View Full Version : Dropping meteors on planets

nottoc

2005-Jun-20, 07:29 PM

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?

PatKelley

2005-Jun-20, 07:32 PM

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.

gopher65

2005-Jun-20, 08:14 PM

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.

Grey

2005-Jun-20, 08:50 PM

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 (http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html#c6), so I don't have to format this equation using ASCII, the relativistic kinetic energy is equal to

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/imgrel/rke2.gif

Ilya

2005-Jun-20, 09:04 PM

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 kg, then KE is 138 megatons. Not enough to destroy a planet, or even a continent. Would flatten a medium-sized country, though.

gopher65

2005-Jun-21, 12:30 AM

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:

publiusr

2005-Jun-22, 05:56 PM

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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