1. Announcements Homework: Chapter 8 # 46, 47, 50 & 51 plus Estimate the size of the meteor that formed Barringer’s Crater in Arizona. Assume it was a pure iron meteor and the rock in the area was silicon. Exam 2 is three weeks away

2. Cratering is an energy transformation problem The impactor has kinetic energy and it is converted into heat, kinetic energy of ejecta and vaporizing the ground and impactor

3. Energy Transformations Total Initial Energy = Total Final Energy Before an impact, the initial energy is only the kinetic energy of the impactor: ½mv². After the impact the energy takes six forms: 1.Heat some of the impactor and ground rock to their melting point 2.Melt some of the impactor and ground rock 3.Heat some of the molten impactor and ground rock to their vaporization temperature 4.Vaporize some of the impactor and ground rock 5.What isn’t melted or vaporized is hurled out with some kinetic energy 6.The ground rock around the edge of the crater is deformed

4. Example A spherical iron meteorite with a radius of 100m impacts the Earth at a speed of 44,000 km /hr. If all the meteorite is vaporized on impact, what volume of ground rock will also be vaporized? How large a hole will this make? For simplicity, assume there is no ejecta. Constants needed: SiliconIron

5. Example Solution Initial Energy The initial energy is just the kinetic energy of the impactor: ½mv 2. To find the mass, find the volume of the sphere and multiply by the density of iron.

6. Example Solution 2 After the impact, there will be a number of energies: Heat to raise temperature of impactor to its melting temperature Heat to raise the ground rock to its melting temperature Heat to melt the impactor Heat to melt the ground rock Heat to raise the temperature of the molten impactor to its vaporization temperature Heat to raise the temperature of the molten ground rock to its vaporization temperature Heat to vaporize the impactor Heat to vaporize the ground rock

7. Example Solution 3 Sum up all the energies after, equate to the initial energy. Now solve for the mass of the ground rock

8. Example Solution 4 Using the constants given with the problem we find the mass of ground rock to be 1.301 x 10 11 kg. This mass was what filled the hemispherical hole that the impact creates so find the radius of the hole. The hole will be about 300 m deep and 600 m wide.

9. Rock dating can be done by radioactive decay analysis P f is the final amount (moles) of an isotope, P 0 is the initial amount (moles) of the isotope and n is the number of half-lives that have elapsed.

10. Example A rock sample is found to contain 2.35 grams of U 235 and 6.21 grams of Pb 207 ? What is the age of the rock? The atomic mass of an isotope is the subscript in the symbol for the element.

11. Example Solution First find the number of moles of U 235 and Pb 207. The atomic mass of U 235 is 235 grams so there are 0.01 moles of it. The atomic mass of Pb 207 is 207 grams so there are 0.03 moles of it. That means there was initially 0.04 moles of U 235. Thus, two half-lives have elapsed. The half-life of U 235 is 713,000,000 years so the rock is 1,426,000,000 year old