- #1

constantinou1

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## Homework Statement

Given the following data, calculate the main sequence lifetime of the Sun (in years), assuming that all the initial mass is hydrogen and all of it is converted into helium.

Mass of the Sun = M = 2x10

^{30}kg

Luminosity of the Sun = L = 4x10

^{32}W

Energy released in the proton-proton reaction (4H [itex]\rightarrow He[/itex]) is E

_{fusion}= 4x10

^{-12}J

Comparing the mass of the final helium-4 atom with the masses of the four protons reveals that 0.7% of the mass of the original protons has been lost.

## Homework Equations

L = [itex]\frac{E}{t}[/itex]

where L is the stars Luminosity, E is the total energy supplied by hydrogen fusion and t is the stars main-sequence lifetime.

The main-sequence lifetime of a star is also given by:

t = [itex]\frac{fMc^2}{L}[/itex]

where f is fraction of of the stars mass converted into energy and c being the speed of light.

## The Attempt at a Solution

- First attempt...

Using the data given, I can just punch it into the equation above, which gave me:

t = [0.007*2x10

^{30}*(3x10

^{8})

^{2}] ÷ 4x10

^{26}

This gives me an answer of 3.15x10

^{18}seconds.

Converting this into years yields...

3.15x10

^{18}x [itex]\frac{1}{60*60*24*365}[/itex]

which gives 99.9x10

^{9}years.

This means the main sequence lifetime of the Sun is about 100 billion years, which can't be correct because in almost every textbook I look in, it states its main sequence lifetime should be 10 billion years. I know that I am by a factor of 10 out somewhere, but I can't seem to find where I have gone wrong.

- Second attempt...

The amount of fusion reactions that take place in the Sun in order to provide the necessary Luminosity is...

L ÷ E

_{fusion}= (4*10

^{26}) ÷ (4x10

^{-12}) = 10

^{38}s

^{-1}

Using the above, I can find the total mass of hydrogen converted into helium per unit time (using Wikipedia to find the mass of hydrogen nuclei)...

R = rate at which hydrogen is converted = 4(1.67x10^-27)(10^38) ≈ 670x10^9 kg s

^{-1}

I multiplied by 4 in the above because 4 hydrogen nuclei are required for 1 fusion reaction.

The lifetime of the Sun can now be found by dividing the mass of the Sun by the rate at which the hydrogen is converted to helium, so...

[itex]\frac{M}{R}[/itex] = 2x10

^{30}÷ 670x10

^{9}≈ 3x10

^{18}s.

Converting this into years yields...

3.1x10

^{18}x [itex]\frac{1}{60*60*24*365}[/itex]

which gives 99.9x10

^{9}years.

This is the exact same answer that I got in the first attempt! Either what I have done is correct or I have gone wrong in the two attempts somewhere, but I know that I must be going wrong somewhere because the answer is out by a factor of 10 and its quite frustrating now.

Thanks in advance for any replies and sorry for the bad layout of the 'attempt at a solution' section.