# Astrophysics Help

1. Oct 9, 2008

### Thoth

hey guys, im new here and thought you might like to help me out. im a first year astronomy student at Curtin University in Western Australia.

my current assignment has asked me to calculate the mass of hydrogen converted in the sun to helium and energy over 10 billion years (10 billion years being its lifespan on the HR diagram).

now, i understand peoples reluctance to help others who dont prove they have done any work themselves. which is lovely, considering youll now be reading all my own calculations and the result im getting!

my problem is that my amount of hydrogen burnt in 10 billion years seems to be only 14% of the mass of hydrogen in the sun now. shouldnt it be MUCH more than 14%? stars are supposed to burn of all of their hydrogen by the time they hit red giant phase.

without further ado, and recognising my use of the letter e in figures is scientific notation (10 to the power of) and nothing to do with natural logs.

FIRSTLY, the amount of seconds in 10 billion years.

60 x 60 = 3600 seconds per hour

3600 x 24 = 86400 seconds per day

86400 x 365 = 31536000 seconds per non leap year
86400 x 366 = 31622400 seconds per leap year.

since 10 billion years will consist of 75% non leap years and 25% leap years:

0.75 x 10 billion years = 7.5 e 9 years
0.25 x 10 billion years = 2.5 e 9 years

therefore:

(2.5e9 x 31622400) + (7.5e9 x 31536000) = 3.16224 e 17 seconds in 10 billion years

SECONDLY since the sun emits 3.84e26 joules per second of energy, the total power emitted over 10 billion years would be:

3.84 e 26 x 3.16224 e 17 = 1.214 e 44 watts (or joules per second) of power in 10 billion years we will need this figure later.

THIRDLY now that we have the total power output of the sun over 10 billion years we can use e=mc^2 to find out how much mass has been converted into energy in that time.

1.214 e 44 = m (3 e 8)^2

rearranging for m, gives m = 1.349 e 27 kilograms of mass converted to energy in the sun over 10 billion years

FOURTH so now we get onto the heavy stuff. knowing the masses of the hydrogen atom and helium atoms, and the reaction in the sun which converts the former to the latter + energy, i should be able to get a figure for the total hydrogen burnt. lets give it ago!

The fusion reaction summed up is 4 Hydrogen + 2electrons ----> 1 Helium + 2 neutrinos + 6 photons. Now, remember this is a summary of the 3 steps, but they arent important. only need to know the input and final output.

the mass of a hydrogen atom is 1.673 e -27
the mass of a helium atom is 6.647 e -27

now there are 4 hydrogens going into the reaction so:

4 x 1.673 e -27 = 6.692 e -27 the mass of hydrogen going into a single full reaction

therefore:

6.692 e -27 = (6.647 e -27 + X) with X being the amount of mass converted to energy per reaction.

rearranging for X gives X = 4.5 e -29 kg mass converted into energy per reaction.

FIFTH so if the total mass converted to energy over 10 billion years is 1.349 e 27, and we know now that the mass being converted to energy per full reaction is 4.5 e -29 then:

1.349 e 27 / 4.5 e -27 = 2.998 e 55 total reactions in 10 billion years

so if 6.692 e -27 kg of hydrogen goes into a single reaction, and there is 2.998 e 55 total reactions then:

6.692 e -27 x 2.998 e 55 = 2.006 e 29 kg of hydrogen converted into helium and energy in 10 billion years

so now you see my problem. if sun right now has 70% hydrogen, making that total hydrogen to be 1.393 e 30 kg of hydrogen in the sun right now, then how can the hydrogen burnt over 10 billion years be only one seventh (14%) of the mass of hydrogen right now?

2. Oct 9, 2008

### granpa

helium is denser then hydrogen. maybe the reaction rate increases as the sun becomes denser.

3. Oct 9, 2008

### Nabeshin

The sun will not burn up all its hydrogen by the time it turns into a red giant. The commonly quoted lifetime for the sun, 10 billion years, is for the sun as a Main Sequence star. Even after it leaves the MS, it will still have plenty of hydrogen.