Question

016. Derive an expression for g acceleration due to gravity below the Earth surface and
above the earth surface.​

Answer 1

for above the earth surface,

$g=G \times \frac{m}{ {r}^{2} } - - (1)$

$g'= G \times \frac{m}{(r + h) ^{2} } - - (2)$

now,

$\frac{g}{g'} = \frac{ G \times m \times (r + h) ^{2} }{G \times m \times {r}^{2} }$

$\frac{g}{g'} = \frac{ {(r+h)}^{2} }{ {r}^{2} }$

$g {r}^{2} = g' {(r + h)}^{2}$

$g = g' (\frac{ {r}^{2} }{ {r}^{2} } + \frac{ {h}^{2} }{ {r}^{2} } + \frac{2hr}{ {r}^{2} } )$

$g=g'(1+ \frac{h}{r} )^{2}$

or

$g'=g(1+ \frac{h}{r} ) ^{-2}$

by binomial theorem,

$g' = g(1 - \frac{2h}{r} )$

for below the earth surface,

let the density of earth be d,

and

volume at surface,

$v = \frac{4}{3} \pi {r}^{3}$

volume at depth h,

$v'= \frac{4}{3} \pi(r-h) ^{3}$

so,

$m=volume \times density$

$m = \frac{4}{3} \pi {r}^{3} \times d$

and

$m'= \frac{4}{3} \pi ({r-h})^{3} \times d$

$g=G \times \frac{m}{ {r}^{2} } - - (1)$

$g'=G \times \frac{m'}{ ({r - h})^{2} } - - (2)$

now putting the values in eq.

$g=G \times \frac{ \frac{4}{3} \pi {r}^{3} \times d }{ {r}^{2} }$

$g=G \times \frac{4}{3} \pi rd$

and,

$g'=G \frac{ \frac{4}{3} \pi {(r - h) }^{3} d}{ {(r-h) }^{3} }$

$g'=G \frac{4}{3} \pi (r-h) d$

on dividing,

$\frac{g'}{g} = \frac{ G \times \frac{4}{3} \pi (r-h)d}{G \times \frac{4}{3} \pi rd}{ }$

$\frac{g'}{g} =\frac{r - h}{r}$

$g' = g(1- \frac{r}{h} )$