Chapter 11.1, Problem 3PP

To determine

The kinetic energy of the comet in Joules, and then to compare the work that is done by Earth in stopping the comet.

Answer to Problem 3PP

  $K{E}_{comet}=2.45\times {10}^{20}\text{J}$ and $5.8\times {10}^4$ bombs are needed to be released from Earth in order to stop the comet.

Explanation of Solution

Given:

The mass of the comet, $m=7.85\times {10}^{11}\text{kg}$

The velocity of the comet with which it strikes Earth, $v=25.0\text{km/s}=2.5\times {10}^4\text{m/s}$

The energy that was released by the largest nuclear weapon, $E=4.2\times {10}^{15}\text{J}$

Formula used:

The kinetic energy of an object of mass $m$ travelling at a speed of $v$ is given by,

  $KE=\frac{1}{2}m{v}^2$

Calculation:

Using the given values, the kinetic energy of the comet is,

  $\begin{array}{c}KE=\frac{1}{2}m{v}^2\\ =\frac{1}{2}\left(7.85\times {10}^{11}\right){\left(2.5\times {10}^4\right)}^2\\ =2.45\times {10}^{20}\text{J}\end{array}$

Then, the ratio of the kinetic energy of comet to the energy provided by the weapon released from Earth is,

  $\begin{array}{c}\frac{KE}{E}=\frac{2.45\times {10}^{20}}{4.2\times {10}^{15}}\\ =5.8\times {10}^4\end{array}$

Conclusion:

Thus, $5.8\times {10}^4$ bombs must be released from Earth in order to stop the comet.