Suppose The Earth Contracts Suddenly To Half Its Present Radius. What Would Be The Duration Of A Day?

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Let us consider the mass and the radius of the earth is \( M \) and \( R \) respectively. When the radii are \( R \) and \( \frac{R}{2} \), the moment of inertia of the earth are \( I_1 \) and \( I_2 \) respectively, \( {\omega}_1 \) and \( {\omega}_2 \) are the angular velocities respectively, \( T_1 \) and \( T_2 \) are the respective time period of the earth.

We know that,

\( {I_1}\cdot{\omega}_1={I_2}\cdot{\omega}_2 \\or,\ \frac{I_1}{I_2}=\frac{\omega_2}{\omega_1} \\or,\ \frac{\frac{2}{5}MR^2}{\frac{2}{5}M{(\frac{R}{2})}^2}= \frac{\omega_2}{\omega_1} \\or,\ 4= \frac{\omega_2}{\omega_1} \\or,\ 4=\displaystyle{\frac{\frac{2\pi}{T_2}}{ \frac{2\pi}{T_1}}} \\or,\ 4=\frac{T_1}{T_2} \\or,\ T_2=\frac{T_1}{4}=\frac{24}{4}=6\ hr \)

So The duration of a day would be 6 hours.

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