Copyright (c) 1996 by Stan Berry In all this talk of lobbing asteroids and chunks of ice at the planet, > I wonder how big a rock you would need to rival a "typical" nuclear > missile in sheer destructive force? Any ideas from you physics > intensive list members? It is surprisingly small. For a nickel iron sphere: Impact Kinetic Radius Velocity Energy 2.5 meter* 10 km/sec 6 kt 5 meter 10 km/sec 50 kt 10 meter 10 km/sec 400 kt 50 meter 10 km/sec 50 Mt 100 meter 10 km/sec 400 Mt 500 meter 10 km/sec 50000 Mt 5 meter 30 km/sec 450 kt 10 meter 30 km/sec 3.6 Mt 50 meter 30 km/sec 450 Mt *Note: 2.5 m radius is 5 meter diameter and this does agree with the numbers shown by benn1@server.uwindsor.ca (John Benn). Ice has about one eighth the density of nickel-iron, and therefore for the same volume and speed would have the one eighth the kinetic energy. But ice would melt more and would be more likely to shatter during reentry. The equations are easily handled by a spreadsheet such as Excel. I will happily E-mail such an Excel sheet to anyone requesting it; please tell me whether you have a Mac or PC, and please be patient through the mailing hiccups. Inputting sizes in METERS and velocity in METERS/SECOND, Volume: Sphere: 4/3*PI*R^3. Cylinder: PI*R^2*H Density: 1000 kg/m^3 for ice or water (1.0 gram/cm^3) 8000 kg/m^3 for nickel-iron (8.0 gram/cm^3) Mass: Density*Volume Kinetic Energy in Joules. 0.5 * Mass * Velocity^2 Since 1 kiloton is defined as 1.0E12 calories, and 1 calorie is 4.19 Joules, divide the Kinetic energy in joules by 4.19E12 to get the Kinetic Energy in kilotons. Note that the energy at impact is for the portion of the meteor that actually reaches the ground. A portion will be lost during re-entry. ______________________________ | berrys@puzzler.nichols.com | | Stan Berry | | Nichols Research Corp. | | Huntsville, AL | |______________________________|