What is the difference in the values of acceleration due to gravity at the polar region and at the equator due to the diurnal motion of the earth ? Radius of the earth=6400 km.

Explanation

Angular velocity of diurnal motion of the earth
`omega=(2pi)/(24xx60xx60)rad *s^(-1)`
Radius of the earth =`6400 km =6.4xx10^8 cm`
Acceleration due to gravity at latitude `theta`,
`g^.=g(1-(omega^2R)/(g)cos^2 theta)`
`=g-omega^2R cos ^2 theta`
In the polar regions `theta =90^@ , i.e., cos theta =0`.
Therefore the acceleration due to gravity is `g_1=g`.
At the equator `theta =0^@ and cos theta =1` , thus the acceleration due to gravity is `g_2=g-omega^2R`.
So, the difference in the value of acceleration due to gravity in these two regions
`g_1-g_2=omega^2R=((2pi)^2xx6.4xx10^8)/((24xx60xx60)^2)=3.38 cm*s^(-2)`.