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The power to destroy all atoms in the Milky Way!

TheUnshakableOne

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VS Battles
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Overall point of this thread: How much power is needed to destroy all the atoms in the Milky Way Galaxy.

Is it possible to calc this by finding all the atoms in all the stars in the Milky Way and then calc atomic destructions from there? Then tack on a "Likely slightly higher due to the calc not including all plantoids, and smaller objects in the Milky Way?"
 
Last edited:
This was my attempt at it:

The mass of the Milky Way is about 3e42 kg.

The most common isotopes on the Solar System are Hydrogen-1, Helium-4 and Oxygen-16. I’ll make the assumption that the Milky Way’s composition isn’t too different from this.

Hydrogen-1 is composed of only one proton, meaning it doesn’t add anything to this question.

Helium-4 contributes about 27.5% of the mass of the solar system. Extrapolating to the Milky Way this would be 8.25e41 kg. The mass of an Helium-4 atom is 4*(1.6e-27) kg. Dividing up this is 1.29e68 Helium-4 Atoms in the Galaxy.

Oxygen-16 contributes about 5.92% of the mass of the solar system. Extrapolating to the Milky Way this would be 1.78e41 kg. The mass of an Oxygen-16 atom is 16*(1.6e-27) kg. Dividing up this is 6.95e66 Oxygen-16 Atoms in the Galaxy.

The nuclear binding energy of Helium-4 is 28.3 MeV, while for Oxygen-16 it’s 123.7 MeV

1 MeV is 1.60218e-13 Joules.

(1.29e68*28.3*1.60218e-13) + (6.95e66*123.7*1.60218e-13) = 7.2264967e56 Joules

Very high-end Solar System level.
 
This was my attempt at it:

The mass of the Milky Way is about 3e42 kg.

The most common isotopes on the Solar System are Hydrogen-1, Helium-4 and Oxygen-16. I’ll make the assumption that the Milky Way’s composition isn’t too different from this.

Hydrogen-1 is composed of only one proton, meaning it doesn’t add anything to this question.

Helium-4 contributes about 27.5% of the mass of the solar system. Extrapolating to the Milky Way this would be 8.25e41 kg. The mass of an Helium-4 atom is 4*(1.6e-27) kg. Dividing up this is 1.29e68 Helium-4 Atoms in the Galaxy.

Oxygen-16 contributes about 5.92% of the mass of the solar system. Extrapolating to the Milky Way this would be 1.78e41 kg. The mass of an Oxygen-16 atom is 16*(1.6e-27) kg. Dividing up this is 6.95e66 Oxygen-16 Atoms in the Galaxy.

The nuclear binding energy of Helium-4 is 28.3 MeV, while for Oxygen-16 it’s 123.7 MeV

1 MeV is 1.60218e-13 Joules.

(1.29e68*28.3*1.60218e-13) + (6.95e66*123.7*1.60218e-13) = 7.2264967e56 Joules

Very high-end Solar System level.
wouldn't it be easier to just use all the stars in in the Milky Way?


If we know assume that all the other bodies in the Solar System (Jupiter, Earth, the other planets, the Moon, other moons, asteroids, etc.) are insignificant compared to the Sun we can approximate the number of atoms in the Solar System as 1.2 × 10^56.

Our galaxy, the Milky Way, contains approximately 100 to 400 billion stars. If we take this as 200 billion or 2 × 10^11 stars and assume that our sun is a reasonable average size we can calculate that our galaxy contains about (1.2 × 10^56) × (2 × 10^11) = 2.4 × 10^67 atoms
[1]

30852.2 (j/cc) [2] x 100000000000000000000000000000000000000000000000000000000000000000000 = 3.08522e+72????????????
 
I don't understand the last thing you did. The reason I used mass was because the wikipedia table had the isotopes sorted by mass fraction, so using mass was actually easier at least for me.
 
I don't understand the last thing you did. The reason I used mass was because the wikipedia table had the isotopes sorted by mass fraction, so using mass was actually easier at least for me.
The last thing I did was applying atomic destruction value on the calculation page from vswiki here is the link https://vsbattles.fandom.com/wiki/Calculations CTRL + F "Atomization" to find the "30852.2 (j/cc)"

The "100000000000000000000000000000000000000000000000000000000000000000000" is the standard form of 10^67 when i entered that into a calculator online.
 
I don't think you can use the atomization method like that. You need to multiply it by some volume because it's J/cc.

And also it's not atomization, it's sub-atomization.

What you say should work like this:

Volume of the Sun * sub-atomization Value * Number of Stars in the Galaxy = Energy

1.4e33 cc * 5.4e13 j/cc * 2e11 = 7.56e57 Joules

Which is pretty close to what I got anyway.
 
I don't think you can use the atomization method like that. You need to multiply it by some volume because it's J/cc.

And also it's not atomization, it's sub-atomization.
So there is different levels of atomization? For the verse in particular i am asking for they have 34+ statements of destroying atoms and i believe 1 statement of separating atoms apart.

However, the higher tiers could destroy stuff smaller than atoms
 
Separating atoms apart is Atomization, destroying atoms is Sub-Atomization. Sub-Atomization requires a looot more of energy than Atomization.

"Atomization: Applied only if clearly stated. It describes the energy to separate all atoms in a chemical substance. The value is 30852.2 (j/cc)."

"Subatomic Destruction: Applied only if clearly stated. It describes the energy necessary to destroy all atoms in a substance, by separating the particles in their nucleus. Note that Protons and Neutrons still stay intact. Value is 5.403E13 (j/cc)."
 
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