How far from the Earth's surface does the acceleration due to gravity becomes 16%of its value .find the surface of earh R=6400km
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Acceleration due to gravity at height h is given as
g’ = g [R / (R + h)]^2
4g/100 = g [R / (R + h)]^2
0.04 = [R / (R + h)]^2
0.2 = R / (R + h)
0.2R + 0.2h = R
0.2h = 0.8R
h = 4R
It will be 4% that on surface of earth at height equal to 4R (where R is radius of Earth)
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