Question

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

Answer 1

Let $M$ be mass of the earth$,$ $R$ be the radius of the earth

$g_d$ be gravitational acceleration at depth ${}^{\prime }{d}^{\prime }$ from the earth surface

$g$ be gravitational acceleration on the earth surfaces$.$

$p$ be the density of the earth$.$

${}^{\prime }{p}^{\prime }$ be the point inside the earth at depth ${}^{\prime }{d}^{\prime }$ from earth surfaces$.$

$\therefore CS-CP=d,\therefore CP=R-d$-----------$\left(1\right)$ $($since CS=R$)$

$g=\frac{GM}{R^2}$

$\therefore g=\frac{G\frac{4}{3}\pi R^{3}p}{R^2}$

$\therefore g=\frac{4G\pi Rp}{3}$--------------$\left(2\right)$

$gd=$ acceleration due to gravity at depth ${}^{\prime }{d}^{\prime }$

$g_d=\frac{G\times MassofthespherewithradiusCP}{C{p}^2}$

$\therefore g_d=\frac{G\frac{4}{3}\pi C{P}^3\rho }{C{P}^2}$

$\therefore g_d=\frac{4G\pi CP\rho }{3}$-----------$\left(3\right)$

Dividing eq$.$ $\left(3\right)$ by eq$\left(2\right)$

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

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