An object is taken to height $2 R$ above the surface of earth, the increase in potential energy is $[R$ is radius of earth]
$\frac{m g R}{2}$
$\frac{m g R}{3}$
$\frac{2 m g R}{3}$
$2 m g R$
In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be $(g = 10\,m{s^{ - 2}}$ and radius of earth is $6400 \,kms)$
The two planets have radii $r_1$ and $r_2$ and their densities $p_1$ and $p_2$ respectively. The ratio of acceleration due to gravity on them will be
The value of escape velocity on a certain planet is $2\, km/s$ . Then the value of orbital speed for a satellite orbiting close to its surface is
The mass of planet is $\frac{1}{9}$ of the mass of the earth and its radius is half that of the earth. If a body weight $9\,N$ on the earth. Its weight on the planet would be ........ $N$
In a satellite if the time of revolution is $T$, then $K.E.$ is proportional to