Question 80 of 223

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 : $r\lt {R}_{E}$)

A$1+\dfrac{r}{{R}_{E}}-\dfrac{{r}^{2}}{{R}_{E}^{2}}-\dfrac{{r}^{3}}{{R}_{E}^{3}}$B$1+\dfrac{r}{{R}_{E}}+\dfrac{{r}^{2}}{{R}_{E}^{2}}+\dfrac{{r}^{3}}{{R}_{E}^{3}}$C$1-\dfrac{r}{{R}_{E}}-\dfrac{{r}^{2}}{{R}_{E}^{2}}-\dfrac{{r}^{3}}{{R}_{E}^{3}}$D$1+\dfrac{r}{{R}_{E}}-\dfrac{{r}^{2}}{{R}_{E}^{2}}+\dfrac{{r}^{3}}{{R}_{E}^{3}}$

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Answer: (A) $1+\dfrac{r}{{R}_{E}}-\dfrac{{r}^{2}}{{R}_{E}^{2}}-\dfrac{{r}^{3}}{{R}_{E}^{3}}$

$\dfrac{{g}_{h}}{{g}_{d}}=\dfrac{\dfrac{GM}{{\left({R}_{E}+r\right)}^{2}}}{\dfrac{GM}{{R}_{E}^{3}}\left({R}_{E}-r\right)}=\dfrac{{R}_{E}^{3}}{{\left({R}_{E}+r\right)}^{2}\left({R}_{E}-r\right)}$
$\dfrac{{g}_{d}}{{g}_{h}}=\dfrac{{\left({R}_{E}+r\right)}^{2}\left({R}_{E}-r\right)}{{R}_{E}^{3}}=\dfrac{{R}_{E}^{3}-{R}_{E}{r}^{2}+r{R}_{E}^{2}-{r}^{3}}{{R}_{E}^{3}}=1-\dfrac{{r}^{2}}{{R}_{E}^{2}}+\dfrac{r}{{R}_{E}}-{\left(\dfrac{r}{{R}_{E}}\right)}^{3}$