lists twelve observed times of minimum as heliocentric Julian dates, with their uncertainties.
| No. | Date | Time of minimum (HJD) | Uncertainty (day) |
|---|---|---|---|
| 1 | 22 December 2006 | 2 454 092.4111 | 0.0004 |
| 2 | 23 December 2006 | 2 454 092.5478 | 0.0002 |
| 3 | 23 September 2007 | 2 454 367.3284 | 0.0005 |
| 4 | 23 September 2007 | 2 454 367.4656 | 0.0005 |
| 5 | 15 October 2007 | 2 454 388.5175 | 0.0009 |
| 6 | 15 October 2007 | 2 454 388.6539 | 0.0011 |
| 7 | 26 August 2008 | 2 454 704.8561 | 0.0002 |
| 8 | 5 November 2008 | 2 454 776.4901 | 0.0007 |
| 9 | 3 January 2009 | 2 454 835.2734 | 0.0007 |
| 10 | 15 January 2009 | 2 454 847.3039 | 0.0004 |
| 11 | 15 January 2009 | 2 454 847.4412 | 0.0001 |
| 12 | 16 January 2009 | 2 454 847.5771 | 0.0004 |
(a) Find a first value of the period, assuming that it is constant and that the observations on any one night were continuous. (b) Using
- Five pairs of minima were observed on the same nights, numbers 1 and 2, 3 and 4, 5 and 6, 10 and 11, and 11 and 12. Their differences are 0.1367, 0.1372, 0.1364, 0.1373 and 0.1359 days, with a mean of 0.1367 days, which is half a period. So
days. - With
and days, the epoch of each minimum is rounded to the nearest half. The epochs are 0.0, 0.5, 1005.5, 1006.0, 1083.0, 1083.5, 2240.0, 2502.0, 2717.0, 2761.0, 2761.5 and 2762.0, and the values of O − C, in days, are 0.0000, 0.0000, 0.0136, 0.0141, 0.0142, 0.0139, 0.0290, 0.0322, 0.0345, 0.0354, 0.0360 and 0.0352. - The points lie along a straight line that starts near zero and rises by about 0.035 days over 2762 cycles. By least squares its slope is
days per cycle, uncertain by , and its intercept is 0.0004 days, uncertain by 0.0002. The points scatter about it by 0.0004 days, about as much as the uncertainties of the timings. - The period used was therefore too short by
days, so days and . - The window runs from HJD 2 455 806.2917 to 2 455 806.5833. The start of the window is 6268.47 cycles after the improved
, so the first minimum in it has and falls at HJD 2 455 806.2992. The next two, at = 6269.0 and 6269.5, fall at 2 455 806.4359 and 2 455 806.5726. - Each prediction is uncertain by about 1.2 minutes, from the uncertainties of
and . Taking the heliocentric dates as Universal Time ignores the heliocentric correction, which for this star is a few minutes.
AnswerThe improved ephemeris is
This problem also appears among the olympiad problems of Section D8, Light curves and other time series, which teaches the ideas it uses. Equation and figure numbers refer to the course.