## FANDOM

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I think there is a way to quantify the infamous Uni-Lord Solar System feat by Norrin Rad.

We can just use the Energy to destroy using a center-originating omni-directional blast which is: GBE x 4(A/R)^2

• R is radius of object to be destroyed
• A is distance from center point of blast

I will be assuming that the Solar System he destroyed is the same as our's since that's the one we know the most about.

GBE of the Sun = 5.693e41

Radius of the Sun = 6.957e8 metres

## High End

Assuming the SS was at the edge of the solar system the distance from center of blast then the distance would be 1 ly (the minimum radius of the Solar System's Hill sphere) or or 9.461e+15 metres.

5.693e41 * 4 * (9.461e+15/6.957e8)^2 = 4.2114462e+56 Joules or 4.2114462 TeraFoe

## Mid End

Assuming he was at Uranus's distance then that would be 2.871e12 metres.

5.693e41 * 4 * (2.871e12/6.957e8)^2 = 3.8781406e+49 Joules or 387.81406 KiloFoe

## Low End

Assuming he was at Mercury's distance then that would be 57.91e9 metres.

5.693e41 * 4 * (57.91e9/6.957e8)^2 = 1.5778449e+46 Joules or 157.78449 Foe

Thoughts? I would personally take Mid End, because Low End makes SS hundreds, even thousands, of times weaker than Thor (which isn't right) and High End is...it's fucking 4 trillion supernovae how much more do I have to say?

## EDIT: Proof of Uranus Distance

This scan shows that SS was 7 planets away from the star, so if assume that the Solar System was the same size as ours (which we have to since we have limited knowledge of other star systems) then Uranus would be the one to use.

I'll just post the calc again here.

GBE of the Sun = 5.693e41

Radius of the Sun = 6.957e8 metres

Distance of Uranus = 2.871e12 metres.

5.693e41 * 4 * (2.871e12/6.957e8)^2 = 3.8781406e+49 Joules or 387.81406 KiloFoe

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