Nova Delphini 2013
- Round
- Data analysis
- Points
- 45
- Level
- Silver
- Taught in
- Section 8.6
Adapted from IOAA 2019 (Keszthely), data-analysis problem 1. We have restated it in our own words and kept the original data. The original asked for readings from a light curve and a set of spectra, and for a graph. We give the readings of the official solution instead and leave out those parts, so the points add up to 45 of the original 60. The solution was last checked on 21 October 2026.
The nova V339 Delphini was discovered on MJD 56518.584, where MJD is the Julian date minus 2 400 000.5. Its light curve peaked at visual magnitude
, by Della Valle and Livio (1995). , by Downes and Duerbeck (2000). , by the same authors.
The colour excess towards the nova is
| MJD | 56518.986 | 56519.813 | 56520.843 | 56521.835 | 56522.829 | 56523.827 |
|---|---|---|---|---|---|---|
| Wavelength (Å) | 6527 | 6531 | 6534 | 6537 | 6542 | 6544 |
(a) Find
(b) Find the extinction
(c) Find the distance of the nova and its uncertainty, in kpc. [11 points]
(d) Find the radial velocity of the shell from each spectrum. [14 points]
(e) Assuming that the shell was negligibly small at the first spectrum, find its radius at the last, in astronomical units. [7 points]
(f) Find the angular diameter of the shell at the last spectrum, about 5 days after discovery. [4 points]
- The three fits give
, and . Their mean is and their standard deviation, with , is 0.07. - By (4.14),
, with uncertainty . - The exponent is
, so pc. Its uncertainty is pc. - By the Doppler formula ((S.30)),
, which gives , , , , and km/s. The shell is slowing down. - The area under the speeds by the trapezium rule, over the 4.84 days between the first and last spectra, is
km, or 3.4 au. - The diameter, 6.8 au, at 3220 pc covers
.
Answer(a)
This problem also appears among the olympiad problems of Section 8.6, Interacting binaries and novae, which teaches the ideas it uses. Equation and figure numbers refer to the course.