If the density of the earth is doubled keeping its radius constant,then the acceleration due to gravity will be........ $m/s^2$. $(g = 9.8\,m/s^2)$

Find the gravitational field at a distance of $2000\, km$ from the center of the Earth. (in $m/s^2$)
(Given: $R_{\text{earth}} = 6400\, km$,$r = 2000\, km$,$M_{\text{earth}} = 6 \times 10^{24}\, kg$)

For a sphere of mass $M$ and radius $R$,if the gravitational forces at distances $r_1$ and $r_2$ from the center are $F_1$ and $F_2$ respectively,then:

At which location on Earth is the effective acceleration due to gravity maximum?

The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth,assumed to be a sphere of radius $R$ of uniform density,is as shown in the figure below. The correct figure is:

At what depth does the acceleration due to gravity become $\frac{g}{4}$? ($R$ = radius of the Earth)