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 $r<R_E$)

• 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

The correct option is B $1+\frac{r}{R_E}-\frac{r^2}{R_E^2}-\frac{r^3}{R_E^3}$

Acceleration due to gravity at depth $r$ is $g_{depth}=g(1-\frac{r}{R})$
Acceleration due to gravity at height $r$ is $g_{height}=g{(1+\frac{r}{R})}^{-2}$

Ratio of gravity is given by $\frac{g_{depth}}{g_{height}}=\frac{g(1-\frac{r}{R})}{g{(1+\frac{r}{R})}^{-2}}=(1-\frac{r}{R}){(1+\frac{r}{R})}^2$
$\frac{g_{depth}}{g_{height}}=1+\frac{r}{R}-\frac{r^2}{R^2}-\frac{r^3}{R^3}$