From: Alejandro Dubrovsky (s335984@student.uq.edu.au)
Date: Thu Dec 14 2000 - 15:02:26 MST
* Spike Jones <spike66@attglobal.net> [001214 16:16]:
> So assuming a 1 km^2 frontal area, the absorbed energy
> into the shield is
>
> (.097)*(.5)*(6E-6 kg/km^2 sec) * (3E4 km/sec)^2 * (1km^2)
> =260 kg km^2/sec^2 = 2.6E8 newton meters.
>
> A calorie is about 4.2 nm, so 6.3E7 calories per second.
>
> If the ice shield starts at 20K and the collision energy
> warms it to 273K at about .5 calories per gram, then
> provides heat of fusion, 80 calories per gram, then at
> vacuum vaporization can occur at 273K at 540 cal/g,
> about 3800 calories per gram heat of dissociation when
> the water molecule is torn apart. Then there is some
> further energy absorption possible in ionization, but let
> us ignore that for now, for conservative calcs. That means
> the shield absorbs about 4600 calories per gram, or
> about 1100 joules per gram.
hmmmm.... i think there's something wrong there. 4600 calories are not 1100 joules, but 4600 x 4.2 J
= 19320 J (so that's one less order of magnitude in thickness)
(sorry, too late (8am) and no calculator handy to follow through the res fo the calculations, but it
should come up to about 0.12 cm / day)
>
> (2.6E8 joules per second) / (1100 joules / gram) = 240 kg/sec
> of the shield lost, so since we started with a square km
> of shield, then
>
> (240 kg/sec/km^2)* (m^3/900 kg) * (24*3600 sec/day)*
> (km^2/1E6 M^2 )=
> .023 meters per day, or 2.3 cm/day. For 45 years.
>
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