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  • If be the radius of Earth, then the ratio between the acceleration due to gravity at a depth <img alt="' r '" src="https://entrancecorner.codecogs.com/gif.latex?%27%20r%20%2

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} )
 
Option: 1 1+\frac{r}{R_{E}}-\frac{r^{2}}{R_{E}^{2}}-\frac{r^{3}}{R_{E}^{3}}
Option: 2 1+\frac{r}{R_{E}}+\frac{r^{2}}{R_{E}^{2}}+\frac{r^{3}}{R_{E}^{3}}
Option: 3 1-\frac{r}{R_{E}}-\frac{r^{2}}{R_{E}^{2}}-\frac{r^{3}}{R_{E}^{3}}
Option: 4 1+\frac{r}{R_{E}}-\frac{r^{2}}{R_{E}^{2}}+\frac{r^{3}}{R_{E}^{3}}

Answers (1)

best_answer

g_{h}=\frac{GM}{\left ( R +h\right )^{2}}=\frac{GM}{\left ( R+r \right )^{2}}=\frac{GM}{R^{2}\left ( 1+\frac{r}{R} \right )^{2}}\\             ........(1)

g_{d}=g\left ( 1-\frac{d}{R} \right )=\frac{GM}{R^{2}}\left ( 1-\frac{r}{R} \right )\\

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

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

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

\frac{g_{d}}{g_{h}}= 1+\frac{r}{R}-\frac{r^{2}}{R^{2}}-\frac{r^{3}}{R^{3}}

The correct option is (1)

Posted by

vishal kumar

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