Saturday, March 24, 2012

Quick Estimation of Turbolaser Power

Last post I talked about photon torpedoes, and came up with a range (and a pretty wide one at that) for their max yield. This will go somewhat quicker.

The basis for comparison that I am using is the asteroid that is vaporized by a single turbolaser blast in the asteroid field. As in "not there" any longer--and this is a legitimate claim, just watch Empire Strikes back. Also note that this is well below their actual yield per shot--the bolt continues onwards after vaporizing the rock.


So, let's estimate the size. 20m diameter seems about right (if you think this is too much, please let me know). That gives a volume of 4200 m^3 (a little less, but not too much.) At irons density of 7.87 g/cm^3, this gives 33*10^6 kg of iron. Now, just like last time, I'm going to assume initial temperature of 11 Kelvin (as I did for the asteroid in Pegasus, to be wholly fair). In order to vaporize, it first must increase temperature by 1800 K, melt, then increase temperature by 1323 and then vaporize. Molar heat capacity is 25 J/(mol*K), heat of fusion is 13.8 kj/mol, heat of vaporization is 340kJ/mol. At 55.845 g/mole, we have 5.9*10^8 moles of iron. This means 2.7*10^13 Joules to increase temperature the first time, 2.0*10^13 to increase temperature the second time, 2*10^14 Joules to vaporize, 8*10^12 to melt, for a total of ~2.5*10^14 Joules/shot., .1 megaton.



But, as has been pointed out elsewhere, this is quite low--the asteroid in question is a diameter of 80 meters (using the Millennium Falcon as a point of reference), which would require 16000terrajoules of energy to vaporize--which is 40 megatons.)





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