Deepak Patra
Last Activity: 10 Years ago
Sol. At equator, g’ = g – ω^2R …(1)
Let at ‘h’ height above the south pole, the acceleration due to gravity is same.
Then, here g’ = g (1 – 2h/R) …(2)
∴ g - ω^2 R = g (1 – 2h/R)
or 1 - ω^2R/g = 1 2h/R
or h = ω^2R^2/2g = (7.3 * 10^-5)^2 * (6400 * 10^3)^2/2 * 9.81 = 11125 N = 10Km (approximately)