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If ${R}_{{E}}$ be the radius of Earth, then the ratio between the acceleration due to gravity at a depth $' {r} '$ below and a height $' r '$ above the earth surface is:
(Given : $\left.{r}<{R}_{{E}}\right)$
$1-\frac{{r}}{{R}_{{E}}}-\frac{{r}^{2}}{{R}_{{E}}^{2}}-\frac{{r}^{3}}{{R}_{{E}}^{3}}$
$1+\frac{{r}}{{R}_{{E}}}+\frac{{r}^{2}}{{R}_{{E}}^{2}}+\frac{{r}^{3}}{{R}_{{E}}^{3}}$
$1+\frac{{r}}{{R}_{{E}}}-\frac{{r}^{2}}{{R}_{{E}}^{2}}+\frac{{r}^{3}}{{R}_{{E}}^{3}}$
$1+\frac{{r}}{{R}_{{E}}}-\frac{{r}^{2}}{{R}_{{E}}^{2}}-\frac{{r}^{3}}{{R}_{{E}}^{3}}$
Solution
$g_{u p}=\frac{g}{\left(1+\frac{r}{R}\right)^{2}}$
$g_{d o w n}=g\left(1-\frac{r}{R}\right)$
$\frac{{g}_{\text {down }}}{{g}_{{up}}}=\left(1-\frac{{r}}{{R}}\right)\left(1+\frac{{r}}{{R}}\right)^{2}$
$=\left(1-\frac{{r}}{{R}}\right)\left(1+\frac{2 {r}}{{R}}+\frac{{r}^{2}}{{R}^{2}}\right)$
$=1+\frac{{r}}{{R}}-\frac{{r}^{2}}{{R}^{2}}-\frac{{r}^{3}}{{R}^{3}}$
