# [stars] Nuclear Fuel usage rate

At which rate do Stars burn there fuel, I know there are different stars (giants,dwarfs etc...)

For instance i read that our sun fuses 655million tons of Hydrogen into 650 million tons of Helium. The other 5 million is converted into 400 million watts of energy in the process. How did they get this figure for our star. And what is the rate for others, and how do they get the rate(s) for those stars? ## Answers and Replies

Garth
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Luminosity (power) of Sun = 3.8 1033 ergs/sec ~4 1023 kW

Mass required to be totally converted into energy per second = E/c2

Solar conversion mass/sec ~ 4 1033/1021 ~ 4 1012 gms/sec
~ 4 million tonnes/second

Garth

Chronos
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The sun converts about 600 million tons of hydrogen to 596 million tons of helium every second [the e/c^2 equivalent of 4 million tons of mass to energy every second]:

http://www.squ1.com/index.php?http://www.squ1.com/solar/the-sun.html
http://www.solarviews.com/eng/sun.htm

The basic way of computing this is to measure the average energy of sunlight striking the surface of the earth and factoring in the distance of earth to the sun. That allows you to estimate the energy output per square unit at the surface of the sun.

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SpaceTiger
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The reason we know how much hydrogen is being burned at the center is that we believe the sun to be in approximate equilibrium. To put it another way, we expect that the amount of energy lost per unit time (by radiation --> the luminosity) should equal the amount of energy generated per unit time (by nuclear fusion). If this weren't the case, the temperature and physical size of the sun would be changing rapidly (by cosmic standards) with time.

In reality, the sun can't be in exact equilibrium because its chemical composition is changing; that is, hydrogen is being converted to helium. As a result, there is a slight change in its size and temperature with time, but it's very slow, having changed its size by only about 20% in its lifetime.

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Garth
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Astronuc said:
Putting the mass consumption in perspective - it is in quasi-equilibrium, almost but not quite -

1.9891x1030 kg (~2x1030 kg)

or about (~2x1027 metric tons)

as compared to a consumption rate of hydrogen - 6x108 tons (English or metric? - I don't have time to do the math at the moment
Total solar mass 2x1030 kg being used up at 6 x1011 kg/sec will last ~ 3x1018 secs = 1011 years .
Actually the Sun will go Red Giant in about a tenth of this time, as only the core is available for fuel, so the total lifetime of the solar system as we know it will be 1010 years and the present age is 5x109 years, so we are half way through and have a little way to go!!

Garth

well, wow,Thanks guys. My maths classes won't be starting until the end of this month and i have lost all skills in algebra, so these equations are bit, out there for me. But i'm very sure this is the best way to explain these answers to me, i appreciate it.

Garth
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vincentm said:
My maths classes won't be starting until the end of this month and i have lost all skills in algebra, so these equations are bit, out there for me. But i'm very sure this is the best way to explain these answers to me, i appreciate it.
Well you did ask,
How did they get this figure for our star. And what is the rate for others, and how do they get the rate(s) for those stars?
For other stars you simply exchange the Sun's mass and luminosity for that of the star in question. Note: large stars that convect deep down onto the core will use a greater percentage of their overall mass as fuel and therefore last somewhat longer than otherwise. As it is, larger stars are much more luminous, and therefore have shorter lifetimes, than our Sun. (Thank goodness our Sun isn't larger!)

Garth

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One more question, how is it, in star birth (or when once a star is "born") that it has so much hydrogen to burn? I do know that the Helium once it has been converted from hydrogen (because it is done in such high numbers) that this helps fight the inward pull of gravity from the core of the star? why does the core have such a gravitational force? Forgive me, im just trying to figure out how stars work. I'm currently reading Joseph Silk's book on the big bang theory and i'm at the part of star formation. but posting here helps as well. ok that was more than one more question... Garth
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vincentm said:
One more question, how is it, in star birth (or when once a star is "born") that it has so much hydrogen to burn?
The 'cosmic mix', the recipe of relative abundances for the average present day content of the universe is, ~ 3/4 hydrogen, 1/4 helium and 2% everything else. You can play with the numbers a little but that is basically it. Of the 2% (the Earth & us!) the most common elements are Oxygen, Carbon and Nitrogen (about 1% of total) then Neon and then Iron and Silicon, (the Earth) and then the rest. That is where the Hydrogen came from - it was there is the first place and was the major constituent out of which the Sun was formed. Where did it all come from? When you do the nucleo-synthesis equations for the Big Bang you find it creates 3/4 hydrogen 1/4 helium and very little of anything else, it came from the BB!
I do know that the Helium once it has been converted from hydrogen (because it is done in such high numbers) that this helps fight the inward pull of gravity from the core of the star? why does the core have such a gravitational force? Forgive me, im just trying to figure out how stars work. I'm currently reading Joseph Silk's book on the big bang theory and i'm at the part of star formation. but posting here helps as well.
Asking questions is how you learn...
The gravitational force is so strong because the mass is so concentrated. Stars are fighting a balance between the immense pressures of dense plasma at a temperature of about 13,000,0000K and a gravitational vice that is threatening to squeeze them out of existence, literally into a black hole. What is it that prevents this collapse? The answer is the heat energy created by the nuclear fusion reactor in the stellar core. Once hydrogen is used up then the core contracts until helium fuses into beryllium and the other elements. Once they are used up - the end of the line being Iron then the star contracts into either a white dwarf, (up to 1.4 solar masses), or a neutron star (up to 2 ~3 solar masses) or a black hole.

I hope this helps, I expect you have just read all that in Silk's book!

Garth

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Thanks Garth, that helped alot.

Garth said:
Actually the Sun will go Red Giant in about a tenth of this time, as only the core is available for fuel, so the total lifetime of the solar system as we know it will be 1010 years and the present age is 5x109 years, so we are half way through and have a little way to go!!

Anyone interested in demonstating a calculation in how much more massive the sun would have to be in order to have reached 'Helium Flash' or the 'Red Giant' phase already?

Garth
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Orion1 said:

Anyone interested in demonstating a calculation in how much more massive the sun would have to be in order to have reached 'Helium Flash' or the 'Red Giant' phase already?
I tried to post the relevant equations in tex but it was all screwed up, nevertheless, the result is (and you can find the equations in any good undergrad astrophysics text book such as Carroll)
M = 1.31Msolar

Garth.

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Chronos
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Garth is well versed in these matters. He has my respect. Oh, I also agree with his comments [despite minor technical issues].

Solar Solution...

Mass-Luminosity relation:
$$\frac{t_1}{t_{\odot}} = \frac{m_1 L_{\odot}}{m_{\odot} L_1} = \frac{m_1}{m_1^{3.5}} = \frac{1}{m_1^{2.5}}$$

$$\frac{t_1}{t_\odot} = \frac{1}{m_1^{2.5}}$$

$$\boxed{m_s = \left( \frac{t_{\odot}}{t_1} \right)^{\frac{1}{2.5}} m_{\odot}}$$ $$t_{\odot} = 1 \cdot 10^{10} \; \text{years}$$ - Sol's lifetime
$$t_1 = 4.56 \cdot 10^9 \; \text{years}$$ - Sol's current age

$$\boxed{m_s = 1.369 \cdot m_{\odot}}$$ Last edited:
Garth
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Orion1 said:

Mass-Luminosity relation:
$$\frac{t_1}{t_{\odot}} = \frac{m_1 L_{\odot}}{m_{\odot} L_1} = \frac{m_1}{m_1^{3.5}} = \frac{1}{m_1^{2.5}}$$

$$\frac{t_1}{t_\odot} = \frac{1}{m_1^{2.5}}$$

$$\boxed{m_s = \left( \frac{t_{\odot}}{t_1} \right)^{\frac{1}{2.5}}}$$ $$t_{\odot} = 1 \cdot 10^{10} \; \text{years}$$ - Sol's lifetime
$$t_1 = 4.56 \cdot 10^9 \; \text{years}$$ - Sol's current age

$$\boxed{m_s = 1.369 \cdot m_{\odot}}$$ I worked it out for a star of just half the Sun's lifetime, exact estimates of our Sun's expected lifetime should be taken with caution.
Garth

Should 'Sol's current age', also be taken with equal caution?

Garth
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Orion1 said:

Should 'Sol's current age', also be taken with equal caution?
Well, your figure was the acepted age of the Earth from the age of the oldest meteorites and isotope ages of the oldest rocks - but those rocks are not the originals.
"Astrophysical Quantities" gives the Earth's age as 4.55 +/- 0.05 Gyr.
It gives the Sun's age simply as 5 Gyr. - a little older than the Earth. Solar models will also give an approximate age from the Sun's position across the Main Sequence line on the H-R diagram, but that also depends on the Sun's initial composition.

I think all we can be confident about is the first significant figure.

Garth

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Chronos
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Garth is being the consumate scientist allowing for the full error bars. The age of the earth is tightly constrained by a number of parameters [especially radioactive decay.] It is very logical to guess the sun is slightly older than earth, but not by more than about 10% [another fairly tight constraint].

SpaceTiger
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Garth said:
Once hydrogen is used up then the core contracts until helium fuses into beryllium and the other elements. Once they are used up - the end of the line being Iron then the star contracts into either a white dwarf, (up to 1.4 solar masses), or a neutron star (up to 2 ~3 solar masses) or a black hole.

Just a small thing -- iron is only the end of the line for massive stars. Stars like our sun will end up as white dwarfs composed mainly of carbon, nitrogen, and oxygen.

Solar Solution...

$$\boxed{m_s = \left( \frac{t_{\odot}}{t_1} \right)^{\frac{1}{2.5}} m_{\odot}}$$

$$t_{\odot} = 1 \cdot 10^{10} \; \text{years}$$ - Sol's lifetime (G2 V phase)
$$t_1 = 5 \cdot 10^9 \; \text{years}$$ - Sol's current age (beyond Terra's age)

$$\boxed{m_s = 1.319 \cdot m_{\odot}}$$

What is the duration of Sol's 'red giant' phase?

What is the duration of Sol's 'white dwarf' phase?

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Garth
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Orion1 said:
$$\boxed{m_s = \left( \frac{t_{\odot}}{t_1} \right)^{\frac{1}{2.5}} m_{\odot}}$$

$$t_{\odot} = 1 \cdot 10^{10} \; \text{years}$$ - Sol's lifetime (G2 V phase)
$$t_1 = 5 \cdot 10^9 \; \text{years}$$ - Sol's current age (beyond Terra's age)

$$\boxed{m_s = 1.319 \cdot m_{\odot}}$$
Concur
What is the duration of Sol's 'red giant' phase?
An intelligent 'guestimate' would be ~108 yrs.
What is the duration of Sol's 'white dwarf' phase?
White dwarf = 9 magnitudes fainter than Main Sequence progenitor; a factor of ~ 4,000 less luminous.
Thermal lifetime of Sun at present luminosity ~ 3.107 years (no nuclear fusion)
Lifetime as white dwarf at constant luminosity ~ 3.107 .4.103 ~1011 years (OOM)
Therefore it will exist as a typical WD for about 10 times longer than it will on the Main Sequence. It will actually fade into a red and then brown dwarf over the subsequent eons.

As SpaceTiger pointed out it will not ever reach to iron stage and may finish its life as a carbon star. What is carbon under tremendous temperature and pressure? ............"Lucy in the sky with diamonds"!!

Just a nice thought,

Garth

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SpaceTiger
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Most of what was said here is alright, but there's some potentially confusing terminology being used:

Garth said:
It will actually fade into a red and then brown dwarf over the subsequent eons.

The white dwarf will cool (Carroll & Ostlie derive the time-dependence) and its color will become more red, but the terms "red dwarf" and "brown dwarf" are reserved for objects of a different character. The former are just low-mass main sequence stars and the latter are even lower-mass stars that were unable to fuse hydrogen in their core. Unlike white dwarfs, brown dwarfs are only partially degenerate.

As SpaceTiger pointed out it will not ever reach to iron stage and may finish its life as a carbon star.

Again, a carbon star is quite a different beast. It refers to a red giant with a lot of carbon absorption in its spectrum.

Therefore it will exist as a typical WD for about 10 times longer than it will on the Main Sequence.

Finally, just a small comment on this. The time Garth calculated is a rough approximation of the characteristic cooling time for a white dwarf, but I wouldn't call it a "lifetime", since the white dwarf is still around afterwards. It's just cooler and harder to see.

Garth
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SpaceTiger said:
The white dwarf will cool (Carroll & Ostlie derive the time-dependence) and its color will become more red, but the terms "red dwarf" and "brown dwarf" are reserved for objects of a different character. The former are just low-mass main sequence stars and the latter are even lower-mass stars that were unable to fuse hydrogen in their core. Unlike white dwarfs, brown dwarfs are only partially degenerate. Again, a carbon star is quite a different beast. It refers to a red giant with a lot of carbon absorption in its spectrum.
Yes I was being undisciplined in my use of terminology, just playing with the term "white" dwarf and "carbon", I should have put it in inverted commas. You are correct, it is important not to confuse people.
Finally, just a small comment on this. The time Garth calculated is a rough approximation of the characteristic cooling time for a white dwarf, but I wouldn't call it a "lifetime", since the white dwarf is still around afterwards. It's just cooler and harder to see.
My point exactly.

Garth

SpaceTiger said:
Just a small thing -- iron is only the end of the line for massive stars. Stars like our sun will end up as white dwarfs composed mainly of carbon, nitrogen, and oxygen.
Sorry to resurrect this thread from the graveyard. But why is it that more massive stars face a different fate than stars like that of our sun? Does it have to do with the amount of Hydrogen that is being depleted? Or the amount of heavier elements being burned once it's main sequence of fuel is finished?