Question

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

• A. 5R
• B. 15R
• C. 3R
• D. 4R

Answer 1

Answer: (C)

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
$m{g}^{\prime }=\frac{mg}{{\left(1+\frac{h}{R}\right)}^2}$
$\therefore$ Weight of body at height $h,W=m{g}^{\prime }$
Weight of body at surface of earth, $W=mg$
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 $h=3R$