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If earth contracts to half of its radius...

If earth contracts to half of its radius, what would be the duration of the day?

Text Solution

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We know `I_omega` = constant.
Here I = moment of inertia
`omega` = angular velocity.
Let `I_1, I_2` be two values of moment of inertia for `omega_1` and `omega_2` then from law of conservation of angular momentum, we have
`I_1omega_1 = I_2 omega_2`
`I_1 (2pi)/(T_1) = I_2 (2pi)/(T_2) [because omega = (2pi)/T]`
`implies (I_1)/(T_1) = (I_2)/(T_2)`
`implies T_2 = (I_2)/(I_1) xx T_1 " " .....(i)`
As earth is perfect solid sphere, so `I = 2/5 MR^2`
`implies I_1 = 2/5 MR_1^2 , I_2 = 2/5 MR_2^2`
Also, `R_2 = (R_1)/2implies (I_2)/(I_1) = (2/5)(MR_1^2)/((2/5)MR_1^2) = 2/5 (MR_1^2)/(4) xx 1/(2/5 MR_1^2) implies (I_2)/(I_1) = 1/4`
Putting `(I_2)/(I_1) = 1/4` in equation (i), we have
`T_2 = (T_1)/(4) .` As `T_1 = 24` hour, therefore `T_2 = 24/4 = 6` hour.
So when earth contracts to half of its radius, duration of day will become 6 hour.
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