# Is there a way to compare input and output energy of solar fusion to the input and output energy of man made fusion?

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.

• 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. Jun 27 at 20:03
• 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 Jun 27 at 21:36
• 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. Jun 27 at 21:40
• 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.:) Jun 27 at 21:50
• 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. Jun 28 at 5:27

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".