At what height above the Earth's surface does the acceleration due to gravity decrease by 1%? (Radius of Earth = 6400 km)

64 km
16 km
128 km
32 km

Correct answer: 32 km

Solution

gₕ = g*(1+h/R)⁻² ≈ g*(1-2h/R) for h << R. For 1% decrease: gₕ = 0.99*g => 1-2h/R = 0.99 => 2h/R = 0.01 => h = 0.01*R/2 = 0.005*6400 = 32 km.

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