Question

I f the earth were a perfect sphere of radius 6.37 x 10 6 m rotating about its axis with a period of 1 day (= 8.64 x 10 4 s ), how much would the acceleration due to gravity (g) differ from the poles to the equator ?

Answer 1

Answer:

The difference in the acceleration due to gravity at poles and equator is 3.365 x 10⁻² m/s².

Step-by-step explanation:

Let m be the mass of the object , R be the radius of earth and T is the time period of rotation of earth.

Now at poles we don't feel any movement due to the rotation of earth

Hence the net force acting on the object is F = mg₁  i.e. its weight alone.

But at equator there is also a centripetal force acting on the object due to rotation of earth which is given by mRω²

Hence net force on the object at equator, F = mg₂ + mRω²

Hence comparing both the equation, we get,

mg₁ = mg₂ + mRω²

=> g₁ - g₂ = Rω²

= R(2π/T)²

= 6.37 x 10⁶ x 4 x 3.14² / 8.64² x 10⁸

= 3.365 x 10⁻² m/s²

Hence the difference in the acceleration due to gravity at poles and equator is 3.365 x 10⁻² m/s²