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 4.8 × 106 kg; it's speed was about 18 km/s. Had it entered the atmosphere vertically, it would have hit Earth's surface with about the same speed. (a) Calculate the meteorite's loss of energy (as a positive number, in joules) that would have been associated with the vertical impact. (b) Express the energy as a multiple of the explosive energy of 1 megaton of TNT, which is 4.2 × 1015 J. (c) 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

A) Since the speed in the y direction is about equal to the speed in the x direction, y is about 18 km/s. Convert km/s to m/s so

18 km/s (1000m/km)=18,000 m/s

Kinetic Energy= mv^2/2

(4.8x10^6 kg)(18,000 m/s)^2/2= 7.8 x 10^14 J (ans)

B) Convert previous answer to megatons

7.8 x10^14 J (1 megaton/ 4.2x10^15 J) = 0.1851 megaton (ans)

C) 13 kilotons (1 mt/1000kt)(4.2 x 10^15 J/1 mt)= 5.46x10^13 J

Equivalent no. of bombs = (7.8 x 1014) / (5.46x10^13)

                                             = 14.29 = 14 (ans)

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