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 $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{kg}$; its speed was about $\mathbf{15}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{s}$. 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 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 $\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}\mathbf{}\mathbf{J}$. (c) The energy associated with the atomic bomb explosion over Hiroshima was equivalent to $\mathbf{13}\mathbf{}\mathbf{kilotons}$ of TNT. To how many Hiroshima bombs would the meteorite impact have been equivalent?

Short answer

1. The meteorite’s loss of kinetic energy that would have been associated with the vertical impact is $-4.5\times {10}^{14}\mathrm{J}$.
2. The energy loss as a multiple of the explosive energy of 1 megaton of TNT is $-0.1\mathrm{megaton}\mathrm{TNT}$.
3. The number of Hiroshima bombs that would have been equivalent to the meteorite impact is 8.

Step-by-step solution

Given data

1. Mass of the meteorite $\mathrm{m}=4\times {10}^6\mathrm{kg}$.
2. Speed ${\mathrm{v}}_{\mathrm{i}}=15\mathrm{km}/\mathrm{s}=15\times {10}^3\mathrm{m}/\mathrm{s}$.
3. $1\mathrm{megaton}\mathrm{of}\mathrm{TNT}=4.2\times {10}^{15}\mathrm{J}$.
4. Energy associated with the atomic bomb explosion is data-custom-editor="chemistry" $13\mathrm{kilotons}\mathrm{of}\mathrm{TNT}$.

Understanding the concept

Using the formula of kinetic energy, we can find the meteorite’s loss of kinetic energy that would have been associated with the vertical impact. Using the energy loss and relation between 1 megaton of TNT and joule, we can find the energy loss as a multiple of multiple of the explosive energy of 1 megaton of TNT.

Formula:

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$

(a) Calculate the meteorite’s loss of kinetic energy (in joules)

We can calculate the kinetic energy with the expression,

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$

Initial kinetic energy will be,

${\mathrm{K}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2$

Substitute the values in the above expression, and we get,

role="math" localid="1660904026139" $\begin{array}{rcl}{\mathrm{K}}_{\mathrm{i}}& =& \frac{1}{2}\left(4\times {10}^6\mathrm{kg}\right){\left(15\times {10}^3\frac{\mathrm{m}}{\mathrm{s}}\right)}^2\\ & =& 4.5\times {10}^{14}.\left(1\mathrm{kg}\times 1\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}\times \frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\times \frac{1\mathrm{N}}{1\mathrm{kg}.{\displaystyle \frac{\mathrm{m}}{{\mathrm{s}}^2}}}\right)\\ & =& 4.5\times {10}^{14}\mathrm{J}\end{array}$

Final kinetic energy will be,

${\mathrm{K}}_{\mathrm{f}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2$

Here v_f is the final speed, whose value is zero.

Substitute the values in the above expression, and we get,

${\mathrm{K}}_{\mathrm{f}}=0\mathrm{J}$

Therefore, the change in kinetic energy can be calculated as,

data-custom-editor="chemistry" $∆\mathrm{K}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}$

Substitute the values in the above expression, and we get,

data-custom-editor="chemistry" $$\begin{gathered} ∆\mathrm{K}=0\mathrm{J}-4.5\times {10}^{14}\mathrm{J} \\ ∆\mathrm{K}=-4.5\times {10}^{14}\mathrm{J} \end{gathered}$$

Therefore, the meteorite’s loss of kinetic energy that would have been associated with the vertical impact is $-4.5\times {10}^{14}\mathrm{J}$.

(b) Express the energy as a multiple of the explosive energy of 1 megaton of TNT 

Given,

$1\mathrm{megaton}\mathrm{of}\mathrm{TNT}=4.2\times {10}^{15}\mathrm{J}$.

Therefore, the energy loss in terms of megatons of TNT can be calculated as,

$$\begin{gathered} ∆\mathrm{K}=-\left(4.5\times {10}^{14}\mathrm{J}\right)\left(\frac{1\mathrm{megaton}\mathrm{TNT}}{4.2\times {10}^{15}\mathrm{J}}\right) \\ ∆\mathrm{K}=-0.1\mathrm{megaton}\mathrm{TNT} \end{gathered}$$

Therefore, the energy loss as a multiple of the explosive energy of 1 megaton of TNT, which is $4.2\times {10}^{15}\mathrm{J}$, is $-0.1\mathrm{megaton}\mathrm{TNT}$.

(c) Calculate how many Hiroshima bombs would the meteorite impact have been equivalent 

Given,

$1\mathrm{Megaton}=1000\mathrm{kilotons}$.

The energy associated with the atomic bomb explosion is $13\mathrm{kilotons}\mathrm{of}\mathrm{TNT}$.

We can calculate the number of bombs as,

$\begin{array}{rcl}\mathrm{N}& =& \frac{0.1\times 1000\mathrm{kilotons}\mathrm{of}\mathrm{TNT}}{13\mathrm{kilotons}\mathrm{TNT}}\\ & =& 7.69\\ & & ~8\end{array}$

Therefore, the number of Hiroshima bombs that would have been equivalent to the meteorite impact is 8.