Question

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

Answer 1

Answer: 0.25

Here, $R=6400km;x=16km$
The value of acceleration due to gravity at a depth $x$,
$g^{\prime }=g\left(1-\frac{x}{R}\right)=g\left(1-\frac{16}{6,400}\right)=\frac{399}{400}g$
$\therefore g-g^{\prime }=g-\frac{399}{400}g$
$=\frac{1}{400}g=2.5\times 10^{-3}g$
If $m$ is mass of the body, then $mg$ and $mg$ ' 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
$=\frac{mg-m{g}^{\prime }}{mg}\times 100$
$=\frac{g-g^{\prime }}{g}\times 100$
$=\frac{2\times 5\times 10^{-3}g}{g}\times 100=0.25\%$