The acceleration due to gravity on the earth’s surface at the poles is g and angular velocity of the earth about the axis passing through the pole is ω. An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is (h << R, where R is the radius of the earth)

Solution

Physics sir · 4 steps

Rotation trims ω² R off g at the equator; climbing h trims 2gh/R at the pole. Set them equal.

1. At the equator. The spring balance reads the effective weight. At the equator the object moves in a circle of radius R, so part of gravity goes into the centripetal force. This gives g_(eq) = g - ω² R.
2. Above the pole. The pole lies on the rotation axis, so rotation has no effect there. At height h ≪ R, the binomial approximation gives gₕ = g(1 + h/R)⁻² ≈ g(1 - 2h/R).
3. Equal weights. The mass is the same in both places, so equal weights mean equal effective g. Set the two values equal and cancel g from both sides.

   g - ω² R = g - 2gh/R ⇒ h = R²ω²/2g

4. So. The height is h = R²ω²/2g, which is option (a).

The idea to keep. Near the surface, height lowers g by 2gh/R and depth lowers it by gd/R. Rotation lowers it by ω² R cos² λ, which is largest at the equator and zero at the poles.

Answer from the NTA final answer key. Solution by PrepWiser.