A rocket is projected in the vertically upwards direction with a velocity kve where $v_e$ is escape velocity and $k < 1$. The distance from the centre of earth upto which the rocket will reach, will be
$R_e\,(1 -k^2)$
$\frac{{\left( {1 - {k^2}} \right)}}{{{R_e}}}$
$\sqrt {R_e}\, (1 -k^2)$
$\frac{{{R_e}}}{{\left( {1 - {k^2}} \right)}}$
If the gravitational potential on the surface of earth is $V_0$, then potential at a point at height half of the radius of earth is ..........
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
A skylab of mass $m\,kg$ is first launched from the surface of the earth in a circular orbit of radius $2R$ (from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius $3R$ . The minimum energy required to shift the lab from first orbit to the second orbit are
If the change in the value of ' $g$ ' at a height ' $h$ ' above the surface of the earth is same as at a depth $x$ below it, then ( $x$ and $h$ being much smaller than the radius of the earth)
The kinetic energy needed to project a body of mass $m$ from the earth's surface (radius $R$ ) to infinity is