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Old 20-June-2005, 08:29 PM
nottoc nottoc is offline
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Default Dropping meteors on planets

Phil's review of Revenge of the Sith talked about dropping meteors on planets. In few books by C.J. Cherryh she mentions the notion that a meteor weighing just a few pounds and travelling at 90% the speed of light would essentially render a planet uninhabitable (if not destroy it). When I read that I essentially just accepted the notion. Even a small object travelling at that speed would have a huge amount of energy.

But, now I'm wondering if it is a correct notion. Any thoughts?
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Old 20-June-2005, 08:32 PM
PatKelley PatKelley is offline
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Just calculate the amount of energy to get a few pounds of matter up to 90% of lightspeed, and you'll have a good picture of the energy range you are talking about.
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Old 20-June-2005, 09:14 PM
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gopher65 gopher65 is offline
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But how do you calculate that? I did it for an object travelling at non-relativistic velocities (IIRC, at 0.1c a 50kg mass would be 5 megatonnes, but that is just off the top of my head), but once you get up over ~.5-.6c that would be completely wrong.
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Old 20-June-2005, 09:50 PM
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Grey Grey is offline
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Quote:
Originally Posted by gopher65
But how do you calculate that? I did it for an object travelling at non-relativistic velocities (IIRC, at 0.1c a 50kg mass would be 5 megatonnes, but that is just off the top of my head), but once you get up over ~.5-.6c that would be completely wrong.
From here, so I don't have to format this equation using ASCII, the relativistic kinetic energy is equal to
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Old 20-June-2005, 10:04 PM
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Ilya Ilya is offline
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Quote:
Originally Posted by gopher65
But how do you calculate that? I did it for an object travelling at non-relativistic velocities (IIRC, at 0.1c a 50kg mass would be 5 megatonnes, but that is just off the top of my head), but once you get up over ~.5-.6c that would be completely wrong.
Relativistic kinetic energy is:

[1/sqrt(1 - v^2) - 1] * mc^2

where mc^2 is the object's mass-energy equivalence (9 * 10^16 joule/kg)

So, a 50 kg mass at 0.9c:

[1/sqrt(1 - 0.81) - 1] * 50 * 9 * 10^16 = 1.3 * 50 * 9 * 10^16 = 5.8 * 10^18 joule = 1,380 megatons

using conventional definition of megaton as 4.184 x 10^15 joules

If "few pounds" are 5 pounds, then KE is 138 megatons. Not enough to destroy a planet, or even a continent. Would flatten a medium-sized country, though.
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Old 21-June-2005, 01:30 AM
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cool thanks. I wrote it into a little program on my calculator so I won't forget. Errr.... just in case I ever have to led a planetary assault :roll:
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Old 22-June-2005, 06:56 PM
publiusr publiusr is offline
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Or using a non-nuclear method of digging a sea-level replacement for the Panama Canal--Shoemaker-Levy machine-gun style.
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