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 3.2 × 106 kg; it's speed was about 17 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?

Answer 1

a) $4.62\cdot 10^{14} J$

b) 0.110 megatons

c) 8.46 bombs

Explanation:

a)

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

$\Delta K=K_f-K_i$

Which can be rewritten as:

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where:

$m=3.2\cdot 10^6 kg$ is the mass of the meteorite

$v=0$ is the final speed of the meteorite

$u=17 km/s = 17,000 m/s$ is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

$\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J$

So, the energy lost by the meteorite is $4.62\cdot 10^{14} J$

b)

The energy equivalent to 1 megaton of TNT is

$E_{TNT}=4.2\cdot 10^{15} J$

Here the energy lost by the meteorite is

$E=4.62\cdot 10^{14} J$

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

$\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110$

So, the energy lost by the meteorite corresponds to 0.110 megatons.

c)

The energy of one atomic bomb explosion in Hiroshima is equal to

$E'=13 kt$ (13 kilotons)

which corresponds to

$E'=0.013 Mt$ (0.013 megatons)

Here the energy of the meteorite is equal to

$E=0.110 Mt$ (0.110 megatons)

Therefore, we can find how many Hiroshima bombs are equivalent to teh meteorite impact by using the following rules of three:

$\frac{1 bomb}{0.013 Mt}=\frac{x bombs}{0.110 Mt}\\x=\frac{1\cdot 0.110}{0.013}=8.46$

So, 8.46 bombs.