Energy to destroy matter of the observable universe

Introduction

We want to calculate the value required to destroy all regular matter (in other words, excluding antimatter, dark matter, dark energy and exotic matter) of the observable universe. To do this, we will be assuming a celestial object at the edge of the observable universe, and calculating the energy of a center-originating omni-directional blast, required to destroy said celestial object.

Energy Formulas used:

• GBE = 0.6 x G(M^2)/R
  • G = Universal gravitational constant
  • M is mass of the celestial object in kilograms
  • R is radius of object in metres
• Energy to destroy using a center-originating omni-directional blast = GBE x 4(π)(A/R)²

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

Low End

Assuming Sun at the edge of the observable universe.

Mass = 1.9888 x 10³⁰ kg

Radius = 6.955 x 10⁸ m

GBE = 0.6 x G(M²)/R =  0.6 x (6.67 x 10^-11) x (1.9888x10³⁰)²/(6.955x10⁸) = 2.2759x10⁴¹ J

Energy needed to destroy it (located at edge of observable universe) from center-originating omni-directional explosion, from center of universe = GBE x 4(π)(A/R)²

=> (2.2759x10⁴¹) x 4(π) x ((46.6x10⁹)(9.46x10¹⁵)/(6.955x10⁸))² = 1.149x10⁷⁸ J

High End

Assuming neutron star (this) at the edge of the observable universe.

Star in question

Mass  = 2.10 solar masses = 2.01 x 1.9888 x 10^30 kg = 3.997488x10^30 kg

Radius = 13 km (mean radius) = 13000 m

GBE = 0.6 x G(M²)/R =  0.6 x (6.67 x 10^-11) x (3.997488x10³⁰)²/(13000) = 4.92x10⁴⁶ J

Energy needed to destroy it (located at edge of observable universe) from center-originating omni-directional explosion, from center of universe = GBE x 4(π)(A/R)²

=> (4.92x10⁴⁶) x 4(π) x ((46.6x10⁹)(9.46x10¹⁵)/(13000))² = 7.11x10⁹² J

Conclusion

Energy required to destroy the observable universe:

• Low end: 1.149x10⁷⁸ Joules
• High end: 7.11x10⁹² Joules

Note: This is not the value required for Universe level. This is an intermediate value, an anecdote, for those interested.