Question

Derive an expression for acceleration due to gravity at a depth 'd' below the earth's earth surface.

Answer 1

Step 1: Consideration

Consider M as the mass of the earth, R be the radius of the earth, g_d be acceleration due to gravity at depth ′ d ′ from the earth's surface, and g be gravitational acceleration on the earth's surfaces, $\rho$ be the density of the earth.′ p ′ be the point inside the earth at depth ′ d ′ from earth surfaces.

Step 2: Derivation

$$\begin{gathered} \therefore CS-CP=d \\ andCP=R-d......\left(1\right) \end{gathered}$$ $(\sin ceCS=R)$

We know that $g=GM/R^2$ Where G is the universal gravitational constant

$$\begin{gathered} \therefore g=\frac{G\rho V}{R^2}=G\frac{4}{3R^2}\rho \pi R^3(\because \rho =M/VandM=\frac{4}{3}\pi R^3) \\ \therefore g=\frac{4G\pi R\rho }{3}————–\left(2\right) \end{gathered}$$

$g_d=accelerationduetogravityatdepth\prime d\prime$

$g_d=G\times \frac{MassofthespherewithradiusCP}{{\left(CP\right)}^2}$

$g_d=\frac{Gm}{{(R-d)}^2}$

$\therefore g_d=\frac{G\rho 4\pi {(R-d)}^3}{3{(R-d)}^2}——–\left(3\right)$

Dividing eq. (3) by eq(2)

$\frac{g_d}{g}=\frac{R-d}{R}$

$\therefore g_d=g(1-d/R)$

The above relation is the required expression.