The potential energy of a satellite of mass $m$ and revolving at a height $R_e$ above the surface of earth where $R_e =$ radius of earth, is
$-m\, g\, R_e$
$\frac{{ - m\,g\,{R_e}}}{2}$
$\frac{{ - m\,g\,{R_e}}}{3}$
$\frac{{ - m\,g\,{R_e}}}{4}$
If the gravitational acceleration at surface of Earth is $g$ , then increase in potential energy in lifting an object of mass $m$ to a height equal to half of radius of earth from surface will be
The height at which the weight of a body becomes $1/16^{th}$, its weight on the surface of earth (radius $R$), is
In order to shift a body of mass $m$ from a circular orbit of radius $3R$ to a higher radius $5R$ around the earth, the work done is
A body of mass $m$ falls from a height $R$ above the surface of the earth, where $R$ is the radius of the earth. What is the velocity attained by the body on reaching the ground? (Acceleration due to gravity on the surface of the earth is $g$)
Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ respectively are released in free space with initial separation between their centres equal to $12\,R$. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is