Calculate the change in weight of a 20 kg body, when it is taken from equator to pole of earth.

Explanation

At equator `g_(e) = g - R omega^(2)`
At pole `g_(P) = g`
Therefore,
`mg_(e) - mg_(P) = m(g_(e) - g_(P))`
`=20(g - R omega^(2) - g)`
`= 20 R omega^(2)` (Ignoring negative sign)
`mg_(e) - mg_(P) = 20 xx R omega^(2) = 20 xx R xx (4pi^(2))/(T^(2))`
`= (20 xx 6,400 xx 1,000 xx 4 xx 3.14 xx 3.14)/((24 xx 3,600)^(2))`
= 0.676 N