Gravity Reduce the gravity again

For the value of $g$ below the surface of Earth, we have

$g'=g\left( 1-\frac { d }{ R } \right)$

where $g'$ is the effective gravitational acceleration

$g$ is acceleration of gravity at surface of Earth = $9.8 m/s^2$

$d$ is the depth below the surface and

$R$ is the Radius of Earth = $6400 km$

Since effective gravity is asked as $36 percent$ we get

$\frac { 36 }{ 100 } g=g\left( 1-\frac { d }{ 6400 } \right)$

$\frac { 36 }{ 100 } =\left( 1-\frac { d }{ 6400 } \right)$

$\frac { d }{ 6400 } =1-\frac { 36 }{ 100 }$

$\frac { d }{ 6400 } =\frac { 64 }{ 100 }$

$\frac { d }{ 64 } =64$

$d=64\times 64$

$d=4096$

By expressing newton's law of gravitational acceleration as a function of density instead of mass, one finds that the relationship in this case is the product: (radius) (density) (constant terms). Therefore just multiply the radius of the earth by .36 and subtract the result from the radius itself to find the correct answer.

One could prove that there's a linear relationship between gravitational force and distance from Earth's center, and find the depth when the gravity at that point is 36% of that on the surface.

g' = g( 1-d/r)

hence d=.64 R

ie d= .64*6400 km