The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$,the radius of the earth,is

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 $(R_e)$?

If the Earth has no rotational motion,the weight of a person on the equator is $W$. Determine the speed with which the Earth would have to rotate about its axis so that the person at the equator will weigh $\frac{3}{4} W$. The radius of the Earth is $6400 \ km$ and $g = 10 \ m/s^2$.

What is the ratio of the acceleration due to gravity on two planets with radii $R_1$ and $R_2$ and densities $\rho_1$ and $\rho_2$?

If the radius of the Earth is $R$,then the height $h$ at which the value of $g$ becomes one-fourth is:

The moon's radius is $1/4$ that of the earth and its mass is $1/80$ times that of the earth. If $g$ represents the acceleration due to gravity on the surface of the earth,then the acceleration due to gravity on the surface of the moon is: