The mass and the radius of the Moon are, respectively, about $\frac{1}{81}$ time and about $\frac{1}{3.7}$ time those of the Earth. Given that the rotational period of the Moon is 27.3 days, compare the rotational kinetic energy of the Earth with that of the Moon.
Q 96.10
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$
\text { Data : } M_M=\frac{1}{81} M_E, R_M=\frac{1}{3.7} R_E, T_M=27.3 \text { days, } T_E=1 \text { day }
$
Let $I_E$ and $I_M$ be the moments of inertia of the Earth and the Moon about their respective axes of rotation, and $\omega_E$ and $\omega_M$ be their respective rotational angular speeds. Assuming the Earth and the Moon to be solid spheres of uniform densities,
$
I_{\mathrm{E}}=\frac{2}{5} M_{\mathrm{E}} R_{\mathrm{E}}^2 \text { and } I_{\mathrm{M}}=\frac{2}{5} M_{\mathrm{M}} R_{\mathrm{M}}^2
$
Kinetic energy of rotation, $E_{\mathrm{rot}}=\frac{1}{2} \mathrm{I} \omega^2$
$
=\frac{1}{2} I\left(\frac{2 \pi}{T}\right)^2=2 \pi^2 \frac{I}{T^2}
$
Therefore, the ratio of the rotational KE of the Earth to that of the Moon is MaharashtraBoardSolutions.Guru
$
\begin{aligned}
\frac{E_{\text {rot (Earth) }}}{E_{\text {rot (Moon) }}} & =\frac{I_{\mathrm{E}}}{I_{\mathrm{M}}} \cdot\left(\frac{T_{\mathrm{M}}}{T_{\mathrm{E}}}\right)^2=\frac{M_{\mathrm{E}}}{M_{\mathrm{M}}} \cdot\left(\frac{R_{\mathrm{E}}}{R_{\mathrm{M}}}\right)^2 \cdot\left(\frac{T_{\mathrm{M}}}{T_{\mathrm{E}}}\right)^2 \\
& =81 \times(3.7)^2 \times(27.3)^2=8.264 \times 10^5
\end{aligned}
$
\text { Data : } M_M=\frac{1}{81} M_E, R_M=\frac{1}{3.7} R_E, T_M=27.3 \text { days, } T_E=1 \text { day }
$
Let $I_E$ and $I_M$ be the moments of inertia of the Earth and the Moon about their respective axes of rotation, and $\omega_E$ and $\omega_M$ be their respective rotational angular speeds. Assuming the Earth and the Moon to be solid spheres of uniform densities,
$
I_{\mathrm{E}}=\frac{2}{5} M_{\mathrm{E}} R_{\mathrm{E}}^2 \text { and } I_{\mathrm{M}}=\frac{2}{5} M_{\mathrm{M}} R_{\mathrm{M}}^2
$
Kinetic energy of rotation, $E_{\mathrm{rot}}=\frac{1}{2} \mathrm{I} \omega^2$
$
=\frac{1}{2} I\left(\frac{2 \pi}{T}\right)^2=2 \pi^2 \frac{I}{T^2}
$
Therefore, the ratio of the rotational KE of the Earth to that of the Moon is MaharashtraBoardSolutions.Guru
$
\begin{aligned}
\frac{E_{\text {rot (Earth) }}}{E_{\text {rot (Moon) }}} & =\frac{I_{\mathrm{E}}}{I_{\mathrm{M}}} \cdot\left(\frac{T_{\mathrm{M}}}{T_{\mathrm{E}}}\right)^2=\frac{M_{\mathrm{E}}}{M_{\mathrm{M}}} \cdot\left(\frac{R_{\mathrm{E}}}{R_{\mathrm{M}}}\right)^2 \cdot\left(\frac{T_{\mathrm{M}}}{T_{\mathrm{E}}}\right)^2 \\
& =81 \times(3.7)^2 \times(27.3)^2=8.264 \times 10^5
\end{aligned}
$
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