COSMOS AND UNIVERSES: The Final Theory of the Cosmos and the Universes by Balslev J
Author:Balslev, J. [Balslev, J.]
Language: eng
Format: epub
Published: 2020-05-03T16:00:00+00:00
M NS = 3.2 M ⊙ = 3.2 · 1.99×1030 kg = 6.37×1030 kg,
where, M ⊙ = 1.99×1030 kg, [120] is the solar mass. Since each neutron releases the energy E n = 1.49×10‒10 J and the mass of the neutron is m n = 1.67×10‒27 kg, [114] the total energy content of a neutron star must be equal to:
E (NS) = 6.37×1030 /1.67×10‒27 · 1.49×10‒10 J = 5.7×1047 J ≈ 3.6×1060 MeV,
as 1 J ≈ 6.242×1012 MeV. Some researchers set the density of the neutron star at the time of the explosion to at least 1.0×1018 kg/m3 . [86] So a cubic meter of such a neutron star must at least release the energy:
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