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User:GiverOfThePeace/Standard Calculation for destroying Earth's Core: Difference between revisions
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Percentage reduction assumed to be 70% if not shown or stated otherwise | Percentage reduction assumed to be 70% if not shown or stated otherwise | ||
((7.6000000000000003e24cc) * (70%)) * (67.5j/cc) = 3.5910000000000001e26 Joules, 85.826959847036320639 Petatons of TNT. '''Multi-Continent level''' | ((7.6000000000000003e24cc) * (70%)) * (67.5j/cc) = '''3.5910000000000001e26 Joules''', '''85.826959847036320639 Petatons of TNT'''. '''Multi-Continent level''' | ||
[[Category:Calculations]] | [[Category:Calculations]] | ||
[[Category:Blog posts]] | [[Category:Blog posts]] | ||
Latest revision as of 15:37, 7 January 2024
Introduction
There's some verses where destroying a planet is done through destroying core causing a chain reaction, but destroying the core itself should net some results, so let's take a look.
Calculation
The Earth core's volume is 7.6e18 m^3 or 7.6000000000000003e24 cc.
The inner core is made of nickel and iron.
Iron's toughness value is 67.5 J/cc.
Formula is: ((Total Volume(V¹) * (Percentage reduction(V²)) * Fracture Toughness(F)
Percentage reduction assumed to be 70% if not shown or stated otherwise
((7.6000000000000003e24cc) * (70%)) * (67.5j/cc) = 3.5910000000000001e26 Joules, 85.826959847036320639 Petatons of TNT. Multi-Continent level
