What is the height from the surface of earth,where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth (in $km$)? $(R_E = 6400 \ km)$

The acceleration due to gravity on the Moon is $\frac{1}{6}$ times that of the Earth. If the ratio of the density of the Earth $(\rho_e)$ to the Moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,what is the radius of the Moon $(R_m)$ in terms of the radius of the Earth $(R_e)$?

The weight of an object on the surface of the Earth is $72 \, N$. What will be the weight of the object at a height of $R/2$ from the surface of the Earth? ($R$ = radius of the Earth)

The value of acceleration due to gravity at Earth's surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$,is close to: (Radius of Earth $= 6.4 \times 10^6\, m$)

If a planet has a mass and radius both half that of the Earth,the acceleration due to gravity at its surface would be ......... $m/s^2$ ($g$ on Earth $= 9.8\, m/s^2$)

If the value of $g$ (acceleration due to gravity) at the Earth's surface is $10 \ m/s^2$,what is its value in $m/s^2$ at the center of the Earth,assuming the Earth is a sphere of radius $R$ meters with uniform mass density?