Radius of the earth is $R$. If a body is taken to a height $3R$ from the surface of the earth than change in potential energy will be

A small ball of mass $'m'$ is released at a height $'R'$ above the Earth surface, as shown in the figure. If the maximum depth of the ball to which it goes is $R/2$ inside the Earth through a narrow grove before coming to rest momentarily. The grove, contain an ideal spring of spring constant $K$ and natural length $R,$ the value of $K$ is ( $R$ is radius of Earth and $M$ mass of Earth)

The height at which the weight of a body becomes $1/16^{th}$, its weight on the surface of earth (radius $R$), is

A body of mass $m$ is lifted up from the surface of the earth to a height three times the radius of the earth. The change in potential energy of the body is

where $g$ is acceleration due to gravity at the surface of earth.

A spherical part of radius $R/2$ is excavated from the asteroid of mass $M$ as shown in the figure. The gravitational acceleration at a point on the surface of the asteroid just above the excavation is

If $v_e$ is escape velocity and $v_0$ is orbital velocity of satellite for orbit close to the earth's surface. Then these are related by