Let ω is the angular velocity of the earth's rotation about its own axis and acceleration due to gravity on the earth's surface at the poles is g . An object weighed by a spring balance gives the same reading at the equator, as well as at a height h above the poles ( h ≪ R ). What is the value of h ?

Step-by-step solution

The value of acceleration due to gravity at the equator is g E = g - R ω 2 and the acceleration due to gravity at a height h above the pole is g h = 1 - 2 h R g . It is given that ∵ m g E = m g h ∴ g E = g h ∴ g - R ω 2 = 1 - 2 h R g = g - 2 h g R ∴ R ω 2 = 2 h g R ∴ h = R 2 ω 2 2 g