From: g_d_pusch_remove_underscores@xnet.com   
   Copy: rstokes1@san.rr.com   
      
   "Roger Stokes"  writes:   
      
   > "Gordon D. Pusch"  wrote in message   
   > news:giwu7tcjlq.fsf@pusch.xnet.com...   
   >...CLIP   
   >> ...An explanation of the physics of a 1kg matter-antimatter "bomb"   
   > detonation   
   >...CLIP   
   >   
   > I have read in other newsgroups of the concept of a "relativistic kill   
   > vehicle" that might be used in future space-based warfare. Disregarding   
   > whether or not the concept might make any military sense, has anyone   
   > calculated the effect of (say) a 1kg mass hitting the Earth's atmosphere   
   > at 0.9c, which would release about as much energy as a 1kg antimatter   
   > explosion?   
      
   The relativistic \gamma corresponding to 0.9 c is around 2.3, so the kinetic   
   energy of the 1 kg relativistic projectile would be 1.3 times its rest energy,   
   or about 1.2e17 joules. If I haven't botched the conversion, the kinetic yield   
   is around 28 megatons (1 megaton = 1e15 calories = 4.18e15 joules).   
      
      
   > Would the explosion occur in the upper atmosphere, or could the projectile   
   > make it to the ground before it was fully vaporized?   
      
   At 0.9 c, the kinetic energy of each baryon in the projectile is about   
   1.3 GeV, which is of the same order of magnitude as the average energy   
   of a cosmic ray at sea level (~2 GeV/baryon); conversely, the relativistic   
   projectile will see every atom of atmosphere that hits it as a "cosmic ray"   
   carrying the same energy per baryon. Since we observe cosmic rays with such   
   energies, they certainly do have some significant probability of penetrating   
   the atmosphere to reach the ground. Unfortunately, the mean range of such   
   particles are off the top of the graph in the Particle Data Book, but   
   extrapolating a bit, it appears to be somewhere between 2000 gm/cm^2   
   and 3000 gm/cm^2; the atmosphere provides about 6800 gm/cm^2 of shielding,   
   so one expects on the order of e^-3.4 to e^-2.3 of the particles to reach   
   the ground, i.e., 3% -- 10%. Hence, most of the energy will be deposited   
   in the atmosphere, but a non-negligible fraction of it will reach ground level.   
      
   >From the projectile's point of view, it sees the atmosphere as a blast of   
   cosmic rays, each baryon of which will deposit energy at a "rate" of about   
   1 MeV/(gm/cm^2). If we assume for simplicity that the projectile is a cube   
   made of iron (7.9 gm/cm^3), it will be about 5 cm on a side, its areal   
   density will be about 40 gm/cm^2, so each atmospheric atom will deposit   
   around 40 MeV of energy in the projectile, or about 0.3% of its total   
   kinetic energy. The total kinetic energy flux of the oncoming column   
   of atmosphere is about 2e19 joules, so 0.3% of it is around 6e16 joules,   
   or 14 megatons. I think we can safely assume the projectile will be   
   vaporized in penetrating the atmosphere... :-/  However, this will not   
   stop the particles it is made out of; as observed earlier, about 3%--10%   
   of its 1.3 GeV/baryon nuclei will make it to the ground, or 0.8--2.8   
   megatons of energy delivered as energetic charged particles. The remainder   
   will be deposited in the atmosphere above the target, but because of   
   relativistic aberration effects, much of that energy will be directed   
   downward, and will still reach the area around "ground zero."   
      
   The final question is how large a circle of ground will be incinerated.   
   relativistic aberration implies that most of the shower of particles   
   the projectile break up into will almost certainly be confined in a   
   cone of opening angle less than 1/(2\gamma) ~= 0.2 radians ~= 11 degrees.   
   As observed earlier, the "thickness" of the atmosphere is about three times   
   the mean range of the particles; since the "half range" of the particles   
   is about 1.44 times the mean range, which in turn is (very roughly) about   
   half the total "thichness" of the atmosphere, we expect about half the   
   nuclei to have hit something by up around the 0.5 bar level. The 0.5 bar   
   level is about 1.44 scale heights above the surface, or around 13 km   
   altitude (the Earth's scale height is around 9 km). Hence, the circle   
   of radioactive devastation is likely to be more than 3 km in radius.   
   Conversely, nearly all the atmosphere is below five scale heights,   
   so the circle of radioactive devastation is likely to be less than   
   10 km in radius.   
      
      
   -- Gordon D. Pusch   
      
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