At what distance from the centre of the earth,the value of acceleration due to gravity $g$ will be half that on the surface? ($R =$ radius of earth)

The weight of $1 \, kg$ becomes $1/6$ on the moon. If the radius of the moon is $1.768 \times 10^6 \, m$,then the mass of the moon will be:

Assuming the Earth to be a sphere of uniform mass density,the weight of a body at a depth $d = R/2$ from the surface of the Earth,if its weight on the surface of the Earth is $200 \, N$,will be $........... \, N$ (Given $R =$ Radius of Earth).

The height at which the weight of a body becomes $\left(\frac{1}{9}\right)^{\text{th}}$ of its weight on the surface of the Earth is $(R = \text{radius of Earth})$: (in $R$)

The acceleration due to gravity near the surface of a planet of radius $R$ and density $d$ is proportional to

Different planets have masses $M_1, M_2, M_3$ and radii $R_1, R_2, R_3$ respectively. The acceleration due to gravity on their surfaces are $g_1, g_2, g_3$ respectively. Based on the given graph,arrange their masses in descending order.