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Re: extinction5



Mr Graydon wrote at January 25:
'On Sat, Jan 25, 2003 at 05:48:25AM +0000, Simonyi scripsit:
> The energy of the extraterestrial object is
> E = m*ve2/2
> m = s*V
> where m - mass, e- exponent, v - velocity, s - specific gravity, V - volume.
> 
> If we want to count the density of the energy then 
> V = 1 m3
> If
> s = 2 g/cm3 = 2,000 kg/m3

This is a low value for specific gravity for a carbonaceous chondrite.

> v = 5,000 m/s

This is impossible.  It can't possibly be less than escape velocity, 11
km/s, minus the rotational velocity of the Earth. 5 km/s is half that
minimal velocity, which puts you more than a factor of four low.

Given that your specific gravity is a factor of two or more low, you're
an order of magnitude off in your energy density calculation.

> then
> E = 2*10e3*25*10e6/2 = 2.5*10e6  KJoule/m3
> 
> The density of the energy of the first A-Bomb was around 1 billion times
> bigger than the density of energy of the extraterestrial object.

Well, yes, but you're comparing apples to kumquats -- we don't really
care about energy density, we care about energy delivered.  That is what
has been calculated to exceed the energy delivered by setting off the
whole nuclear arsenal.'

Mr Cunningham wrote at January 25:
'Some calculations indicate that the impactor delivered a respectable
amount of energy that can be visualized approximately as follows.  The
Soviet Union manufactured the largest hydrogen bombs ever built.  If you
take 7 of these largest of the fusion bombs and strap them into a
bundle, then place duplicates of these bundles of 7 approximately 8
kilometers apart (4.75 miles for us illiterate types), all over the
surface of the earth (land and sea alike), then detonate them
simultaneously, that can be taken as a visual indicator of the amount of
energy delivered by the impactor.  It is enough energy to create a
fireball, and I'm grateful that the actual impact energy was
concentrated rather than dispersed in this manner.  And that much of the
energy was released immediately back into space rather than being
absorbed by the planet.'

Mr Marjanovic wrote at January 25(> sign is from my original letter :
'> MY ANSWER:
>
> Ir and NaCl in solid form and pure Cl2 atmosphere together isn't in any
> fireball or sea.

OK, but individual Ir and Cl atoms and ions are in such a fireball.

> Melted KNO3 isn't in the nature and for the oxidation of Ir (exclude
> Ir-soot) must this form.

Oh. -- But, again, individual Ir atoms flying through an atmosphere that
contains NO and NO2 should still oxidize.

> The energy of the first A-Bomb was equal 50,000 t TNT (tri-nitro-toluol).
> TNT can produce 1 Kcal/g or 1 million Kcal/t heat. This A-Bomb produced
> 5*10e4*10e6 = 5*10e10 Kcal = 2*10e11 KJoule
> (where e - exponent).
>  The volume of the source of nuclear energy was around 100 cm3.
> The density of the energy was 2*10e15 KJoule/m3
>
> The energy of the extraterestrial object is
> E = m*ve2/2
> m = s*V
> where m - mass, e- exponent, v - velocity, s - specific gravity, V -
volume.
>
> If we want to count the density of the energy

Mmm... no. We want to calculate the amount of energy, not its density, as
long as it's certain that the fireball consists of plasma = has a
temperature of around 10,000 °C or higher.

> then
> V = 1 m3
> If
> s = 2 g/cm3 = 2,000 kg/m3

Not more? Concrete has 2,200 kg/m³, glass has 2,500 kg/m³... a chondrite is
supposed to contain mainly silicates, so perhaps it's a good idea to assume
2,500 kg/m³.

> v = 5,000 m/s

Sounds like a good guess. It might be a bit low, though.

OK... if we give the meteorite a diameter of 12 km, then its volume is
around 7.24*10e12 m³ (or 7,240 km³), therefore its mass would be around
1.81*10e16 kg (or 18.096 Tt -- yes, teratonnes), so its total energy would
be... m*V²/2... 4.74*10e41 J... in other words, the energy of 2.37*10e27
Hiroshima bombs. I repeat: 2.37*10e27. No typo: 2.37*10e27. :-)

The energy from the first A-bomb was applied to the entire area of
Hiroshima. The incredible energy I've just calculated was applied to an area
of... let's unrealistically assume that the meteorite managed to touch the
ground with the entire surface of one of its hemispheres before it
evaporated... a circle that is 12 km in diameter has an area of almost
_226.2 km²_. (I don't know the size of Hiroshima, but I bet it is much, much
bigger.) A simple division shows that, under these simplistic assumptions,
the energy of nearly 1.048*10e24 Hiroshima bombs was applied to every square
kilometer of the impact site.

Therefore I claim that the energy density that resulted from the impact was
much bigger than the one that resulted from the nuking of Hiroshima.

> The density of the energy of the first A-Bomb was around 1 billion times
> bigger than the density of energy of the extraterestrial object. According
> this fact the temperature in the fireball produced an A-Bomb is much, much
> more higher than the temperature of the surface of the object. (And the
> temperature of H-Bomb is much, much more higher than A-Bomb!)

So what? The temperature only has to be 10,000 degrees. It need not be a
million.

> But not only this is different. The process of the explosion of an A-Bomb
> is produce from the nuclear energy first heat and from the heat produce
> other type of energy. The collision produce first different form of
> mechanical energy (deformation, cracking)

Where will this kinetic energy go when the movement is stopped so suddenly?
It will transform into the kinetic energy of individual atomic nuclei and
electrons -- it will become heat. LOTS of heat.

> and the remaind part of energy
> will be heat and from the heat will take off chemical reactions. This
cause
> a smaller temperature, too.
>
> Summerized this: The collision don't produce a fireball, a plasma with
> 10,000 centigrades or same.

Oh yes it does. Of course it does. Because the ENTIRE impact energy is
delivered to an area equal to the cross-sectional area of the meteorite. Why
do you think that not too small impacts produce craters that are much, much
bigger than themselves?!? Don't you know the story of the man who discovered
that the Meteor Crater in Arizona was an impact crater, spent years digging
for a huge mass of nickel, iron and other valuable metals, never found more
than a few tiny nuggets, and was told a few years before his death that
practically the entire meteorite must have _evaporated_? And we're talking
about a TINY crater here!

Funnily enough, in your calculations of the impact energy you have neglected
the impact (the sudden stopping of 18.960 Tt that moved at 5 km/s) itself.
:o)´'


My answer:

If the specific gravity is 4 g/cm3 (what is too much according Mr
Marjanovic) and the velocity is 11 km/s then the energy density of the
first A-Bomb is only (?) 150 million times bigger than the extraterrestrial
object.

Why is important the energy density and why isn't the total energy? The
temperature of the object determines of the energy density and isn't the
total energy.

A well known example: The energy of the sunshine don't cause any
catastrophe on sailboats. It was same at Roman fleet, too. But Mr
Archimedes could set Roman boats fire at Syracuse with sunshine. He didn't
give any energy. He increased only the energy density of the sunshine with
his optics.  

The maximum of the temperature in the first A-Bomb was around 2 million
Celsius therefore the maximum of the temperature of the extraterrestrial
object was around 3600 Celsius. (If the total kinetic energy of the object
converts to heat and this heat is heating the object only. What is quite
impossible! I will discuss with the amount of energy and processes of the
collision in an other letter.)

And Mr Marjanovic didn't count good because:
The volume of a sphere is 4*R^3*Pi/3 = D^3*Pi/6
where ^ - power.
If D = 12 km then V is around 900 km3 and isn't 7,200.
In the equation of the energy can find the v (velocity) isn't V )volume)
therefore the magnitude of the amount of the energy isn't 10e41 J but 10e22.