[meteorite-list] Impact Question

From: John Hendry <pict_at_meteoritecentral.com>
Date: Mon, 25 Apr 2011 10:56:13 -0700
Message-ID: <C9DAE248.B982%pict_at_pict.co.uk>

Sterling,

On 24/04/2011 23:28, "Sterling K. Webb" <sterling_k_webb at sbcglobal.net>
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
Received on Mon 25 Apr 2011 01:56:13 PM PDT