Answer 1

1. The Earth can be considered to be a sphere made of a large number of concentric uniform spherical shells.
2. When an object is on the surface of the Earth it experiences the gravitational force as if the entire mass of the Earth is concentrated at its center.
3. The acceleration due to gravity on the surface of the Earth is, g = `"GM"/"R"^2`
4. Assuming that the density of the Earth is uniform, mass of the Earth is given by 
   M = volume × density = `4/3 pi"R"^3rho`
   ∴ g = `("G" xx 4/3 pi"R"^3rho)/"R"^2 = 4/3pi"R"rho"G"`    ....(1)
5. Consider a body at a point P at the depth d below the surface of the Earth as shown in the figure.
   
   Here the force on a body at P due to the outer spherical shell shown by the shaded region, cancel out due to symmetry.
   The net force on P is only due to the inner sphere of radius OP = R – d.
6. Acceleration due to gravity because of this sphere is,
   `"g"_"d" = "GM'"/("R - d")^2` where,
   M' = volume of the inner sphere × density
   ∴ M' = `4/3 pi("R - d")^3 xx rho`
   ∴ g_d = `("G" xx 4/3pi("R - d")^3rho)/("R - d")^2`
   ∴ g_d = `"G" xx 4/3pi("R - d")rho`     ...(2)
7. Dividing equation (2) by equation (1) we get,
   `"g"_"d"/"g" = ("R - d")/"R"`
   ∴ `"g"_"d"/"g" = 1 - "d"/"R"`
   ∴ `"g"_"d"/"g" = "g"(1 - "d"/"R")`

This equation gives acceleration due to gravity at depth d below the Earth’s surface.