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 3.4 × 106 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. (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)
The velocity of the meteorite just before hitting the ground is:
$v=20 km/s=20000 m/s$
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
$\Delta K = \frac{1}{2}mv^2= \frac{1}{2}(3.4 \cdot 10^6 kg)(20000 m/s)^2=6.8 \cdot 10^{14}J$

(b) 1 megaton of tnt is equal to $1 MT=4.2 \cdot 10^{15}J$
To find to how many megatons the meteorite energy loss $\Delta E$
corresponds, we can set the following proportion
$1 MT: 4.2 \cdot 10^{15}J=x: \Delta E$
And so we find
$x= \frac{\Delta E}{4.2 \cdot 10^{15}J} = \frac{6.8 \cdot 10^{14}J }{4.2 \cdot 10^{15}J} =0.162 MT$
So, 0.162 megatons.

(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of $\Delta E=0.162 MT=162 kT$. So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
$1:13 kT=x:162 kT$
And so we find
$x= \frac{162 kT}{13 kT}=12.46$
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.