Help with Life expectancy of Main Sequence stars.

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If the nuclear fusion reaction of converting 4 H ! He occurs at an
efficiency of 0.7%, and that mass is converted into energy according
to the equation E = mc2, then estimate the Main Sequence lifetime
of the Sun (spectral type G2) in years if the luminosity of the Sun is
3.83×1033 ergs s−1. Assume the Sun’s core (10% of the total mass) is
converted from H into He. The Sun’s mass is M⊙ = 1.9891 × 1033 g.

t=1/M^2.5

t=1/(91.9891x10^32)^2.5
t= the wrong answer.

What are we doing wrong?
 
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I have no particular knowledge relating to your question. However, your expression for t leads to a t with dimension g^-2.5. Since t is supposed to be in years, I presume that there must be, at a minimum, a conversion constant of some sort.
 
That could be the wrong equation altogether...
 
The life expectancy of a main sequence star is inversely proportional to it's mass - i.e., large stars live fast and die hard, tiny brown dwarfs live dang near forever.
 
Assuming that this is just a homework assignment, what you must do is use Einstein’s equation to determine the amount of mass you get from 3.83×10^33 ergs/second(or rather 3.83 x 10^33 erg/s=mass x c^2, and solve for the mass). BTW, according to the value in the Wiki, this should be 3.85 x 10^33 ergs/sec, but its your homework :). Then divide the Sun’s core mass (which is described as 10% of the value you are given or .1989 x 10^33 grams) by this figure. This is how many seconds it takes to convert the core’s H into He. Finally, just convert seconds to years.
However, the Sun isn’t just going to fuse itself out of existence. It will eventually become a White Dwarf star and remain so for perhaps more than 10^100 years.
 
Last edited:
There's a nice "car" analogy to this problem. Your gas tank holds 20 gallons. You burn 2 gallons per hour. How long until you run out of gas? It's really the same question.
 
Fusion in small stars is a much more efficient process compared to large stars.
 

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