How long the sun would shine if it converts all it's Hydrogen to Helium

AI Thread Summary
The discussion revolves around calculating how long the sun will shine based on its hydrogen conversion to helium. Participants clarify that the mass of hydrogen in the sun and the rate of hydrogen conversion are the key factors for this calculation. Initial calculations suggested a lifespan of 7.85 years, which was deemed incorrect due to missing zeros. A more accurate approach involves considering that only about 10% of the sun's hydrogen will be used for fusion, leading to a revised estimate of approximately 7.85 billion years, aligning closely with current scientific estimates. The consensus emphasizes using mass rather than energy for a more straightforward solution.
b_o3
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Hello, I've got this problem:

The sun converts 1.91exp19 kg of Hydrogen to Helium per year, given that the sun started out with 75% Hydrogen by mass, computer how long the sun would shine if it converts all it's Hydrogen to Helium.

I computed the total energy of the sun as 3.81exp41 joules.

I tried subtracting the total energy of the sun from 75% of the energy from Hydrogen (computed using E=mc2) but that doesn't really add up. I computed the luminosity of the sun = 3.9exp26. I'm a little lost with this question can someone help me please.

thanks,
 
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sorry I forgot to mention, mass of sun = 2exp30 kg, radius of sun = 7exp8 m. ... should i use GM2/R to get energy of the sun?
 
I found total energy of sun contributed by Hydrogen = 2.14exp41 Joules...
 
Why are you worrying about the energy of the Sun?

You know the mass of the Sun
You know what percentage of that is hydrogen.
You know how much hydrogen is used up in a year.

That's all you need to know to solve this problem.
 
hows that going to work ?
 
u know the mass of Sun ,hence u know the mass of hydrogen in it.

u know the mass of hydrogen used every year ,so u know how many years u can use it!
 
b_o3 said:
hows that going to work ?

Simular problem:

Your gas tank holds 10 gals.
It is half full.
Your car uses 1/25 of a gal for every mile you drive.

How many miles can you drive on the gas in your tank?
 
so I found mass of H in sun = 1.5 exp30 kg, I divided that by rate of Hydrogen conversion which is equal to 1.91 exp 19 kg per year, and I got 7.85 years is this the correct answer?
 
so I found mass of H in sun = 1.5 exp30 kg, I divided that by rate of Hydrogen conversion which is equal to 1.91 exp 19 kg per year, and I got 7.85 years is this the correct answer?
 
  • #10
b_o3 said:
so I found mass of H in sun = 1.5 exp30 kg, I divided that by rate of Hydrogen conversion which is equal to 1.91 exp 19 kg per year, and I got 7.85 years is this the correct answer?

7.85 years?! Aren't you missing some zeros?
 
  • #11
but wait hows this right... the sun shines when there's energy, so I should probably use the total energy available to compute how long it will shine .. right? When I do that I get 3.93 exp 12

thanks,
 
  • #12
What does your intuition tell you about that answer? Will the sun really really run out of hydrogen in the year 2015?

Check your division. Also, the sun realistically won't burn all its hydrogen, but only about 10% of it since only the central regions get hot enough for fusion to occur.

Try 10% of 75% of the sun's mass divided by the mass it burns per year, and see if your answer is in rough agreement with the estimated main sequence lifetime of the sun from a source such as wikipedia.
 
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  • #13
b_o3 said:
but wait hows this right... the sun shines when there's energy, so I should probably use the total energy available to compute how long it will shine .. right? When I do that I get 3.93 exp 12

thanks,

No, stay away from energy. It's a simple as Janus' explanation.
 
  • #14
how would it run out of hydrogen in 2015... i said it was 3.93 exp 12. But now the problem is I am not sure whether to use the total energy of the sun to divided by rate of conversion..or should i use the total mass of Hydrogen
 
  • #15
okay cool then .. I got 7.85 exp 10 .. which seems about right
 
  • #16
a little optimistic, but within an order of magnitude or so of current estimates using much more sophisticated analysis.
 
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