Assuming that the earth is a sphere of radius RE with uniform density, the distance from its centre at which the acceleration due to gravity is equal to (g/3) (g = the acceleration due to gravity on the surface of earth) is

Question

Assuming that the earth is a sphere of radius RE with uniform density, the distance from its centre at which the acceleration due to gravity is equal to (g/3) (g = the acceleration due to gravity on the surface of earth) is (A)  (RE/2)  (B)  2 (RE/3)  (C) (RE/3) (D)  (RE/4)

Tardigrade Admin · Accepted Answer

For an h depth below from the surface we know that, Given: g'=g'=g(1-h / R) g / 3=g(1-h / R) ⇒ h=2 R / 3 (depth from top) Hence, from centre h'=R-h=RE / 3

Q. 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

$\frac{R _{E}}{2}$

$2 \frac{R _{E}}{3}$

$\frac{R _{E}}{3}$

$\frac{R _{E}}{4}$

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
$h^{'} = R - h = R_{E} /3$

Q. Assuming that the earth is a sphere of radius RE​ with uniform density, the distance from its centre at which the acceleration due to gravity is equal to 3g​ (g = the acceleration due to gravity on the surface of earth) is

2RE​​

23RE​​

3RE​​

4RE​​