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 $r < R_{E}$ )

1 + \frac{r}{R_{E}} + \frac{r^{2}}{R_{E}^{2}} + \frac{r^{3}}{R_{E}^{3}}

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1 + \frac{r}{R_{E}} - \frac{r^{2}}{R_{E}^{2}} - \frac{r^{3}}{R_{E}^{3}}

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1 - \frac{r}{R_{E}} - \frac{r^{2}}{R_{E}^{2}} - \frac{r^{3}}{R_{E}^{3}}

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1 + \frac{r}{R_{E}} - \frac{r^{2}}{R_{E}^{2}} + \frac{r^{3}}{R_{E}^{3}}

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Solution

The correct option is B $1 + \frac{r}{R_{E}} - \frac{r^{2}}{R_{E}^{2}} - \frac{r^{3}}{R_{E}^{3}}$
Acceleration due to gravity at depth $r$ is $g_{d e p t h} = g (1 - \frac{r}{R})$
Acceleration due to gravity at height $r$ is $g_{h e i g h t} = g {(1 + \frac{r}{R})}^{- 2}$

Ratio of gravity is given by $\frac{g_{d e p t h}}{g_{h e i g h t}} = \frac{g (1 - \frac{r}{R})}{g {(1 + \frac{r}{R})}^{- 2}} = (1 - \frac{r}{R}) {(1 + \frac{r}{R})}^{2}$
$\frac{g_{d e p t h}}{g_{h e i g h t}} = 1 + \frac{r}{R} - \frac{r^{2}}{R^{2}} - \frac{r^{3}}{R^{3}}$

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If RE 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<RE)

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