The escape velocity of a body from earth's surface is $v_e$ . The escape velocity of the same body from a height equal to $R$ from the earth's surface will be
$\frac{{{v_e}}}{{\sqrt 2 }}$
$\frac{{{v_e}}}{2}$
$\frac{{{v_e}}}{{2\sqrt 2 }}$
$\frac{{{v_e}}}{4}$
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
A body of mass $m$ is situated at a distance equal to $2R$ ($R-$ radius of earth) from earth's surface. The minimum energy required to be given to the body so that it may escape out of earth's gravitational field will be
Three identical bodies of equal mass $M$ each are moving along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each body is
Two spheres of masses $m$ and $M$ are situated in air and the gravitational force between them is $F$. The space around the masses is now filled with a liquid of specific gravity $3$. The gravitationalforce will now be
If $R$ is the radius of earth and $g$ is the acceleration due to gravity on the earth's surface. Then mean density of earth is ..........