<p>Roughly <span class="math-container">$25,000$</span>% for main sequence nuclear fusion.</p>
<p>The internal energy of the Sun is <em>roughly</em>
<span class="math-container">$$U \sim \frac{3M_\odot k_B T}{m_p}\ , $$</span>
where <span class="math-container">$T \sim 10^7$</span> K is an average interior temperature and <span class="math-container">$2M_\odot/m_p$</span> is the number of protons and electrons in the Sun, assuming it is ionised hydrogen.</p>
<p>If we take that as the &quot;initial state&quot; that we have to reach to maintain sustained fusion, then we know that the Sun will produce a solar luminosity of power for <span class="math-container">$\tau \sim 10^{10}$</span> years, almost entirely produced by nuclear fusion.</p>
<p>Thus in your terms, the efficiency of main sequence nuclear fusion is
<span class="math-container">$$ E \sim \frac{L_\odot \tau m_p}{3M_\odot k_B T}\ .$$</span>
Putting the numbers in, I get <span class="math-container">$E \sim 250$</span> or <span class="math-container">$25,000$</span>%. This would be reduced by a factor of two if you consider the heat radiated during its initial gravitational contraction as &quot;wasted&quot;.</p>
<p>Of course you then have a shorter giant phase that produces much more luminosity that would increase this by another order of magnitude at least.</p>

CC BY-SA 4.0
2023-06-28T13:56:40.667Z
Thank you for the elegant math ProfRob...like PM 2Ring pointed out. To make the comparison is like comparing an apple to an orange. By the way the analysis above is mind staggering. To say the sun is a beast would be an understatement.
Sedumjoy
CC BY-SA 4.0
2023-06-28T07:37:26.020Z
2023-06-28T07:44:14.357Z
2023-06-28T07:44:14.357Z
ProfRob
<p>The Livermore fusion experiment was said to be 2 megajoules of energy in and 3 megajoules of energy out. However upon closer inspection the facility used 300 megajoules of energy. So man made uses 300 megajoules  and we get 3 megajoules. Can we say Input = 300, output =3.</p>
<p>Now the question is if it is possible to compare the sun to this process. We may need the sum of gravitational potential energy and  we know the power output at the core is about 300 watts per cubic meter The gravitational potential binding energy is ≈3GM2/5R≈2.3×10^41 joules. The other problem is I don't know the average power output. The 300 watts is the core. The other assumption I make is to use the gravitational binding energy to be the activation energy for the proton-proton fusion.</p>
<p>I am wondering if the sun does better than our 300 to 3 megajoules.</p>

CC BY-SA 4.0
2023-06-27T20:03:25.733Z
I don't think this notion of efficiency works well.  Efficiency seems best applied to consumers of energy, ie machines that are powered, but waste some of their power.  Human fusion efforts have not managed to break even, On the other hand, the sun clearly generates a lot of power The sun wins.  I don't think the binding energy is a useful comparison.
James K
CC BY-SA 4.0
2023-06-27T21:36:53.147Z
I don't see that there is an "input energy" with the sun, The core is compressed and heated under it's own gravity, not an external force. So the sun is 0:lots
James K
CC BY-SA 4.0
2023-06-27T21:40:23.073Z
Thank you.!  The word efficiency was not needed and causes confusion. You may also be right about the binding energy...I will need to think about that too.
Sedumjoy
CC BY-SA 4.0
2023-06-27T21:50:14.687Z
Humm...good point. The core is compressed and heated under it's own gravity. I think I was thinking gravity as a type of energyn but again I could be wrong here. I did make quite a few assumptions. I might be trying to compare a apple to an orange myself here.:)
Sedumjoy
CC BY-SA 4.0
2023-06-27T21:52:22.023Z
I will think about it for a while yet but may end up deleting the question once I straighten it out in my head.
Sedumjoy
CC BY-SA 4.0
2023-06-28T05:16:39.497Z
I mentioned the Sun's average power output per cubic metre in this answer to your older question: https://astronomy.stackexchange.com/a/51320/16685 but it's not really relevant, because fusion only happens in the core. IMHO, there's not much point comparing solar fusion to our fusion experiments because they use *very* different reactions.
PM 2Ring
CC BY-SA 4.0
2023-06-28T05:27:39.427Z
Gravitational collapse generates a *lot* of heat. But for fusion you need high temperature and high pressure. So the proto-sun had to *lose* a lot of that heat before the core got dense enough for p-p fusion to start.
PM 2Ring
CC BY-SA 4.0
2023-06-28T13:43:17.223Z
@PM 2Ring. I see your point now. Thank you.  The method between stellar fusion and man made fusion is so radically different that the question looses meaning...  Suffice it to say stellar fusion is exceptionally remarkable.
Sedumjoy
<star><nucleosynthesis><fusion>
Is there a way to compare input and output energy of solar fusion to the input and output energy of man made fusion?
Sedumjoy
# Is there a way to compare input and output energy of solar fusion to the input and output energy of man made fusion?

(Asked by: username_0 on 2023-06-27)

The Livermore fusion experiment was said to be 2 megajoules of energy in and 3 megajoules of energy out. However upon closer inspection the facility used 300 megajoules of energy. So man made uses 300 megajoules and we get 3 megajoules. Can we say Input = 300, output =3.

Now the question is if it is possible to compare the sun to this process. We may need the sum of gravitational potential energy and we know the power output at the core is about 300 watts per cubic meter The gravitational potential binding energy is ≈3GM2/5R≈2.3×10^41 joules. The other problem is I don't know the average power output. The 300 watts is the core. The other assumption I make is to use the gravitational binding energy to be the activation energy for the proton-proton fusion.

I am wondering if the sun does better than our 300 to 3 megajoules.

**Tags:** star, nucleosynthesis, fusion

> **Comments:**
> **username_1**: I don't think this notion of efficiency works well.  Efficiency seems best applied to consumers of energy, ie machines that are powered, but waste some of their power.  Human fusion efforts have not managed to break even, On the other hand, the sun clearly generates a lot of power The sun wins.  I don't think the binding energy is a useful comparison.  
> **username_1**: I don't see that there is an "input energy" with the sun, The core is compressed and heated under it's own gravity, not an external force. So the sun is 0:lots  
> **username_0**: Thank you.!  The word efficiency was not needed and causes confusion. You may also be right about the binding energy...I will need to think about that too.  
> **username_0**: Humm...good point. The core is compressed and heated under it's own gravity. I think I was thinking gravity as a type of energyn but again I could be wrong here. I did make quite a few assumptions. I might be trying to compare a apple to an orange myself here.:)  
> **username_0**: I will think about it for a while yet but may end up deleting the question once I straighten it out in my head.  
> **username_2**: I mentioned the Sun's average power output per cubic metre in this answer to your older question: https://astronomy.stackexchange.com/a/51320/16685 but it's not really relevant, because fusion only happens in the core. IMHO, there's not much point comparing solar fusion to our fusion experiments because they use *very* different reactions.  
> **username_2**: Gravitational collapse generates a *lot* of heat. But for fusion you need high temperature and high pressure. So the proto-sun had to *lose* a lot of that heat before the core got dense enough for p-p fusion to start.  
> **username_0**: @username_2. I see your point now. Thank you.  The method between stellar fusion and man made fusion is so radically different that the question looses meaning...  Suffice it to say stellar fusion is exceptionally remarkable.  

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## 1 answers

### Answer by username_3 on 2023-06-28. Score: 2 | Accepted answer

Roughly $25,000$% for main sequence nuclear fusion.

The internal energy of the Sun is *roughly*
$$U \sim \frac{3M\_\odot k\_B T}{m\_p}\ , $$
where $T \sim 10^7$ K is an average interior temperature and $2M\_\odot/m\_p$ is the number of protons and electrons in the Sun, assuming it is ionised hydrogen.

If we take that as the "initial state" that we have to reach to maintain sustained fusion, then we know that the Sun will produce a solar luminosity of power for $\tau \sim 10^{10}$ years, almost entirely produced by nuclear fusion.

Thus in your terms, the efficiency of main sequence nuclear fusion is
$$ E \sim \frac{L\_\odot \tau m\_p}{3M\_\odot k\_B T}\ .$$
Putting the numbers in, I get $E \sim 250$ or $25,000$%. This would be reduced by a factor of two if you consider the heat radiated during its initial gravitational contraction as "wasted".

Of course you then have a shorter giant phase that produces much more luminosity that would increase this by another order of magnitude at least.

> **Comments:**
> **username_0**: Thank you for the elegant math username_3...like PM username_2 pointed out. To make the comparison is like comparing an apple to an orange. By the way the analysis above is mind staggering. To say the sun is a beast would be an understatement.  

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