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 : \(\left.{r}<{R}_{{E}}\right)\)

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Correct Answer

(D) \(1+\frac{{r}}{{R}_{{E}}}-\frac{{r}^{2}}{{R}_{{E}}^{2}}-\frac{{r}^{3}}{{R}_{{E}}^{3}}\)

Detailed explanation

\(g_{u p}=\frac{g}{\left(1+\frac{r}{R}\right)^{2}}\)

\(g_{d o w n}=g\left(1-\frac{r}{R}\right)\)

\(\frac{{g}_{\text {down }}}{{g}_{{up}}}=\left(1-\frac{{r}}{{R}}\right)\left(1+\frac{{r}}{{R}}\right)^{2}\)

\(=\left(1-\frac{{r}}{{R}}\right)\left(1+\frac{2 {r}}{{R}}+\frac{{r}^{2}}{{R}^{2}}\right)\)

\(=1+\frac{{r}}{{R}}-\frac{{r}^{2}}{{R}^{2}}-\frac{{r}^{3}}{{R}^{3}}\)

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