A body reads 49 N on a spring balance at the north pole. What will the same balance read at the equator, given g = G*M/R² = 9.8 m/s² and Earth's radius R = 6400 km?

49 N
48.83 N
49.83 N
49.17 N

Correct answer: 48.83 N

Solution

At the pole there is no rotational effect, so W_pole = m*g = 49 N, giving m = 5 kg. At the equator the centripetal term reduces the reading: W_eq = m*(g - omega²*R). omega = 2*pi/(86400 s) = 7.27*10⁻⁵ rad/s. omega²*R = (7.27e-5)² * 6.4e6 = 5.29e-9 * 6.4e6 = 0.0338 m/s². So W_eq = 5*(9.8 - 0.0338) = 5*9.766 = 48.83 N.

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