If the diurnal motion of the earth stops due to some reason, what would be the percentage change in the weight of a body at the equator ? Radius of the earth `=6400 km and g=9.8 m*s^(-2)`.

Explanation

Due to diurnal motion of the earth, the apparent weight of a body at the equator
`W^.=W-momega^2R`
When the earth stops rotating ,`omega=0` and hence `W^.=W`. In this case, increase in weight of the object,
`Delta W=W-W^.=W-(W-momega^2R)=momega^2R`
`therefore` Percentage increases in weight
`=(DeltaW)/(W)xx100=(momega^2R)/(mg)xx100=((2pi)/T)^2xxR/gxx100 `
`=(4pi^2)/((24xx60xx60)^2)xx(6400xx10^3)/(9.8)xx100=0.3454%`.