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Question:

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

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

     

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

  4. $1+\frac{r}{R_{E}}-\frac{r^{2}}{R_{E}^{2}}-\frac{r^{3}}{R_{E}^{3}}$


Correct Option: , 4

Solution:

$g_{\text {up }}=\frac{g}{\left(1+\frac{r}{R}\right)^{2}}$

$g_{\text {down }}=g\left(1-\frac{r}{R}\right)$

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

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

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

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