[meteorite-list] Impact Question

From: Sterling K. Webb <sterling_k_webb_at_meteoritecentral.com>
Date: Mon, 25 Apr 2011 18:32:52 -0500
Message-ID: <20D0B3B9A4704392B0752583EDFC0755_at_ATARIENGINE2>

John, List,

Well, I was calculating for rock, not iron. Rock
melts more easily, of course. My source (John S.
Lewis, Physics and Chemistry of the Solar System)
says it takes about 1.2 x 10^10 ergs to melt one
gram of "average" rock. Since a joule is 10^7 ergs,
it would take 1200 joules to melt a gram of rock,
or the kinetic energy per gram achieved at 1550 m/s.

OK, I used my calculator for that one, but when I was
clicking away at the Post, I didn't even use "back
of the envelope" figures, just the back of my head.
The figure was worse than approximate, more like
order-of-magnitude rough, just to be sure that
melt velocities were a fraction of impact velocities.

"Rock" covers a lot of ground, with a high variability
of melting points, different for each. Likewise, the
"boiling" or vapor point of rock tends to be niot greatly
about the melting point. Basalts melt at 1210 K.
and boil at 1450 K., for example.

The point is that the velocities associated with the
energy necessary to vaporize a rock or even an iron
are small compared to the velocities of the encounters.
Further, the bigger (and more massive) the incoming
object, the less effect the atmosphere will have on it.
If it's big enough, there is little or no retardation.

Partly, it's a matter of time. The time for atmospheric
heating of a 10 meter rock to commence and reach
high temperatures is very short, a few 100 milliseconds,
so retardation starts at high altitudes. The retardation
is called compressibility drag, "pushing" asgainst the
atmosphere. The ability of an object to overcome this
drag is proportional to its "loading," the mass per unit
cross section.

Whatever the loading of a 10 meter rock, the loading
of a 100 meter rock is ten times greater (1000 volumes
divided by 100 cross-sections); the loading of a 1000
meter rock, 100 times greater, a 10,000 meter rock,
1000 times greater.

To a Chicxuluber, the Earth's atmosphere is effectively
a vacuum that offers so little resistance as to not even
be mentioned in the footnotes. Not that there's time
to write footnotes.

A Chicxuluber transits the atmosphere in 2 to 4
seconds, and when the surface of impactor strikes
the surface of the Earth, its backside is still in the
stratosphere, moving at Full Tilt Boogie while the
frontside is stopping dead in its tracks.

A great deal of the converted kinetic energy goes
into the shock wave that starts at the contact and
flashes back into the still-oncoming rock. Shock waves
have no inherent limitations.

The stellar explosion we call a supernova is the result,
not of nuclear reactions, but a gravity-driven shock
wave (when the radiation pressure fails and the material
of the expanded star falls back in on the center).

(Answering other objections here...)

Unlike nuclear reactions which convert a small fraction
of a mass into energy, the conversion of kinetic energy
is purely mechanical. The term "conversion" may not
be the best, but that's the term they use in physics,
with the context of "transforming" energy from one
form to another: motion into heat which vaporizes
rock which in turn imparts kinetic energy to the rock
vapor molecules which expand, and so on.

We tend to think of nuclear reaction devices as the
ultimate weapons, of course, but imagine if you had
one of those railgun thingees and could speed up a
10 kilogram steel slug to 920 kilometers per second.
OK, I know that's really fast, but in a vacuum with lots
of energy on hand (about 1200 gigawatt-hours) and
a long enough railgun, all possible in orbit somewhere,
why not?

Unless my calculator is confused, that's a kinetic
energy very close to 4,183,999,999,994,176 joules,
or one megaton, stored in a 10 kg slug, a mass which
if it were plutonium (and standing still) could "only"
produce an explosion of 0.00002 megatons.

Potent stuff, kinetic energy.


Sterling K. Webb
----------------------------------------------------------------------------
----- Original Message -----
From: "John Hendry" <pict at pict.co.uk>
To: "'meteorite list'" <meteorite-list at meteoritecentral.com>
Cc: "Sterling K. Webb" <sterling_k_webb at sbcglobal.net>
Sent: Monday, April 25, 2011 12:56 PM
Subject: Re: [meteorite-list] Impact Question


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 aboveS

Assuming physical properties for say pure ironS

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 thisS
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 07:32:52 PM PDT