Chapter 8, Problem 46P

(a)

To determine

To find: The value of daily rotation about its axis

Answer to Problem 46P

The value is $K{E}_{daily}=2.56\times {10}^{29}\text{J}$ .

Explanation of Solution

Given information:

Estimate the kinetic energy of the Earth with respect to the Sun as the sum of two terms

Formula used:

  $K{E}_{daily}=\frac{1}{2}I{\omega }^2$

Here,

  $KE$ is the kinetic energy,

  $I$ is the moment of inertia,

  $\omega$ is the angular velocity.

Calculation:

The mass of the Earth $M=5.98\times {10}^{24}\text{kg}$ and radius $R=6.37\times {10}^6\text{m}$ and the distance between the Earth and the Sun is $1.5\times {10}^{11}\text{m}$

For the daily rotation about its axis, treat the Earth as a uniform sphere, with an angular frequency of one revolution per day. Hence the rotational kinetic energy about its own axis

  $\begin{array}{l}K{E}_{daily}=\frac{1}{2}I{\omega }^2\\ \text{or}\\ K{E}_{daily}=\frac{1}{2}\times \left(\frac{1}{5}M{R}^2\right){\left(\frac{2\pi }{T}\right)}^2\end{array}$

Since the Earth is consider as uniform sphere, its rotational inertia is $I=2/5M{R}^2$ and the angular frequency in terms of time period $\omega =2\pi /T$ . For daily rotation of the Earth it takes time $T=24\times 60\times 60\mathrm{s}$

  $\begin{array}{l}=\frac{1}{5}\left[\left(5.98\times {10}^{24}\text{kg}\right){\left(6.37\times {10}^6\text{m}\right)}^2{\left(\frac{2\pi }{24\times 60\times 60}\right)}^2\right]\\ \text{or}\\ K{E}_{daily}=2.56\times {10}^{29}\text{J}\end{array}$

Conclusion:

The value is $K{E}_{daily}=2.56\times {10}^{29}\text{J}$ .

(b)

To determine

To find: yearly revolution about the Sun

Answer to Problem 46P

The value is $KE=2.6\times {10}^{33}\text{J}$ .

Explanation of Solution

Given information:

Estimate the kinetic energy of the Earth with respect to the Sun as the sum of two terms,

Formula used:

  $K{E}_{yearly}=\frac{1}{2}\left(M{R}_1^2\right){\left(\frac{2\pi }{T_1}\right)}^2$

Here,

  $KE$ is the kinetic energy,

  $M$ is the mass,

  $R_1$ is the radius,

  $T$ is the Time taken.

Calculation:

Distance of the Earth from the Sun $R_1=1.5\times {10}^{11}\text{m}$ . Time taken by the Earth to revolve around the Sun $T_1=365\text{days}$

Or $T_1=\left(365\times 24\times 60\times 60\right)\mathrm{s}$

Hence Rotational kinetic energy of the Earth about the Sun is

  $\begin{array}{l}K{E}_{yearly}=\frac{1}{2}\left(M{R}_1^2\right){\left(\frac{2\pi }{T_1}\right)}^2\\ K{E}_{yearly}=\left[\frac{\text{1}}{\text{2}}\text{×5}{\text{.98×10}}^{\text{24}}\text{kg×}{\left(\text{1}{\text{.5×10}}^{\text{11}}\text{m}\right)}^{\text{2}}\left(\frac{\text{2π}}{\text{365×24×60×60}}\right)\right]\\ K{E}_{yearly}=2.66\times {10}^{33}\text{J}\end{array}$

As $K{E}_{yearly}>K{E}_{daily}$

The total kinetic energy

  $\begin{array}{l}KE=K{E}_{yearly}+K{E}_{daily}\\ =\left(2.6\times {10}^{33}\text{J}\right)\\ KE=2.6\times {10}^{33}\text{J}\end{array}$

The $KE$ clue to the daily motion is about $10,000$ smaller than that due to the yearly motion.

Conclusion:

The value is $KE=2.6\times {10}^{33}\text{J}$