What is the effect on the value of $g$ at the equator if the Earth stops its rotation?

The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is $.......... \, km$ : [Radius of earth $R = 6400 \, km$ and $\sqrt{3} = 1.732$]

The weight of a body on the earth is $400\,N$. Then weight of the body when taken to a depth half of the radius of the earth will be ............ $N$.

The mass of the moon is $7.34 \times 10^{22} \ kg$. If the acceleration due to gravity on the moon is $1.4 \ m/s^2$,calculate the radius of the moon. (Given: $G = 6.667 \times 10^{-11} \ N \cdot m^2/kg^2$)

At what depth below the surface of the earth will the acceleration due to gravity be half of its value at $1600 \ km$ above the surface of the earth?

There is a planet which is $8$ times more massive and $27$ times denser than the Earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surfaces of the planet and the Earth respectively,then: