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(A) According to the law of conservation of angular momentum, $L = I\omega = 	ext{constant}$.Since the Earth is a sphere, its moment of inertia is $I

Vedclass · Accepted Answer

$18$

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(A) According to the law of conservation of angular momentum, $L = I\omega = 	ext{constant}$.Since the Earth is a sphere, its moment of inertia is $I = \frac{2}{5}MR^2$.Thus, $I \propto R^2$.Since $L = I\omega = I \cdot \frac{2\pi}{T}$, we have $I_1 \cdot \frac{1}{T_1} = I_2 \cdot \frac{1}{T_2}$, which implies $T \propto I \propto R^2$.If the radius $R$ becomes $R' = \frac{R}{2}$, then the new time period $T'$ is $T' = T \cdot (\frac{R'}{R})^2 = T \cdot (\frac{1}{2})^2 = \frac{T}{4}$.Given the current duration of a day $T = 24 	ext{ hours}$, the new duration is $T' = \frac{24}{4} = 6 	ext{ hours}$.The decrease in the duration of the day is $\Delta T = T - T' = 24 - 6 = 18 	ext{ hours}$.

If the radius of the Earth shrinks to half of its current value, the duration of the day will decrease by how many hours?

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