At approximately what height above Earth's surface does the gravitational acceleration fall to one-third of its surface value? (Radius of Earth R = 6400 km, sqrt(3) = 1.732)

3840 km
4685 km
2133 km
4267 km

Correct answer: 4685 km

Solution

Setting gₕ = g/3: R²/(R+h)² = 1/3, so (R+h)² = 3R², giving R+h = R*sqrt(3). Therefore h = R*(sqrt(3)-1) = 6400*(1.732-1) = 6400*0.732 = 4684.8 km ≈ 4685 km.

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