How many times faster than its present speed the earth should rotate so that the apparent weight of an object at equator becomes zero ? Given radius of the earth = `6.37 xx 10^(6)` m . What would be the duration of the day in that case ?

Answer

17

Explanation

At the equator, we have
`g_(lambda)=g-Romega_(1)^(2)ormg_(lambda)=mg-mRomega_(1)^(2)`
As the apparent weight at equator is zero,
So `mg_(lambda)=0`
`therefore mg-mRomega_(1)^(2)=0`
`therefore omega_(1)=sqrt((g)/(R))=sqrt((9.8)/(6.37xx10^(6)))=1.241xx10^(-3)"rad s"^(-1)`
But the present speed of the earth is
`omega=(2pi)/(T)=(2pi)/(24xx60xx60)=7.27xx10^(-5)"rad s"^(-1)`
`therefore (omega_(1))/(omega)=(1.241xx10^(-3))/(7.27xx10^(-5))=17.06~=17`.
or `omega_(1)=17omega`
i.e. the earth should rotate 17 times faster than its present speed.
New duration of the day `=(24)/(17)=1.412h`.