Question

find the value of acceleration due to gravity at the depth of our by to below the Earth surface​

Answer 1

Effect of depth on value of acceleration due to gravity.

Consider earth to be a homogeneous sphere of radius R and mass M with centre at O. Let g be the value of acceleration due to gravity at a point A on the surface of earth. Then

$\sf{g=\dfrac{GM}{R^2}}$

If $\sf{\rho}$ is the uniform density of the earth, then

$\sf{M=\dfrac{4}{3}\pi\:R^3\rho}$

$\sf{\therefore\:g=\dfrac{G\times\dfrac{4}{3}\pi\:R^3\rho}{R^2}}$

$\sf{\implies\:g=\dfrac{4}{3}\pi\:GR\rho.....(1)}$

Let g' be the acceleration due to gravity at point B at a depth d below the surface of earth.

The distance of the point B from the centre of the earth is (R-d). Tue earth can be supposed to be made of a smaller sphere of radius (R-d) and a spherical shell of thickness d.

The body at B is inside the spherical shell of thickness d. The force on body of mass m at B due to spherical shell is zero.

The body at B is outside the surface of smaller sphere of radius (R-d). The force on the body of mass m at B is due smaller sphere of earth of radius (R-d) is just as if the entire mass M' of the smaller sphere of earth is concentrated at the centre O.

$\sf{\therefore\:g'=\dfrac{GM'}{\left(R-\right)^2}and}\\\sf{M'=\dfrac{4}{3}\pi\left(R-d\right)^3\rho}$

$\sf{g'=\dfrac{G\times\dfrac{4}{3}\pi\left(R-d\right)^3\rho}{\left(R-d\right)^2}}$

$\sf{\implies\:g'=\dfrac{4}{3}\pi\:G\left(R-d\right)\rho.....(2)}$

Dividing (2) by (1), we get

$\sf{\dfrac{g'}{g}=\dfrac{\dfrac{4}{3}\pi\:G\left(R-d\right)\rho}{\dfrac{4}{3}\pi\:GR\rho}}\\\sf{=\dfrac{R-d}{R}}$

$\boxed{\sf{g'=g\left(1-\dfrac{d}{R}\right)}}$

From this we can say that value of acceleration due to gravity decreases with depth.