The earth has a mass of `6 xx 10^(24) kg` and a radius of `6.4 xx 10^(6)m`. Calcualte 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) = 24 hrs, T_(2) = 30 hrs`.
`I = (2)/(5)MR^(2) = (2)/(5) xx 6 xx 10^(24) xx (6.4 xx 10^(6))^(2)`
`= 9.83 xx 10^(37)kg m^(2)`
Work done = Decrease in rotational `KE`
`W = (1)/(2)I (omega_(2)^(2) - omega_(1)^(2)) = (I)/(4) xx 4pi^(2) [(1)/(T_(2)^(2))-(1)/(T_(1)^(2))]`
`=(2)/(5) MR^(2) xx 2pi^(2) ((T_(1)^(2) - T_(2)^(2))/(T_(1)^(2) T_(2)^(2)))`
`= 9.83 xx 10^(37) xx 2 xx (22)/(7) xx (22)/(7)`
`xx{((24 xx 60 xx 60)^(2) - (30 xx 60 xx 60)^(2))/((24 xx 60 xx 60)^(2) (30 xx 60 xx 60)^(2))}`
`W =- 9.36 xx 10^(28) J`