Chapter 8, Problem 58P

(a)

To determine

The angular momentum of the Earth.

Answer to Problem 58P

The angular momentum of the Earth when it is rotating about its axis is $7.1\times {10}^{33}{\text{kg-m}}^{\text{2}}\text{/s}$

Explanation of Solution

Given:

Mass of Earth, $M=6.0\times {10}^{24}kg$

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

Distance of Earth from Sun, $r=1.5\times {10}^{8}km=1.5\times {10}^{11}m$

Time of rotation about its axis, $T=24hours=86400s$

Earth is rotating about its axis.

Formula used:

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

Angular speed, $\omega =\frac{2\pi }{T}$

Angular Momentum, $L=I\omega$

Calculation:

Substituting values for moment of inertia,

  $\begin{array}{l}$ I=\frac{2}{5}(6.0\times {10}^{24}){(6.4\times {10}^6)}^2$I=9.8\times {10}^{37}kg-m^2\end{array}$

Substituting values for angular speed,

  $\begin{array}{l}\omega =\frac{2\pi }{86400}\\ \omega =7.27\times {10}^{-5}\text{rad/s}\end{array}$

Substituting values for angular momentum,

  $\begin{array}{l}L=(9.8\times {10}^{37})(7.27\times {10}^{-5})\\ L=7.1\times {10}^{33}{\text{kg-m}}^{\text{2}}\text{/s}\end{array}$

Conclusion:

Angular momentum of earth while rotating about its axis is $7.1\times {10}^{33}{\text{kg-m}}^{\text{2}}\text{/s}$

(b)

To determine

The angular momentum of Earth.

Answer to Problem 58P

The angular momentum of Earth when it is revolving around the Sun is $2.7\times {10}^{40}{\text{kg-m}}^{\text{2}}\text{/s}$

Explanation of Solution

Given:

Mass of Earth, $M=6.0\times {10}^{24}kg$

Orbital radius, $r=1.5\times {10}^{8}km=1.5\times {10}^{11}m$

Time period of revolution, $T=365days=365\times 86400=3.15\times {10}^{7}s$

Earth is revolving around the Sun.

Formula used:

Moment of inertia of Earth about its axis, $I=M{r}^2$

Angular speed, $\omega =\frac{2\pi }{T}$

Angular Momentum, $L=I\omega$

Calculation:

Substituting values for moment of inertia,

  $\begin{array}{l}$ I=(6.0\times {10}^{24}){(1.5\times {10}^{11})}^2$I=1.35\times {10}^{47}kg-m^2\end{array}$

Substituting values for angular speed,

  $\begin{array}{l}\omega =\frac{2\pi }{3.15\times {10}^7}\\ \omega =2\times {10}^{-7}\text{rad/s}\end{array}$

Substituting values for angular momentum,

  $\begin{array}{l}L=(1.35\times {10}^{47})(2\times {10}^{-7})\\ L=2.7\times {10}^{40}{\text{kg-m}}^{\text{2}}\text{/s}\end{array}$

Conclusion:

Angular momentum of earth while revolving around Sun is $2.7\times {10}^{40}{\text{kg-m}}^{\text{2}}\text{/s}$