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.

What should be the angular speed of the earth, so that a body lying on the equator may appear weightlessness $(g = 10\,m/s^2, R = 6400\,km)$

Suppose, the acceleration due to gravity at the Earth's surface is $10\, m\, s^{-2}$ and at the surface of Mars it is $4.0\, m\, s^{-2}$. A $60\, kg$ pasenger goes from the Earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of figure best represents the weight (net gravitational force) of the passenger as a function of time?

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 

If potential energy of a body of mass $m$ on the surface of earth is taken as zero then its potential energy at height $h$ above the surface of earth is [ $R$ is radius of earth and $M$ is mass of earth]

A geostationary satellite is orbiting the earth at a height of $6\,R$ above the surface of  earth ($R$ is the radius of earth). The time period of another satellite at a height of $2.5\,R$ from the surface of the earth is :-