Find the acceleration of a particle placed on the surface of the earth at the equator, due to the earth rotation. The radius of earth is 6400 km and time period of revolution of the earth about its axis is 24 h.

Explanation

Given, radius of earth `R_(e)=6400"km"=6400xx10^(3)m`
Time period `T=24h=24xx60xx60s`
Angular speed of the earth
`omega=(2pi)/(T)=(2pi)/(24xx60xx60)"rad s"^(-1)`
Acceleration of particle,
`a_(c)=omega^(2)R_(e)=((2pi)/(24xx60xx60))^(2)xx(6400xx10^(3))=0.0033ms^(-2)`