Question

A body is weight by spring balance to be 1 kg atta not to how much it will wait at the equator account for the earth rotation only

Answer 1

Sol. Let g’ be the acceleration due to gravity at equation & that of pole = g

g’ = g ω^2 R

= 9.81 – (7.3 * 10^-5)^2 * 6400 * 10^3

= 9.81 – 0.034

= 9.776 m/s^2

Mg’ = 1 kg * 9.776 m/s^2

= 9.776 N or 0.997 kg

The body will weight 0.997 kg at equator