The height in terms of radius of the earth $(R)$,at which the acceleration due to gravity becomes $g/9$,where $g$ is acceleration due to gravity on earth's surface,is . . . . . . .

An astronaut on a strange planet finds that the acceleration due to gravity is twice that on the surface of the Earth. Which of the following could explain this?

$A$ research satellite of mass $200 \,kg$ circles the earth in an orbit of average radius $3R/2$,where $R$ is the radius of the earth. Assuming the gravitational pull on a mass of $1 \,kg$ on the earth's surface to be $10 \,N$,the pull on the satellite will be ........ $N$.

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

If the density of a small planet is the same as that of the Earth,while the radius of the planet is $0.2$ times that of the Earth,the gravitational acceleration on the surface of that planet is .......... $g$.

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 $\left[\cos (0^{\circ})=\sin (\frac{\pi}{2})=1, \sin (0^{\circ})=\cos (\frac{\pi}{2})=0\right]$