Chapter 8, Problem 65P

To determine

The percent change in angular speed of Earth.

Answer to Problem 65P

The percent change in angular speed of Earth is $2.7\times {10}^{-14}\%$

Explanation of Solution

Given:

Mass of asteroid, $M_1=1.0\times {10}^{5}kg$

Speed of asteroid, $v=30\text{km/s = 3}\times {\text{10}}^4\text{m/s}$

Mass of Earth, $M_2=5.98\times {10}^{24}kg$

Radius of Earth, $R=6.4\times {10}^{6}m$

Angular speed of earth, ${\omega }_1=\frac{2\pi }{86400}=7.27\times {10}^{-5}\text{rad/s}$

Formula used:

Moment of inertia of asteroid, $I_2=M_1{R}^2$

Angular momentum of asteroid, $L=mvr$

Moment of inertia of Earth, $I=\frac{2}{5}M{R}^2$

Angular momentum of Earth, $L=I\omega$

Law of conservation of angular momentum,

  $\begin{array}{l}{L}_1=L_2\\ I_2{\omega }_1+M_{1}vR=(I_1+I_2){\omega }_2\end{array}$

Percent change in angular speed,

  $\%\omega =\frac{{\omega }_2-{\omega }_1}{{\omega }_1}\times 100$

Calculation:

Substituting values in law of conservation of angular momentum,

  $\begin{array}{l}\frac{2}{5}{M}_2{R}^2{\omega }_1+M_{1}vR=\left(\frac{2}{5}{M}_2{R}^2+M_1{R}^2\right){\omega }_2\\ \text{Since}{M}_1<<M_2,M_1{R}^{2}canbeneglected\\ {\omega }_2=\frac{\frac{2}{5}{M}_2{R}^2{\omega }_1+M_{1}vR}{\frac{2}{5}{M}_2{R}^2}\\ {\omega }_2={\omega }_1+\frac{5M_{1}v}{2M_{2}R}\\ {\omega }_2-{\omega }_1=\frac{5M_{1}v}{2M_{2}R}\\ {\omega }_2-{\omega }_1=\frac{5({10}^5)(3\times {10}^4)}{2(5.98\times {10}^{24})(6.4\times {10}^6)}\\ {\omega }_2-{\omega }_1=1.96\times {10}^{-21}\text{rad/s}\end{array}$

Substituting values in percent change,

  $\begin{array}{l}\%\omega =\frac{1.96\times {10}^{-21}}{7.27\times {10}^{-5}}\times 100\\ \%\omega =2.7\times {10}^{-14}\%\end{array}$

Conclusion:

The percent change in angular speed of Earth is $2.7\times {10}^{-14}\%$