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Q.
The height at which the weight of a body becomes $\left(\frac{1}{16}\right)^{\text {th }}$, its weight on the surface of earth (radius $R$ ), is
AIPMTAIPMT 2012Gravitation
Solution:
Acceleration due to gravity at a height $h$ from the surface of earth is
$g^{\prime}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}$
where $g$ is the acceleration due to gravity at the surface of earth and $R$ is the radius of earth.
Multiplying by $m$ (mass of the body) on both sides in (i), we get
$m g^{\prime}=\frac{m g}{\left(1+\frac{h}{R}\right)^{2}}$
$\therefore $ Weight of body at height $h, W=m g^{\prime}$
Weight of body at surface of earth, $W=m g$
According to question, $W^{\prime}=\frac{1}{16} W$
$\therefore \frac{1}{16}=\frac{1}{\left(1+\frac{h}{R}\right)^{2}}$
$\left(1+\frac{h}{R}\right)^{2}=16$
or $ 1+\frac{h}{R}=4$
or $\frac{h}{R}=3$
or $h=3 R$