Question

A body is weighed by a spring balance to be 1.000 Kg at the north pole.How much will it weigh at the equator? Account for the earth's rotation only.

Answer 1

Let g' be the acceleration due to gravity at equation & that of pole = g

$$\omega_{earth}= 7.3 \times 10^{-5}$$

$$ g' = g - \omega^2 R $$

$$g' = 9.81 - (7.3 \times 10^{-5} )^2 \times 6400 \times 10^3 $$

$$g' = 9.81 - 0.034$$

$$g' =9.776 m/s^2 $$

$$ mg = 1 kg \times 9.77 m/s^2 $$

$$ = 9.776\ N\ or\ 0.997\ Kg$$

The body will weigh 0.007 Kg at equator