Q. 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 below the surface of the earth in $ms^{- 2}$ is about
A) 8.5
B) 9.5
C) 9.8
D) 11.5
Explanation
(b) Given, $g_{h} = 9 = \frac{gR^{2}}{(R + R/20)^{2}} = \frac{20 \times 20}{21 \times 21}g$
or $g = \frac{9 \times 21 \times 21}{20 \times 20}$
Now, $g_{d} = g\left( 1 - \frac{d}{R} \right)$
$= \frac{9 \times 21 \times 21}{20 \times 20}\ \left\lbrack 1 - \frac{R/20}{R} \right\rbrack = 9.5ms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.