Okay guys, here is the gist
DB's justification for the disk being at a Lagrange point still stands. 1,500,000 kilometers.
The disk had to be larger enough to eclipse the Earth. However, that is simply not enough. The shadow of an eclipse is called an Umbra, and it is only about 160 km
en.wikipedia.org
In order to do what this did, where the whole planet was put into darkness. We have to assume the umbra matched the diameter of the whole planet, and that the distance was the lagrange point instead of the distance from the Moon to Earth.
A person already engineered a way to calculate it:
He came up with this formula which can accurately calculate the size of an eclipse's umbra:
x = 3474 - 2[380,000Tan(0.25deg)]
Where 3474 is the diameter of the moon and 380,000 is the distance from Earth to the Moon and x is the umbra size
so I just have to plug in the new numbers. I will be generous, DB pointed out the planet is large, so I will use the largest rocky planet:
2.37x the diameter of Earth or 30,198.54 kilometers.
https://spacecalcs.com/calcs/lagrange-points/ = 4,211,935 km
And the distance I will replace 380,000 with 4,211,935
so: 30,198.54 = D - 2[4,211,935Tan(0.25deg)]
D = 30,198.54 + 2[4,211,935Tan(0.25deg)] =
66,954.84 kilometers I got ~25,832 km from Earth size but I am being generous here.
G1 established the thickness ratio was
= 0.0152505447, so it has a thickness of 1,021.10 kilometers.
I will still use stainless steel here rather than glass like Speedy is suggesting. It won't matter.
pi(66,954.84 km)^2 * 1,021.10 kilometers * 7.75 g/cm^3 = 1.11e28 kilograms
Now, let us assume the kinetic energy is equal to it crossing its diameter in one second, or 66,954.84 km/s
The relativistic kinetic energy calculator finds the kinetic energy of an object traveling at a speed similar to the speed of light.
www.omnicalculator.com
I get
2.59(10)^41 joules or 61.90 quettatons of tnt I am being generous here with the velocity involved, and frankly just assuming it created an umbra as large as the largest rocky planet and also calculating the lagrange point based on that.
There is no reason to believe, as long as we don't have any animated media of it (Which won't come out for years), that it exploded
that violently. We have no time frame, no clear image of the scale of its destruction, etc, etc. Violent fragmentation is still the best way to go without a proper timeframe or image of its destruction:
pi(66,954.84 km)^2 * 1,021.10 kilometers * 597 joules/cm^3 =
8.59(10)^30 joules or 2.05 zettatons of tnt There is no need for the Sun Disc to be as large as it stated. In fact, in the original calculation at 1.5 million km away a 700,000 km disc would take up way more space and appeal way larger in the scan we see. This makes way more sense. It's not ohio sized, but it's not star sized.
Going by this, Bardock should have won. The Solar Disc was grotesquely inflated in size allowing for relativistic kinetic energy to go brrrr and wank Omniman to almost supernova level. This works better in the context of the series for sure.