If R E be the radius of Earth, then the ratio between the acceleration due to gravity at a depth r below and…

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

Solution

$\frac{g_h}{g_d}=\frac{\frac{GM}{{\left(R_E+r\right)}^2}}{\frac{GM}{R_E^3}\left(R_E-r\right)}=\frac{R_E^3}{{\left(R_E+r\right)}^2\left(R_E-r\right)}$
$\frac{g_d}{g_h}=\frac{{\left(R_E+r\right)}^2\left(R_E-r\right)}{R_E^3}=\frac{R_E^3-R_E{r}^2+r{R}_E^2-r^3}{R_E^3}=1-\frac{r^2}{R_E^2}+\frac{r}{R_E}-{\left(\frac{r}{R_E}\right)}^3$

Asked in: JEE Main 2021 (31 Aug Shift 2)

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