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
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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