Question

Find the percentage decrease in weight of a body, when taken $16\, km$ below the surface of the earth. Take radius of the earth as $6400\, km$.

Answer 1

Here, $R=6400\, km ; x=16\, km$
The value of acceleration due to gravity at a depth $x$,
$g'=g\left(1-\frac{x}{R}\right)=g\left(1-\frac{16}{6,400}\right)=\frac{399}{400} g$
$\therefore g-g'=g-\frac{399}{400} g$
$=\frac{1}{400} g=2.5 \times 10^{-3} g$
If $m$ is mass of the body, then $m g$ and $m g$ ' will be respectively weight of the body on the surface of earth and at a depth of $m$ below the surface of the earth. Then,
$\%$ decrease in the weight of the body
$=\frac{m g-m g'}{m g} \times 100$
$=\frac{g-g'}{g} \times 100$
$=\frac{2 \times 5 \times 10^{-3} g}{g} \times 100=0.25 \%$