A person moves from a point 16 km below the Earth's surface to a point 16 km above the Earth's surface. Taking the Earth's radius as 6400 km and using the usual first-order approximations for small height and small depth, what is the approximate magnitude of the percentage change in the acceleration due to gravity?
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Question

A person moves from a point 16 km below the Earth's surface to a point 16 km above the Earth's surface. Taking the Earth's radius as 6400 km and using the usual first-order approximations for small height and small depth, what is the approximate magnitude of the percentage change in the acceleration due to gravity?

Options

G1 Physics · Accepted Answer

For a small height h above the Earth's surface, g(above) ≈ g[1-2h ÷ R]. For a small depth h below the surface, g(below) ≈ g[1-h ÷ R]. Therefore the magnitude of the change is |Δg| = g(below)-g(above) = gh ÷ R. Hence, |Δg| ÷ g = h ÷ R = 16 ÷ 6400 = 0.0025. Percentage change: 0.0025×100 = 0.25%. Correct option: B.