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    $\begingroup$ "Fast" and "10,000 ly away" don't really mesh. Let alone manipulating a black hole and using stars as ammunition after traveling 10,000 ly. This is a K2 civilization, not a K3 one. $\endgroup$ Commented Jul 13, 2022 at 20:53
  • $\begingroup$ @Yakk Yes, I acknowledged that this wasn't going to be fast. The reason I posted this answer is because the other answers either would not actually sterilize a planet, or would require the attackers to have a prolonged presence in the victim's star system without being intercepted, which is not feasible. A K2 civilization has enough energy to nudge celestial bodies, it will just be a tedious process. This is the only way to achieve what OP requested from a safe distance. $\endgroup$ Commented Jul 14, 2022 at 15:01
  • $\begingroup$ The weapon involves converting the mass-energy of a star into a beam and controlling said beam. That is 2*10^30 kg, which is 10^47 J. A K2 civilization, by definition, is limited to 4*10^26 W; the weapon you are using is 10^20 times the capabilities of the civilization in question, or (by definition) requires 10^20 seconds of effort to pull off. This is 10^12 years, or many times the lifespan of the universe. In short, your answer reduces to "become a K3 or K4 civilization and swat them". $\endgroup$ Commented Jul 15, 2022 at 1:07
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    $\begingroup$ The difficulty here is hand waved in the "control the black hole", "redirect a star". And you could argue that "having a weapon capable of producing X energy" isn't the same as controlling it. But my basic issue is that this looks like a K3 civilization (or above) weapon, moreso than a K2 weapon. $\endgroup$ Commented Jul 15, 2022 at 1:08
  • $\begingroup$ @Yakk You are not pulling an E=mc² on a star. You are only using enough energy to push it into a black hole at the right angle to spin the black hole towards the target planet. However, doing the math on it, I'll admit that you are still correct in that a K2 civilization cannot move the star on a timescale as short as a few millennia. $\endgroup$ Commented Jul 15, 2022 at 14:59