At what height above the earth's surface does the acceleration due to gravity fall to $1\%$ of its value at the earth's surface (in $,R$)?

The mass of the earth is $81$ times that of the moon and the radius of the earth is $3.5$ times that of the moon. The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is

The variation of acceleration due to gravity $g$ with distance $d$ from the centre of the earth is best represented by ($R =$ Earth's radius)

The ratio of the height $h$ above the surface of the Earth to the depth $d$ below the surface of the Earth,for which the gravitational accelerations are the same (assuming small heights),is:

The difference in the acceleration due to gravity at the pole and equator is ( $g=$ acceleration due to gravity,$R=$ radius of earth,$\theta=$ latitude,$\omega=$ angular velocity,$\cos 0^{\circ}=1, \cos 90^{\circ}=0$ ).

The radii of two planets are respectively $R_1$ and $R_2$ and their densities are respectively $\rho_1$ and $\rho_2$. The ratio of the accelerations due to gravity at their surfaces is