If we took all the heavy water on Earth and used it for fusion, how long would it last if it were consumed at the current solar fusion rate?
How long would all the hydrogen on Earth last if it were used to power fusion at the rate our sun currently consumes hydrogen?
How much longer if we also used all of Earth’s helium as fuel?
Most of the energy released by the Sun is from the proton-proton chain. Although the reaction does breed and then fuse deuterium as part of the chain, it starts with plain single proton/no neutron hydrogen. Heavier stars are dominated by the CNO cycle, and don’t use deuterium hardly at all.
I don’t even have WAG for how long the hydrogen on and in the Earth could sustain fusion, but the amount of helium on Earth (occurring from natural alpha decay of heavy radioactive elements) is insignificant. You couldn’t power the Sun for a New York minute for the amount of helium found on Earth.
Stranger
I calculate that the hydrogen on earth could drive the sun’s p-p chain for 3000 years. No time right now to show my work, but you can probably work out my assumptions from what I typed into Google:
(0.009 * mass of earth) / (1 g / Avogadro’s number) / 4 * 26.7 MeV / 4e26 watts
3000 years seems way too long, since the Earth is roughly a million times smaller in volume than the Sun, only a tiny fraction of that is hydrogen, and the Sun will only last another ~5 billion years at which point it will have consumed a substantial fraction of its hydrogen.
Ok, thanks for the replies. You’ve given me some ideas on how to calculate this. The Earth is 4.6 billion years old and is middle aged. So let’s suppose it started out nearly all hydrogen and can burn for a total of 10 billion years. We just divide by 333,000 (the Sun / Earth mass ratio) and multiply by the Earth’s hydrogen mass ratio.
The Earth is roughly 1% water and 1/9th of water (by mass) is hydrogen. (Since hydrogen has an atomic weight near 1 and oxygen has an atomic weight of about 16 and water is h2o). So let’s figure hydrogen makes up 0.1% of the Earth’s mass. I am going to assume most of the hydrogen is tied up in water.
Therefore, 10 billion years divided by 333,000 times 0.1% = about 30 years.
Check your 1% figure for the Earth’s water. I estimate it is about 50x smaller than that. If the average depth of the ocean is d ~= 2 miles, the area ~70%, then the volume percentage is 30.7d/R, where R is the Earth’s radius. This yields a volume percentage of ~4/4000 or ~0.1%. Since the density of the Earth is about 5x that of water, I estimate that the weight percent of water is ~0.02%.
Of course the Earth’s crust also contains hydrogen in the form of hydrated rocks, hydrocarbons, etc, but I suspect the oceans still hold the vast majority of the Earth’s hydrogen atoms.
Ok, thanks. This site confirms the 0.02% figure. So that lowers the estimate to 0.6 years or about 7 months.
Addressing the question of how much total hydrogen we have, and how long that would last:
I see numbers that say 700 billion tons (that is 7 x 10^14 kg) of hydrogen are converted every second by the sun.
Hydrogen is 0.87% of the Earth’s mass (mostly found in compounds rather than elemental form) (cite). This is a higher number than just the amount of hydrogen in water. The Earth weighs 6 x 10^24 kg so it contains about 5 x 10^22 kg of hydrogen.
Dividing these things out gives 7 x 10^7 seconds of fuel. That’s still only 2.25 years to power the sun.
Interesting side note: The human body produces more heat per cubic metre than the sun.
I don’t have a cite, I saw this on QI. Can anyone verify?
Just found a cite: http://bencraven.org.uk/adam_and_eve.html
The wonders of statistics.
There are 2 popes per square kilometre in the Vatican.
Oh, no. This is how movies like The Matrix get started.
This NASA cite says 600 million tons. That brings your answer up to 2600 years.
This agrees with my completely different approach:
This leads to the Google calculation:
(0.009 * mass of earth) / (1 g / Avogadro’s number) / 4 * 26.7 MeV / 4e26 watts
= 2700 years.