Answer in Brief

**Answer the following question in detail.**

Obtain the formula for the acceleration due to gravity at the depth ‘d’ below the Earth’s surface.

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#### Solution

- The Earth can be considered to be a sphere made of a large number of concentric uniform spherical shells.
- 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.
- The acceleration due to gravity on the surface of the Earth is, g = `"GM"/"R"^2`
- 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) - 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. - 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) - 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.

Is there an error in this question or solution?

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