1. Crudely describe where the center of mass is located relative to the two stars that make up a binary system: all other factors being equal, the centre of mass is located closer to the centre of the more massive star. 2. Star
Mass of Star 1
Position of Star 1
Mass of Star 2
Position of Star 2
System A System B System C
6 4 3
0 0 0
6 2 6
10 6 6
Position of Centre Of Mass 5 2 4
3. If a small O2 spectral type star is completely obscured by a large M4 spectral type star, the drop in total luminosity of the binary system would be best described as A 4. L= 4πR 2σT 4 σ = StefanBoltzmann constant: 5.67 × 10−8 W·m2·K4 Star
Luminosity (L ⊙)
Radius (R ⊙)
Temperature(T ⊙)
A B C
9 16 4
3 1 .5
1 2 2
5. A grandfather clock starts exactly at noon. Imagine that you make measurements of the height of the tip of the minute hand and create a periodic graph of your data (much like a light curve). If you make an observation at 2:40 pm, what would be the phase of this observation? Explain: 16/24: 0.66 The minute hand would be at 66% of it’s cycle. 6. Use the vertical red cursor to advance the system to a phase of 0.03. What is the physical significance of this phase? The smaller star has just eclipsed the larger and is beginning to move away as it does so, the flux value begins to rise as the larger star’s light is no longer obscured. 7. Which star is closer to the Earth during the deeper eclipse? Why? The cooler, 6000K star. The deeper eclipse is caused by the fact that this star is less luminous, and is fully eclipsing the brighter 8000K star by obscuring it from the perspective of the earth. 8. What is the depth of the secondary eclipse in visual flux? 0.78 ± 2 9. Kepler’s second law describes the behavior of planets in elliptical orbits around the sun. What behavior do you see here similar in nature to Kepler’s second
law? (If you increase the eccentricity to 0.6 the effect will be even more pronounced.) Kepler’s 2 nd law: the law of equal areas. In the area in which both stars orbits overlap, there is a noticeable acceleration in the speed at which the stars travel along their orbital vectors. Kepler’s second law states that regardless of their distance from the sun, planets must cover the same distance. In a highly eccentric orbit, this means that apparent velocity variations are most obvious at the planet’s apsides. This behaviour is similar because in kepler’s example, the planets are orbiting the sun as their centre of mass. In this instance, the stars are orbiting a centre of mass which is not the sun, however they obey the same gravitational laws. 10. Decrease the inclination from 90°. Estimate the inclination at which the eclipses become partial. 70.20°±0.5° 11. Continue to decrease the inclination. At what inclination do the eclipses cease to occur? 47.75°±0.5° 12. Decrease the inclination from 90°. Estimate the inclination at which the eclipses become partial. 80.60°±0.2° 13. Continue to decrease the inclination. At what inclination do the eclipses cease to occur? 80.50°±0.2° 14. Based on your experiences with Examples 7 and 8, can you summarize some general guidelines for what types of binary systems are likely to be eclipsing binaries? Yes. The closer to 90 in the inclination of the system orientation as viewed from our frame of reference, and the greater the differentiation in radius between the two stars, the higher the chances that the system will be an eclipsing binary. The smaller the separations between stars, The smaller the inclination required from earth to create an eclipsing boundary. Question 15: RT CrB – Adjust the radius of star 1. a) Best Value of Star 1 Radius: 2.4 solar radii b) Explanation of parameter effect on light curve: Both stars are of similar luminosity, because of this when one eclipses the other, the total luminosity falls by around 50%
Question 16: V478 Cyg – Adjust the inclination. a) Best Value of inclination: 80 degrees b) Explanation of parameter effect on light curve: Both stars are of similar luminosity, and so when one star eclipses 50% of the other, 75% of the sum of the combined star luminosity is visible. Overall apparent luminosity decreases by around 25% Question 17: V477 Cyg – Adjust the separation. a) Best Value of separation: 10 Solar radii b) Explanation of parameter effect on light curve: Star 1 has a significantly higher temperature and will therefore be more luminous. Therefore when Star 1 is obscured the dip in the light curve will be greater than when star 2 is obscured. Question 18: DI Her – Adjust the eccentricity. a) Best Value of eccentricity: 0.49 b) Explanation of parameter effect on light curve: Star 1 has a slightly higher temperature and mass than star 2, and therefore is more luminous. Due to the eccentricity of the system’s orbits, the eclipse of star 1 follows sooner after that of star 2 than it would if the stars had less eccentric orbits. When star 1 is eclipsed this decreases system luminosity more than when star 2 is eclipsed. Question 19: Ag Phe – Adjust the temperature of Star 2. a) Best Value of Star 2 Temperature : 5150K b) Explanation of parameter effect on light curve: Star 1 has a higher temperature, therefore when obscured there is a greater dip in system luminosity than when star 2 is eclipsed. However, star 2 is much smaller than star 1. This implies that when Star 2 blocks Star 1 it is not a total eclipse. Therefore the decrease in system luminosity caused by the eclipse of star 1 is not as great as it would have been if star 2 were able to completely remove star 1’s luminosity from the total system luminosity. Question 20: Rz Cas – Adjust the temperature of star 2 and the inclination a) Best Value of Star 2 temperature: 5000K b) Best Value of inclination = 82.40 Question 21: AF Gem – Adjust the temperature and radius of star 2. a) Best Value of Star 2 temperature: 6100K b) Best Value of Star 2 radius: 2.1 Solar radii