Question:

The value of acceleration due to gravity at Earth's surface is 9.8ms^-2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9ms^-2, is close to :(Radius of earth =6.4 × 10^6m)

1.6×106m

6.4×106m

9.0×106m

2.6×106m

Correct option is D. 2.6×106m
(G/(R+h)^2) = (G/R^2)*(1/2)
(R+h)^2 = 2R^2
R+h = √2R
h = (√2 - 1)R ≈ 2.6 × 10^6m