Weighing a binary from its photometry
- Round
- Theory
- Points
- 20
- Level
- Silver
- Taught in
- Section 8.9
Adapted from IOAA 2022 (Kutaisi), theory problem 6. We have restated it in our own words and kept the original data. The official solution prints the temperature of star A as 5978 K, a slip for 5796 K, which its later steps use. It also rounds the difference of magnitudes to 0.48 and the ratio of fluxes to 0.65, which gives masses of 11.4 and 29.3 solar masses, and names the larger mass M_B where it means M_A. The numbers do not describe a real binary, since stars of these masses and temperatures would be far larger and brighter. The solution was last checked on 23 October 2026.
A binary 89 pc away has a circular orbit seen edge on, with a period of 100 days. A space telescope of diameter
(a) Find the temperatures
(b) Find the distance
(c) Find the total mass of the two stars. (2 points)
(d) The table gives the combined light of both stars in two configurations, with the stars side by side and with B passing in front of A. The extinction in U towards the system is 1.4 magnitudes per kiloparsec, the bolometric correction is
| Configuration | U | U − B | B − V | BC |
|---|---|---|---|---|
| Side by side | 6.39 | 0.2 | 0.1 | 0.1 |
| B in front of A | 6.86 | 0.25 | 0.12 | 0.175 |
- (a) By Wien's law (4.11),
K and K. - (b) By (5.7), the telescope splits two points that are at least
rad apart, which is 9.16 milliarcseconds. This is the of (5.7), which we call here because is needed for the orbit. Let be the angle turned by the stars since they were furthest apart on the sky. They are then apart, which is less than for two spells of 19 days in each period, centred on and . In 19 days the stars turn through , so they merge at . - At that moment
. The distance is 89 pc, or au, so au and - (c) By Kepler's third law (8.1), with the period of 0.2738 years,
- (d) The extinction of
magnitudes is the same in both configurations, so it cancels in the difference of the bolometric magnitudes, which is all the problem needs. Since and , with side by side and in the transit, - So the transit lets through
of the light (4.2). Side by side, the light is proportional to (4.12). In the transit, B hides a disc of area of star A, which leaves . With and , so . - The masses are in the ratio of density times volume,
With the total of 40.7 solar masses, solar masses and solar masses.
Answer(a)
This problem also appears among the olympiad problems of Section 8.9, Problems for Chapter 8, which teaches the ideas it uses. Equation and figure numbers refer to the course.