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The value of acceleration due to gravity at Earth's surface is 9.8 m s−2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 m s−2 is close to: (Radius of earth = 6.4 × 10⁶ m)

1. 6.4 × 10⁶ m
2. 2.6 × 10⁶ m
3. 3.6 × 10⁶ m
4. 9.0 × 10⁶ m

Correct answer: 2.6 × 10⁶ m

Solution

g' = g (R/(R+h))^2. Setting g'=g/2 gives (R+h)/R = sqrt(2), so h = R(sqrt(2)-1) = 6.4x10^6 * 0.414 = 2.6x10^6 m.

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