On August 10, 1972, a large meteorite skipped across the atmosphere above the we

Question

On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much like a stone skipped across water. The accompanying fireball was so bright that it could be seen in the daytime sky and was brighter than the usual meteorite trail. The meteorite's mass was about 4.7 times 10^6 kg; it's speed was about 20 km/s. Had it entered the atmosphere vertically, it would have hit Earth's surface with about the same speed. Calculate the meteorite's loss of energy (as a positive number, in joules) that would have been associated with the vertical impact. Express the energy as a multiple of the explosive energy of 1 megaton of TNT, which is 4.2 times 10^15 J. The energy associated with the atomic bomb explosion over Hiroshima was equivalent to 13 kilotons of TNT. To how many Hiroshima bombs would the meteorite impact have been equivalent?

Explanation / Answer

Loss in energy = change in kineric enrgy = 1/2mv2 = 1/2*4.7*10^6*(20000)^2=9.4*10^14J

B)=9.4*10^14/(4.2*10^15)=0.22

C)energy of one meteorite impact/(13/1000*1 megaton TNT) =9.4*10^14/(13/1000*4.2*10^15)=17.21

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