|Jun15-12, 07:03 PM||#1|
Pair-instability supernova calculation
Usual supernova 6 billion times brighter then ancestor star. Pair-instability supernova 100 times brighter then supernova 600 billion times brighter then ancestor star. If ancestor star apparent magnitude is 0 then apparent magnitude of pair-instability supernova from that star is = - 2.5* lg(6*10^11) = -29.5 Is my calculation correct?
|Jun15-12, 09:56 PM||#2|
Yes, it is correct. I assure you, but the concept is not
the apparent magnitude depends upon the distance also. You also need to compare the distance of the ancestor star from you with the distance of supernova from you. Something like this....
Let the distance from you to ancestor star be d, and distance from you to supernova be x.
Then the apparent magnitude will be -2.5*log((6*10^11)*d^2/x^2), since it is inversely proportional to the square of distance.
That is why you are getting such an unusually high magnitude. In fact, higher than the Sun. It is -26.7.
|Jun17-12, 12:36 AM||#3|
It's hard to imagine that star located several thousand light earth from solar system with so small apparent diameter just 0.00001'' can potentially create on Earth Mercury weather
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