91Ó°ÊÓ

1. The diameter of the crater-1 is, ${\mathrm{d}}_1=1\mathrm{km}$.
2. The diameter of the crater-2 is, ${\mathrm{d}}_2=5\mathrm{km}$.
3. Kinetic energy for explosion-1, ${\mathrm{E}}_1=1\mathrm{megaton}\mathrm{of}\mathrm{TNT}=4.2×{10}^{15}\mathrm{J}$.
4. Energy associated with each atomic bomb explosion is $13\mathrm{kilotons}\mathrm{of}\mathrm{TNT}$.
5. The diameter of the crater is proportional to the energy of the explosion raised to 1/3 power.

Using the given relation between the diameter oftheimpact and the energy of the explosion, we can find the kinetic energy E associated with the impact in terms of megatons of TNT and in terms of the Hiroshima bomb explosion.

Formula:

$\mathrm{d}={\mathrm{E}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

The relationship between the diameter and the energy is expressed as,

$\mathrm{d}={\mathrm{E}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Let's assume for diameter d₁, the energy of the explosion is E₁, and for diameter d₂, the energy of the explosion is E₂.

Therefore, the above expression can be written now as,

$$\begin{gathered} \frac{{\mathrm{d}}_1}{{\mathrm{E}}_1^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}=\frac{{\mathrm{d}}_2}{{\mathrm{E}}_2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} \\ \frac{{\mathrm{d}}_1}{{\mathrm{d}}_2}=\frac{{\mathrm{E}}_1^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{{\mathrm{E}}_2^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \end{gathered}$$

Solving further as,

$\begin{array}{rcl}\frac{{\mathrm{d}}_1}{{\mathrm{d}}_2}& =& {\left(\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ {\left(\frac{{\mathrm{d}}_1}{{\mathrm{d}}_2}\right)}^3& =& \frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}\\ {\mathrm{E}}_2& =& {\left(\frac{{\mathrm{d}}_2}{{\mathrm{d}}_1}\right)}^3{\mathrm{E}}_1\end{array}$

Substitute the values in the above expression as,

$\begin{array}{rcl}{\mathrm{E}}_2& =& \left(1\mathrm{megaton}\mathrm{of}\mathrm{TNT}\right){\left(\frac{50\mathrm{km}}{1\mathrm{km}}\right)}^3\\ & =& 1.25×{10}^5\mathrm{megatons}\mathrm{of}\mathrm{TNT}\\ & ≈& 1×{10}^5\mathrm{megaton}\mathrm{of}\mathrm{TNT}\end{array}$

Therefore, thekinetic energy associated with that impact in terms of megatons of TNT is

$1×{10}^5\mathrm{megatons}\mathrm{of}\mathrm{TNT}$.

For energy associated with the explosion in terms of TNT for explosion-2 is,

${\mathrm{E}}_2=1.25×{10}^5\mathrm{megatons}\mathrm{of}\mathrm{TNT}$

But, for 1 Hiroshima bomb explosion,

$1\mathrm{Hiroshima}\mathrm{Bomb}=13\mathrm{kilotons}\mathrm{of}\mathrm{TNT}$

We know that,

$1\mathrm{kilotons}=1000\mathrm{tons}$

We can write the energy associated with 1 bomb as,

$1\mathrm{Hiroshima}\mathrm{bomb}=13000\mathrm{tons}\mathrm{of}\mathrm{TNT}$

Therefore kinetic energy in terms of Hiroshima bomb equivalents can be calculated as,

$\begin{array}{rcl}{\mathrm{E}}_2& =& 1.25×{10}^5×{10}^6.\left(1\mathrm{tons}\mathrm{of}\mathrm{TNT}×\frac{1\mathrm{Hiroshima}\mathrm{bomb}}{13000\mathrm{tons}\mathrm{of}\mathrm{TNT}}\right)\\ & =& 9.62×{10}^6\mathrm{Hiroshima}\mathrm{bombs}\\ & ≈& {10}^7\mathrm{Hiroshima}\mathrm{bombs}\end{array}$

Therefore, the kinetic energy associated with that impact in terms of Hiroshima bomb equivalent is ${10}^7\mathrm{Hiroshima}\mathrm{Bombs}$.