If R_{E} is the radius of the Earth,then the | vedclass

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(D) The acceleration due to gravity at a height $r$ above the Earth's surface is given by $g_{up} = \frac{g}{(1 + \frac{r}{R_{E}})^{2}}$.The accelerat

Vedclass · Accepted Answer

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

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(D) The acceleration due to gravity at a height $r$ above the Earth's surface is given by $g_{up} = \frac{g}{(1 + \frac{r}{R_{E}})^{2}}$.The acceleration due to gravity at a depth $r$ below the Earth's surface is given by $g_{down} = g(1 - \frac{r}{R_{E}})$.The ratio of acceleration due to gravity at depth $r$ to that at height $r$ is:$\frac{g_{down}}{g_{up}} = \frac{g(1 - \frac{r}{R_{E}})}{\frac{g}{(1 + \frac{r}{R_{E}})^{2}}} = (1 - \frac{r}{R_{E}})(1 + \frac{r}{R_{E}})^{2}$.Expanding the expression:$= (1 - \frac{r}{R_{E}})(1 + \frac{2r}{R_{E}} + \frac{r^{2}}{R_{E}^{2}})$$= 1 + \frac{2r}{R_{E}} + \frac{r^{2}}{R_{E}^{2}} - \frac{r}{R_{E}} - \frac{2r^{2}}{R_{E}^{2}} - \frac{r^{3}}{R_{E}^{3}}$$= 1 + \frac{r}{R_{E}} - \frac{r^{2}}{R_{E}^{2}} - \frac{r^{3}}{R_{E}^{3}}$.

If $R_{E}$ is the radius of the Earth,then the ratio between the acceleration due to gravity at a depth $r$ below and a height $r$ above the Earth's surface is: (Given: $r < R_{E}$)

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