Sterling, On 24/04/2011 23:28, "Sterling K. Webb" <[email protected]> wrote:
<snip> >It takes a little over a joule to melt a gram of rock; that's >the kinetic energy of that gram traveling at the sedate >velocity of a mere 2100 m/s. A good-sized, high-speed >impactor would turn to plasma with close to 100% >efficiency. <snip> I followed all but the above Assuming physical properties for say pure iron Specific Heat Capacity for iron = 460 J/kg/degK (http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html) Melting point of iron = 1530 deg Celcius = 1803 Kelvin (http://www.muggyweld.com/melting.html) Assuming incoming temperature of impactor is 200 Kelvin (http://en.wikipedia.org/wiki/Asteroid_belt) Then to raise the 1 gram impactor to its melting point requires a temperature increase of 1603 K and the energy required to do this should be roughly this 1603 x 0.001 x 460 = 737 Joules. So a typical value would be more like one *Kilojoule* to melt a gram of meteorite if I have my sums right (stone would be higher, maybe around twice as much as iron) Considered as kinetic energy, 1000 Joules would represent a velocity of sqrt[1000/(0.5*0.001)] = 1414 m/s which is ballpark consistent with your velocity estimate, but the energy you quote is a tad on the light side is it not? Regards, John ______________________________________________ Visit the Archives at http://www.meteoritecentral.com/mailing-list-archives.html Meteorite-list mailing list [email protected] http://six.pairlist.net/mailman/listinfo/meteorite-list
