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 $4 \times 10^{6} kg$ ; its speed was about $15 km s^{- 1}$ . 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 kinetic energy (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 $4.2 \times 10^{15} J$ . (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?

Accepted Answer

Given,

Mass of meteorites, $m = 4 \times 10^{6} kg$

Its initial speed before impact, $u = 15 km s^{- 1}$

$4.2 \times 10^{15} J$

(a) Meteorite's collides with Earth's surface with its initial speed and penetrated Earth. Finally, it's speed becomes zero $(v = 0)$ .

Now, loss in kinetic energy of meteorite,

$∆ K = K_{i} - K_{f}$

Here, $K_{i} =$ Initial kinetic energy

$K_{f} =$ Final kinetic energy

$\Rightarrow ∆ K = \frac{1}{2} m u^{2} - \frac{1}{2} m v^{2}$

$\Rightarrow ∆ K = \frac{1}{2} \times 4 \times 10^{6} \times {(15 \times 10^{3})}^{2} - 0$

$\Rightarrow ∆ K = 4.5 \times 10^{14} J$ .

(b) Since, $4.2 \times 10^{15} J$ of energy is contained in $1$ megaton of $TNT$ , so $∆ K = 4.5 \times 10^{14} J$ can be expressed,

$∆ K = 4.5 \times 10^{14} \times \frac{1 megaton of TNT}{4.2 \times 10^{15} J}$

$∆ K \approx 0.1$ megaton of $TNT$ .

(c) Energy associated with atomic bomb $= 13$ kiloton of $TNT = 13 \times 10^{- 3}$ megaton.

So, number of Hiroshima bomb that are equivalent to the meteorite impact are,

$N = \frac{0.1 megaton \times 1 bomb}{13 \times 10^{- 3} megaton} \approx 8 bombs$

$N = 8$ bombs.

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