Calculate the angular momentum and rotational kinetic energy of earth about its own axis. How long could this amount of energy supply one kilowatt power to each of the $3.5 \times 10^9$ persons on earth? (Mass of earth = $6.0 \times 1024kg$ and radius = $6.4 \times 10^{24}km)$.
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Here we assume the earth to be a solid sphere. We know that the moment of inertia of a solid sphere about its axis is $\text{I}=\frac{2}{5}\text{MR}^2=\frac{2}{5}\times(6.0\times10^{24}\text{kg})$
$\times(6.4\times10^6\text{m})^2$
$=9.8\times10^{37}\text{kg}-\text{m}^2$ In onhe day (= $24 \times 60 \times 60sec$.) the earth completes one revolution. i.e., traces an angle of $2\pi$ radian. Hence its angular velocity is given by $\omega=\frac{2\pi}{24\times6\times60}$
$=7.27\times10^{-5}\text{rad/sec.}$
$\therefore$ Angular momentum, $\text{I}\omega=(98\times10^{37}\text{kg}-\text{m}^2)(7.27\times10^{-5}\text{s}^{-1})$
$=7.1\times10^{33}\text{kg}-\text{m}^2\text{sec.}$ The rotatinal energy $\frac{1}{2}\text{I}\omega^2=\frac{1}{2}(9.8\times10^{37}\text{kg}-\text{m}^2)(7.27\times10^{-5}\text{s}^{-1})^2$
$=2.6\times10^{29}\text{joule.}$ power supplied by this energy, $\text{p}=\frac{\text{Energy}}{\text{Time}}=\frac{2.6\times10^{29}}{\text{t}}\text{watt}$
$=\frac{2.6\times10^{29}}{10^3\text{t}}\text{kilowatt.}$ Power required by $3.5 \times 10^9$ persons = $3.5 \times 10^9 \times 1$ kilowatt. $\therefore\frac{2.6\times10^{29}}{10^3\text{t}}=3.5\times10^9$
$\text{or }\text{t}=\frac{2.6\times10^{29}}{10^3\times3.5\times10^9\times365\times24\times60\times60}\text{year}$
$=2.35\times10^6\text{years.}$
$\times(6.4\times10^6\text{m})^2$
$=9.8\times10^{37}\text{kg}-\text{m}^2$ In onhe day (= $24 \times 60 \times 60sec$.) the earth completes one revolution. i.e., traces an angle of $2\pi$ radian. Hence its angular velocity is given by $\omega=\frac{2\pi}{24\times6\times60}$
$=7.27\times10^{-5}\text{rad/sec.}$
$\therefore$ Angular momentum, $\text{I}\omega=(98\times10^{37}\text{kg}-\text{m}^2)(7.27\times10^{-5}\text{s}^{-1})$
$=7.1\times10^{33}\text{kg}-\text{m}^2\text{sec.}$ The rotatinal energy $\frac{1}{2}\text{I}\omega^2=\frac{1}{2}(9.8\times10^{37}\text{kg}-\text{m}^2)(7.27\times10^{-5}\text{s}^{-1})^2$
$=2.6\times10^{29}\text{joule.}$ power supplied by this energy, $\text{p}=\frac{\text{Energy}}{\text{Time}}=\frac{2.6\times10^{29}}{\text{t}}\text{watt}$
$=\frac{2.6\times10^{29}}{10^3\text{t}}\text{kilowatt.}$ Power required by $3.5 \times 10^9$ persons = $3.5 \times 10^9 \times 1$ kilowatt. $\therefore\frac{2.6\times10^{29}}{10^3\text{t}}=3.5\times10^9$
$\text{or }\text{t}=\frac{2.6\times10^{29}}{10^3\times3.5\times10^9\times365\times24\times60\times60}\text{year}$
$=2.35\times10^6\text{years.}$
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