- #1
hasan_researc
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I want to make an order-of-magnitude estimate of the Sun's present-day luminosity.
For this, I want to use just the following two pieces of information in a simple idealised model of (the mechanism of energy production in a star):
1) the mass of a helium nucleus is about 1 percent less than the mass of four hydrogen nuclei.
2) the Sun’s hydrogen-burning lifetime is believed to be 10 Gyr.
This is my attempt at the problem:
The Sun is passing through its mid-life at the moment.
So, I am assuming that half of the hydrogen nuclei have to converted to helium.
Therefore, (0.5)(mass of all H nuclie) + (0.5)(mass of all He nuclie) = (present) Mass of the Sun
"Mass of a helium nuclues = (0.99)(4*Mass of a hydrogen nucleus)" implies that the mass of all helium nuclei in the Sun is 0.8*1030 kg.
So, the intial mass of the Sun = (2*Present mass of the hydrogen nuclie) = 2.481030 kg.
So, total number of reactions (that will have occurred by the time the Sun dies) =
initial mass of Sun / 4 = 0.6*1030.
Energy release per reaction = (delta-m)(c2) = (4*(1-0.99))(c2)
= 6.0*10-12 J.
So, (average) luminosity = Total energy released / energy release per reaction = 3.3*1026 W.
I would be grateful if anyone could point out any flaws in the argument.
Also, I am not sure whether I have arrived at a reasonably good answer with incorrect methods.
For this, I want to use just the following two pieces of information in a simple idealised model of (the mechanism of energy production in a star):
1) the mass of a helium nucleus is about 1 percent less than the mass of four hydrogen nuclei.
2) the Sun’s hydrogen-burning lifetime is believed to be 10 Gyr.
This is my attempt at the problem:
The Sun is passing through its mid-life at the moment.
So, I am assuming that half of the hydrogen nuclei have to converted to helium.
Therefore, (0.5)(mass of all H nuclie) + (0.5)(mass of all He nuclie) = (present) Mass of the Sun
"Mass of a helium nuclues = (0.99)(4*Mass of a hydrogen nucleus)" implies that the mass of all helium nuclei in the Sun is 0.8*1030 kg.
So, the intial mass of the Sun = (2*Present mass of the hydrogen nuclie) = 2.481030 kg.
So, total number of reactions (that will have occurred by the time the Sun dies) =
initial mass of Sun / 4 = 0.6*1030.
Energy release per reaction = (delta-m)(c2) = (4*(1-0.99))(c2)
= 6.0*10-12 J.
So, (average) luminosity = Total energy released / energy release per reaction = 3.3*1026 W.
I would be grateful if anyone could point out any flaws in the argument.
Also, I am not sure whether I have arrived at a reasonably good answer with incorrect methods.