$A$ body weighs $63 \; N$ on the surface of the earth. What is the gravitational force (in $N$) on it due to the earth at a height equal to half the radius of the earth?

Acceleration due to gravity on the moon is $\frac{1}{6}$ of the acceleration due to gravity on the earth. If the ratio of densities of earth $(\rho_e)$ and moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,then the radius of the moon $R_m$ in terms of $R_e$ will be:

Consider Earth to be a sphere of radius $R_e$ rotating about its own axis with angular speed $\omega$. If $g_{E}$ and $g_{P}$ are the accelerations due to gravity at the equator and the poles respectively,then $(g_{P}-g_{E})$ is given by $\left[\cos (0^{\circ})=\sin (\frac{\pi}{2})=1, \sin (0^{\circ})=\cos (\frac{\pi}{2})=0\right]$

At what height above the Earth's surface does the acceleration due to gravity become $1\%$ of its value at the Earth's surface (in $R$)? (Radius of Earth $= R$)

As we go from the equator of the earth to the pole of the earth,the value of acceleration due to gravity

Two equal masses $m$ and $m$ are hung from a balance whose scale pans differ in vertical height by $h$. The error in weighing in terms of density of the earth $\rho$ is: