If RE 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 < RE)
(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 answer: (4) \(1 +\frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}\)
Explanation:
gup = \(\frac{g}{\Big(1 + \frac{r}{R}\Big)^2}\)
gdown = g\(\Big(1 - \frac{r}{R}\Big)\)
\(\frac{g_{down}}{g_{up}}\) = \(\Big(1 - \frac{r}{R}\Big)\)\(\Big(1 + \frac{r}{R}\Big)^2\)
= \(\Big(1 - \frac{r}{R}\Big)\)\(\Big(1 + \frac{2r}{R} + \frac{r^2}{R^2}\Big)\)
= \(1 +\frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}\)