Chapter 7, Problem 12Q

(a)

To determine

The kinetic energy of an asteroid having radius $2\text{ km}$ with an average density of $2500{\text{ kg/m}}^3$ striking the Earth with $25\text{ km/s}$.

Answer to Problem 12Q

Solution:

The kinetic energy of the asteroid is found out to be $3.27\times {10}^{15}\text{ J}$.

Explanation of Solution

Given data:

The velocity of the asteroid hitting the Earth is $25\text{ m/s}$.

The radius of the asteroid is $2\text{ km}$.

The density of the asteroid is $2500{\text{ kg/m}}^3$

Formula used:

The mass of the asteroid can be calculated by the following expression:

$m=\rho \cdot V$

Here, $m$ represents the mass of the asteroid, $\rho$ represents the density of the asteroid, and $V$ represents the volume of the asteroid.

The expression for the volume of a sphere is:

$V=\frac{4}{3}\pi r^3$

Here, $r$ represents the radius.

Conversion formula from kilometer to meter is:

1 km = 1000 m

The kinetic energy is calculated by the below expression:

$K=\frac{1}{2}m{v}^2$

Here, $K$ represents the kinetic energy, $m$ represents the mass of the body, and $v$ represents the velocity of the body.

Explanation:

Recall the expression for calculating the volume.

$V=\frac{4}{3}\pi r^3$

Substitute 1 km for $r$ and also use the conversion formula.

$\begin{array}{c}V=\frac{4}{3}\pi {\left(1\text{ km}\times \left(\frac{1000\text{ m}}{1\text{ km}}\right)\right)}^3\\ =\frac{4}{3}\pi {\left(1000\text{ m}\right)}^3\end{array}$

Recall the expression of mass.

$m=\rho \cdot V$

Substitute $\frac{4}{3}\pi {\left(1000\text{ m}\right)}^3$ for $V$ and $2500\text{}{\text{ kg/m}}^{\text{3}}$ for $\rho$.

$\begin{array}{c}m=\rho \cdot V\\ =\rho \cdot \left(\frac{4}{3}\pi r^3\right)\\ =\left(2500\text{}{\text{ kg/m}}^{\text{3}}\right)\left(\frac{4}{3}\pi {\left(1000\text{ m}\right)}^3\right)\\ =1.047\times {10}^{13}\text{ kg}\end{array}$

The mass of asteroid is $1.047\times {10}^{13}$ kg.

Recall the expression for the kinetic energy.

$K=\frac{1}{2}m{v}^2$

Substitute $25\text{ km/s}$ for $v$, $1.047\times {10}^{13}\text{ kg}$ for $m$, and use the conversion formula:

$\begin{array}{c}K=\frac{1}{2}m{v}^2\\ =\frac{1}{2}\left(1.047\times {10}^{13}\text{ kg}\right){\left(25\text{ km/s}\times \frac{1000\text{ m}}{1\text{ km}}\right)}^2\\ =3.27\times {10}^{21}\text{ J}\end{array}$

Conclusion:

Thus, the kinetic energy of the asteroid is found out to be $3.27\times {10}^{21}\text{ J}$.

(b)

To determine

The comparison between the energy released by the impact of asteroid mentioned in sub-part (a) to the energy released by a 20-kiloton nuclear weapon, which was similar to the nuclear weapon dropped on Hiroshima.

Answer to Problem 12Q

Solution:

The energy released during nuclear destruction is 40 times smaller than the energy released during the asteroid strike.

Explanation of Solution

Given data:

The energy released by the nuclear weapon is 20-kilotons.

Formula used:

1 kiloton of TNT emits $4.2\times {10}^{12}\text{ J}$ of energy.

The expression of energy released in joules is:

$E_{\text{Joules}}=E_{\text{kN}}\times \left(4.2\times {10}^{12}\text{}\text{J}\right)$

Explanation:

From sub-part (a), the value of kinetic energy of the asteroid is $3.27\times {10}^{15}\text{ J}$.

Recall the expression of energy released in joules.

$E_{\text{Joules}}=E_{\text{kN}}\times \left(4.2\times {10}^{12}\text{}\text{J}\right)$

Substitute $20\text{ kN}$ for $E_{\text{kN}}$.

$\begin{array}{c}{E}_{\text{Joules}}=\left(20\text{}\text{ kN}\right)\times \left(4.2\times {10}^{12}\text{}\text{J}\right)\\ =8.4\times {10}^{13}\text{ J}\end{array}$

Refer to the value of energy released by the impact of an asteroid that is $3.27\times {10}^{21}\text{ J}$.

The energy released in nuclear attack is $8.4\times {10}^{13}\text{ J}$.

Taking the ratio of both the above energies,

$\begin{array}{c}\frac{3.27\times {10}^{21}\text{ J}}{8.4\times {10}^{13}\text{ J}}=3.89\times {10}^7\\ \simeq 4\times {10}^7\end{array}$

Conclusion:

Therefore, the kinetic energy of the asteroid striking the Earth’s surface is $4\times {10}^7$ times greater than the energy released during the nuclear attack in Hiroshima.