Q19E - University Physics with Modern Physics

Question

Meteor crater. About 50,000 years ago, a meteor crashed into the earth near present-day Flagstaff Arizona. Measurements from 2005 estimate that this meteor had a mass of about 1.4 X 10⁸ kg (around 15,000 tons) and hit the ground at a speed of 12 km/s. (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a 1.0-megaton nuclear bomb? (A megaton bomb releases the same amount of energy as a million tons of TNT, and 1.0 ton of TNT releases 4.184 X 10⁹ J of energy).

Step-by-Step Solution

Answer

(a)  $1.01\times {10}^{16}\mathrm{J}$

 

(b)  ${\mathrm{K}}_{\mathrm{m}}=2.4 {\mathrm{K}}_{\mathrm{MB}}$

Step 1: Identification of the given data

 The given data is listed below as-

• The mass of the meteor is $\mathrm{m}=1.4\times {10}^8\mathrm{kg}$  
• The speed of the meteor hitting the ground is, $\mathrm{V}=12\mathrm{km}/\mathrm{s}=12000\mathrm{m}/\mathrm{s}$   
• The kinetic energy of one ton of TNT is ${\mathrm{K}}_{\mathrm{TNT}}=4.184\times {10}^9\mathrm{J}$   

Step 2: Significance of the kinetic energy

The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed V. Therefore, kinetic energy on the particle is given by-

 $\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mV}}^{\mathbf{2}}$

The kinetic energy is a scalar and it is always positive or zero.

Step 3: Determination of kinetic energy delivered by meteor to the ground (a)

The amount of kinetic energy which meteor deliver to the ground will be

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

Here, m is the mass of the meteor, and V is the speed of the meteor hitting the ground.

For, $\mathrm{m}=1.4\times {10}^8\mathrm{kg}$ , and $\mathrm{V}=12\mathrm{km}/\mathrm{s}=12000\mathrm{m}/\mathrm{s}$   

The kinetic energy will be

 $$\begin{gathered} \mathrm{K}=\frac{1}{2}{\mathrm{mV}}^2 \\ =\frac{1}{2}\times 1.4\times {10}^8\mathrm{kg}\times {\left(12\times {10}^3{\mathrm{ms}}^{-1}\right)}^2 \\ =100.8\times {10}^{14}\mathrm{J} \\ =1.01\times {10}^{16}\mathrm{J} \end{gathered}$$$$\begin{gathered} \mathrm{K}=\frac{1}{2}{\mathrm{mV}}^2 \\ =\frac{1}{2}\times 1.4\times {10}^8\mathrm{kg}\times {\left(12\times {10}^3{\mathrm{ms}}^{-1}\right)}^2 \\ =100.8\times {10}^{14}\mathrm{J} \\ =1.01\times {10}^{16}\mathrm{J} \end{gathered}$$

Thus, the kinetic energy delivered by meteor to the ground is $1.01\times {10}^{16}\mathrm{J}$.

Step 4: Comparison of energy released by meteor that hits the ground to the energy released by 1.0-megaton nuclear bomb (b)

The energy released by 1.0 ton of TNT   

Therefore, energy released by one million tons of TNT  

For Comparison of energy released by meteor that hits the ground to the energy released by 1.0-megaton nuclear bomb divide the meteor’s kinetic energy by the energy of the megaton bomb.

Therefore, 

 $$\begin{gathered} \frac{{\mathrm{K}}_{\mathrm{M}}}{{\mathrm{K}}_{\mathrm{MB}}}=\frac{1.01\times {10}^{16}\mathrm{J}}{4.184\times {10}^{15}\mathrm{J}} \\ \frac{{\mathrm{K}}_{\mathrm{M}}}{{\mathrm{K}}_{\mathrm{MB}}}=2.4 \end{gathered}$$ 

Hence, 

 ${\mathrm{K}}_{\mathrm{M}}=2.4 {\mathrm{K}}_{\mathrm{MB}}$

Thus, the energy released by a meteor is 2.4 times energy of the megaton bomb.