Question

Assuming that the earth is a sphere of radius $R_{E}$ with uniform density, the distance from its centre at which the acceleration due to gravity is equal to $\frac{g}{3}$ ($g$ = the acceleration due to gravity on the surface of earth) is

• A. $\frac{{{R}_{E}}}{2}$
• B. $2\,\,\frac{{{R}_{E}}}{3}$
• C. $\frac{R_{E}}{3}$
• D. $\frac{{{R}_{E}}}{4}$

Answer 1

Answer: (C)

$\frac{R_{E}}{3}$

Answer 2

For an $h$ depth below from the surface
we know that,
Given: $g'=g'=g(1-h / R)$
$g / 3=g(1-h / R)$
$\Rightarrow h=2 R / 3$ (depth from top)
Hence, from centre
$h'=R-h=R_{E} / 3$