Question

a mine located at a depth R/3 below the surface of earth where R is the radius of the earth the acceleration due to gravity at the mine is

Answer 1

heya mate here, is ur answer  ::::---->>>

we know that,

$g=GM/R^2$............................................(i)

when the Radius is R/3.

Than acceleration due to gravity, $g' = GM/(R/3)^2$

which is equal to$9*GM/R^2$..........................(ii)

thus from eqn (i) and (ii)

we get g'=9*g

thus the new acceleration due to gravity will be 9 times more intense than original once,.

hope it helps