Mirroring many late night caffeine-fueled arguments among Sci-Fi fans, a University of Leicester researcher asks the question:
U= 3GMp2
——
5Rp
Where G is the Gravitational Constant(6.673×10-11), Mp is planet mass, and Rp is the planet’s radius. Using Earth’s mass and radius, the required energy comes out to 2.25 x 1032 Joules. Using Jupiter’s data, the energy required goes up to 2 x 1036 Joules.
Boulderston asserts that (according toStar Wars lore) the Death Star is powered by a ‘hypermatter’ reactor, possessing the energy output of several main-sequence stars. Given that the power output of our Sun is about 3 x 1026 Joules per second, it’s a reasonable assumption the Death Star’s reactor could power the superlaser.
Despite using a simplified model of a planet, Boulderstone states the simplified model is reasonable to use since the Death Star’s main power reactor has the energy output equal to several main-sequence stars. Even if Earth’s exact composition were used in the equation above, the required energy to destroy a planet would only be affected by a few orders of magnitude – well within the Death Star’s power budget.
Boulderstone reiterated that the energy required to destroy a Jupiter-sized planet would put considerable strain on the Death Star. To destroy a planet like Jupiter, all power from essential systems and life support (no re-routing from the auxiliary EPS conduits – that’s aStar Trek hack!) would be required, which is not necessarily possible.
Boulderstone’s conclusion is that the Death Star could indeed destroy Earth-like planets, given its main power source. While the Death Star could destroy an Earth-sized planet, a Jupiter-sized planet would be a tough challenge, and the Galactic Empire would need to resort to using a Suncrusher to destroy stars.
If you’d like to read Boulderstone’s paper, you can access it at: physics.le.ac.uk
Source www.universetoday.com
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