Question

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 \(\omega\). 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)

• A. \(\frac{R^2\omega^2}{2g}\)
• B. \(\frac{R^2\omega^2}{g}\)
• C. \(\frac{R^2\omega^2}{8g}\)
• D. \(\frac{R^2\omega^2}{4g}\)

Hint:

At the equator the effective g is reduced by the centrifugal term: \(g_{eq} = g - \omega^2R\). At height h (h << R), \(g_h \approx g\left(1-\frac{2h}{R}\right)\). Equate the two.

Answer 1

The Correct Option is A

Weight at the equator (spring balance, rotating frame): \(W_{eq} = m(g - \omega^2R)\), with g the value at the poles (the radius is taken as R).
Weight at a height h above the pole (no rotation effect on the axis): \(W_h = m\,g\left(\frac{R}{R+h}\right)^2 \approx mg\left(1 - \frac{2h}{R}\right)\) for h << R.
Equal weights: \(g - \omega^2R = g - \frac{2gh}{R}\).
\(\frac{2gh}{R} = \omega^2R \Rightarrow h = \frac{R^2\omega^2}{2g}\).
Answer: (1) \(\frac{R^2\omega^2}{2g}\)