Question

(a) What is the rotational kinetic energy of the Earth about its spin axis? Model the Earth as a uniform sphere and use data from the endpapers of this book.

(b) The rotational kinetic energy of the Earth is decreasing steadily because of tidal friction. Assuming the rotational period decreases by$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\text{μs}}$each year, find the change in one day.

Short answer

(a) The earth's rotational kinetic energy about its spin axis is$2.57\times {10}^{29}\mathrm{J}$.

(b) The change in one day is$-1.63\times {10}^{17}\mathrm{J}/\text{day}$.

Step-by-step solution

Rotational kinetic energy

When a rigid body rotates with angular speedaround a fixed axis, its rotational kinetic energy is given by the relation

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

The problem is classified as rotational kinetic energy.

The equation to use:

$\left(\mathrm{i}\right)K{E}_{\mathrm{rot}}=\frac{1}{2}I{\omega }^2$

(ii) Moment of inertia of solid sphere,$I=\frac{2}{5}M{R}^2$.

Calculating the rotational kinetic energy of the Earth

The inertia I instant is determined by

$I=\frac{2}{5}M{R}^2$

WhereM is the sphere's mass and R is the sphere's radius.

We know that,

$\begin{array}{c}\omega =\frac{v}{r}\\ =\frac{\sqrt{2gh}}{R}\end{array}$

As a result, the earth's rotational kinetic energy about its spin axis will be.

$\begin{array}{c}E=\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\omega }^2\\ =\left(\frac{1}{2}\right)\left(\frac{2}{5}\right)\left(5.98\times {10}^{24}\mathrm{kg}\right){\left(6.37\times {10}^6\mathrm{m}\right)}^2{\left(\frac{2\pi }{86400 \mathrm{s}}\right)}^2\\ =2.57\times {10}^{29}\mathrm{J}\end{array}$

Change in the rotational kinetic energy in one day

(b) To get the change in one day we that

$\begin{array}{c}\frac{\mathrm{dE}}{\mathrm{dt}}=\frac{d}{dt}\left[\frac{1}{2}\left(\frac{2}{5}M{R}^2\right){\left(\frac{2\pi }{T}\right)}^2\right]\\ =\frac{1}{5}M{R}^2(2\pi {)}^2\left(-2T^{-3}\right)\frac{\mathrm{dT}}{\mathrm{dt}}\\ =\frac{1}{5}M{R}^2{\left(\frac{2\pi }{T}\right)}^2\left(\frac{-2}{T}\right)\frac{\mathrm{dT}}{\mathrm{dt}}\end{array}$

Substitute the values.

$\begin{array}{rcl}\frac{\mathrm{dE}}{\mathrm{dt}}& =& \left(2.57\times {10}^{29}\mathrm{J}\right)\left(\frac{-2}{86400\mathrm{s}}\right)\left(\frac{10\times {10}^{-6}\mathrm{s}}{3.16\times {10}^7\mathrm{s}}\right)(86400\mathrm{s}/\mathrm{day})\\ & =& -1.63\times {10}^{17}\mathrm{J}/\mathrm{day}\end{array}$