ok, so let's get some numbers first:
Delta Trianguli A: Age 8.5 Ga, Mass 1.0 Sols, Luminosity 1.5 Sols, Radius 1.2 Sols (G0 V).
Delta Trianguli B: Age 8.5 Ga, Mass 0.8 Sols, Luminosity 0.34 Sols, Radius 0.76 Sols (K2 V). (these are from my own stellar evolution tables)
Orbital separation = 0.11 AU, eccentricity = 0.02. ( http://www.solstation.com/stars2/del-tri2.htm
). Star1 = 0.0489 AU from barycentre, Star2 = 0.061 AU from barycentre.
Let's say your planet orbits the binary barycentre at a distance of 1.1 AU. It'd orbit the barycentre with a period of 316.67 days.
Since the planet orbits the pair, you need to calculate the temperature when the stars are at 90° (quadrature) to the planet and when the stars are in a line with the planet.
Let's assume for simplicity's sake that the planet doesn't move (the stars will be orbiting each other much more quickly, once every 10 days).
The planet starts with the stars lined up with it, with the primary closer. The planet is 1.051 AU from the G0 V, and 1.16 AU from the K2 V (which is eclipsed by the G0 V anyway). So it's just getting energy from the G0V, so it's basically a straightforward calculation with the first equation (L = 1.5, D = 1.051) so its blackbody temperature is about 300 K.
Five days later, the stars are in a line again but are reversed - the companion is the closer star to the planet. So here we use the first equation again (L = 0.34, D = 1.0389) and find that the planet's blackbody temperature is about 209 K. That's a big drop, but keep in mind that these eclipses are very short (not entirely sure how to calculate it, but it wouldn't be long - maybe an hour or so?). But also this assumes that the second star completely eclipses the first one, which in this case doesn't actually happen (if you calculate the angular diameter of each star, at this point the more distant primary would be 0.56 degrees and the closer companion would be 0.39 degrees, so some of the primary's radiation is still getting through and the temperature would be higher).
Between those times is when you'd use the second Flux equation. At quadrature (where the stars are at 90° to the planet), both stars are 1.011 AU from the planet. So the Flux from the primary star is (278.66^4) * (L/D²) = 601072003 * (1.5/1.212390123) = 7436616222. For the companion star, the Flux = 6010720039 * (0.34/1.213734568) = 1683765848. Add those together and you get 9120382070. Now the combined blackbody temperature would be (9120382070 ^ 0.25 = ) about 309 K.
The full temperature graph would look something like this:
graph.PNG [ 5.97 KiB | Viewed 71 times ]
Does that make any more sense? You really need a spreadsheet to calculate all this as you need to know where the stars are in their orbits etc. Not sure I can get into too much detail on it as I'm pretty rusty on all this now.
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