NOVA bomb: Difference between revisions
→Theory: For comparative purposes
m (→Theory: the OP forgot to take into account the form in which the energy released by a nuclear detonation in space would take. In space, it would be more dangerous to biologicals and unshielded electronics.) |
(→Theory: For comparative purposes) |
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x/ (4 * 3.14195... * 5,000<sup>2</sup>) = 4 Megatons per square kilometer to fragment a 2km moon which requires 8 Megatons to be fragmented. | x/ (4 * 3.14195... * 5,000<sup>2</sup>) = 4 Megatons per square kilometer to fragment a 2km moon which requires 8 Megatons to be fragmented. | ||
x = 1.2 Petatons. | x = 1.2 Petatons - equivalent to 1.2 billion megatons | ||
Some may find this yield hard to believe for a fusion device, even one as large as the NOVA, but the stated effects to the planet and nearby moon require explosive power of this magnitude. Whitcomb does mention, however, that the lithium triteride cases are compressed to "neutron star density" during the detonation, implying that the warheads themselves boost a second, much larger fusion reaction. | Some may find this yield hard to believe for a fusion device, even one as large as the NOVA, but the stated effects to the planet and nearby moon require explosive power of this magnitude. Whitcomb does mention, however, that the lithium triteride cases are compressed to "neutron star density" during the detonation, implying that the warheads themselves boost a second, much larger fusion reaction. | ||