The diameters of two planets are in the ratio $4:1$. Their mean densities have the ratio $1:2$. The ratio of gravitational acceleration on the surface of the planets will be:

Acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth is reduced to one-third of its original value and mass remains unchanged,then the acceleration due to gravity on the surface of the Earth is . . . . . . $g$.

Assume that the Earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the Earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 \, g$. The time period of the motion of the particle will be (approximately) (take $g = 10 \, m/s^2$, radius of Earth $R = 6400 \, km$):

The height above the surface of the earth where acceleration due to gravity becomes $\frac{g}{9}$ is ( $R$ is the radius of the earth,$g$ is the acceleration due to gravity at the surface).

$A$ body weighs $500 \, N$ on the surface of the earth. How much would it weigh halfway below the surface of the earth (in $, N$)?

The weight of a body on the surface of the earth is $100\,N$. The gravitational force on it when taken at a height,from the surface of earth,equal to one-fourth the radius of the earth is $..........\,N$.