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The height at which the weight of a body becomes $\frac{1}{16}$th, its weight on the surface of earth (radius R), is
A
4R
B
5R
C
15R
D
3R
Detailed Solution
$g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$
$\frac{g}{16} = \frac{g}{\left(1 + \frac{h}{R}\right)^2} \Rightarrow \left(1 + \frac{h}{R}\right)^2 = 16$
$1 + \frac{h}{R} = 4 \Rightarrow h = 3R$
$\frac{g}{16} = \frac{g}{\left(1 + \frac{h}{R}\right)^2} \Rightarrow \left(1 + \frac{h}{R}\right)^2 = 16$
$1 + \frac{h}{R} = 4 \Rightarrow h = 3R$