If the radius of the earth decreases by `(1)/(2)%`, then what will be the change in the length of a day? Assume that the earth is a uniform sphere and its moment of inertia, `I= (2)/(5) "MR"^(2)`, where M and R are the mass and the radius of the earth.

Explanation

If the mass of a solid sphere remains unaltered, then its moment of inertia `prop ("radius")^(2)`. Here the changed radius = `(100-(1)/(2))/(100) R = (129)/(200)R`. So, if the moment of inertia of the earth for its present radius R is `I` and the moment of inertia for its changed radius is `I`., then
`(I)/(I.)= (R^(2))/((199R)/(200))^(2)= ((200)/(199))^(2)`
If the present angular velocity of the earth is `omega` and its changed angular velocity is `omega.` then according to the principle of conservation of angular momentum,
`Iomega = I. omega.`
or, `omega. (Iomega)/(I.)`
or, `(2pi)/(T.)=(I)/(I.)xx(2pi)/(T)`
or, `T.=(I.)/(I).T= ((199)/(200))^(2)xx24= 23.76 `h
`:.` The length of the day will decrease by (24-23.76)= 0.24 h =14 min 24 s.