From: eripe.dk@gmail.com   
      
   On Friday, June 9, 2017 at 4:51:08 AM UTC+7, Mike Van Pelt wrote:   
   > In article ,   
   > 0something0   wrote:   
   > >The only problem with this is that the projectile is moving at   
   > >nearly the speed of light, so when you see it, its already   
   > >there, which means its going to take some handwavium to create   
   > >an early-warning system, whcih would involve FTL communications.   
   >   
   > Unless you're doing some sort of skiffy handwavitronics drive,   
   > the propulsion just about has to be coming from something   
   > external.  At .87c, the kinetic energy of the warhead is mc^2.   
   > I.e., the energy it would have if you had total conversion to   
   > turn it entirely into energy, or if it were pure antimatter.   
   >   
   > (Yeah, if it were antimatter, double the mass for the normal   
   > matter it anihilates with.  But then, around half the energy,   
   > more or less, escapes as neutrinos, so it's a wash.)   
   >   
   > Increase the velocity, and the kinetic energy of the warhead   
   > is some multiples of its mass times c^2.   
   >   
   > You can't carry that much energy.   
   >   
   > So, the propulsion energy just about has to be applied   
   > externally, from the launching site.  You might be able to see   
   > the laser/particle beam/whatever used to push the warhead up to   
   > near-c.  Accelerating it up to that velocity would take a while,   
   > and some of the propulsion energy is likely going to leak past   
   > the warhead, and run well ahead of the warhead, even if the   
   > warhead is going .999c when it arrives. (Think Murcheson's Eye.)   
   >   
   > (How fast would a 1-ton warhead need to be going for its   
   > Bragg peak to put it at the core of the planet when it stops?)   
   >   
   > --   
   > "The urge to save humanity is almost            | Mike Van Pelt   
   >   always a false front for the urge to rule."   | mvp at calweb.com   
   >                   -- H.L. Mencken               | KE6BVH   
      
   Good point!   
      
   And if you are building it as a sufficiently advanced gun, the initial mass of   
   the projectile dont really matter. You could use a small stone and just charge   
   it up with as much energy as you need to break the planet.   
      
   (Well there will be some ablation from the interstellar medium, I don't know   
   how much)   
      
   Another thing is that the energy emmitted by the projectile will depend on its   
   frontal area.   
   Density 0,1 H atom pr cm3 = 1.67E-22 kg/m3   
   Front area 1 m2   
   Speed 1 c   
      
   Assuming complete annihilation this comes to mc^2   
   3e8m/s x 1 x 1,67E-28 x (3E8)^2 = 4,5 kW   
      
   Not a lot...   
      
   It then spreads out in a half sphere, so the intensity at range is   
   4500 W / (2 Pi r^2)   
      
   at 10 AU distance it shines with an intensity of 3,2e-22 W/m2   
      
   Say you converted the whole surface of Mercury to a sensor array, that then   
   become an eye 18,7e13 m2. The total ammount of energy it sees from this point   
   is then 6e-9 W. 9 nano Watts.   
      
   Is that enough to detect it? I donno...   
      
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