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Let omega be the angular speed of Earth's rotation. Assume that, ignoring rotation, g has the same value at the equator and at the poles. An object weighed at the equator gives the same reading as it would at a depth d below the surface at a pole (with d << R). Find d.

1. omega²*R²/g
2. omega²*R²/(2g)
3. 2*omega²*R²/g
4. sqrt(R*g)/g

Correct answer: omega²*R²/g

Solution

Equator weight uses g_eq = g - omega²*R. Depth weight uses g_d = g(1 - d/R). Equating the two effective gravities gives d.

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