Question

If $R_E$ be the radius of Earth, then the ratio between the acceleration due to gravity at a depth ${}^{\prime }{r}^{\prime }$ below and a height ' $r$ ' above the earth surface is : (Given : $r<R_E$ )

• A. $1-\frac{r}{R_E}-\frac{r^2}{R_E^2}-\frac{r^3}{R_E^3}$
• B. $1+\frac{r}{R_E}+\frac{r^2}{R_E^2}+\frac{r^3}{R_E^3}$
• C. $1+\frac{r}{R_E}-\frac{r^2}{R_E^2}+\frac{r^3}{R_E^3}$
• D. $1+\frac{r}{R_E}-\frac{r^2}{R_E^2}-\frac{r^3}{R_E^3}$

Answer 1

Answer: (D)

$g_{\text{up }}=\frac{g}{{\left(1+\frac{r}{R}\right)}^2}$
$g_{\text{down }}=g\left(1-\frac{r}{R}\right)$
$\frac{g_{\text{down }}}{g_{\text{up }}}=\left(1-\frac{r}{R}\right){\left(1+\frac{r}{R}\right)}^2$
$=\left(1-\frac{r}{R}\right)\left(1+\frac{2r}{R}+\frac{r^2}{R^2}\right)$
$=1+\frac{r}{R}-\frac{r^2}{R^2}-\frac{r^3}{R^3}$