Question

The acceleration to gravity at a height 1/20th of the radius of the earth above the earth surface is $9ms^{- 2}$ . Its value at a point at an equal distance below the surface of the earth in $ms^{- 2}$ is about

• A. 8.5
• B. 9.5
• C. 9.8
• D. 11.5

Answer 1

Answer: (B)

At a height $h$ ,
$g_{h}=g\left(1 - \frac{2 h}{R}\right)$
for $h=\frac{R}{20}$ we get
$9=g\left(1 - \frac{1}{10}\right)=\frac{9}{10}g$
At a depth $d$ ,
$g_{d}=g\left(1 - \frac{d}{R}\right)$
for $d=\frac{R}{20}$
$g_{d}=g\left(1 - \frac{1}{20}\right)=\frac{19}{20}g$
$\Rightarrow \frac{g_{d}}{9}=\frac{19}{18}\Rightarrow g_{d}=9.5ms^{- 2}$