The earth has a mass of `6 xx 10^(24) kg` and a radius of `6.4 xx 10^(6)m`. Calculate the amount of work that must be done to slow down its rotation so that duration of day becomes `30` hrs instead of 24 hours.
Moment of inertia of earth `= (2)/(5)MR^(2)`.

Explanation

Here, `M = 6 xx 10^(24) kg , R = 6.4 xx 10^(6) m`,
`W =?, T_(1) = 24h, T_(2) = 30h`
Work done = Increase in `K.E.` of rotation
`W = E_(2) - E_(1) = (1)/(2)I (omega_(2)^(2) - omega_(1)^(2))`
`=(1)/(2) xx ((2)/(5)MR^(2)) [((2pi)/(T_(2)))^(2) - ((2pi)/(T_(1)))^(2)]`
`=(4pi^(2))/(5) xx MR^(2) [(1)/(T_(2)^(2))-(1)/(T_(1)^(2))]`
`W = 4xx(22)/(7) xx (22)/(7 xx 5) xx 6 xx 10^(24) (6.4 xx 10^(6))^(2)`
`[(1)/((30 xx 60 xx 60)^(2))-(1)/((24 xx 60 xx 60)^(2))]`
`W =- 9.36 xx 10^(28) J`
Negative sign indicates that work is done against the rotation of earth in order to slow down its speed.