$A$ body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth,the maximum height attained by the body is:

Maximum height reached by an object projected perpendicular to the surface of the earth with a speed equal to $50\%$ of the escape velocity from the earth's surface is - ($R =$ Radius of Earth)

The condition for a uniform spherical mass $m$ of radius $r$ to be a black hole is [$G=$ gravitational constant and $c=$ speed of light]

$A$ particle released at a large distance from a planet reaches the planet only under gravitational attraction and passes through a smooth tunnel through its centre. If $v_e$ is the escape velocity of a body at the planet,then the particle's speed at the centre of the planet is

If $v_e$ is escape velocity and $v_0$ is orbital velocity of a satellite for an orbit close to the earth's surface,then these are related by:

$A$ spherical uniform planet is rotating about its axis. The velocity of a point on its equator is $V$. Due to the rotation of the planet about its axis,the acceleration due to gravity $g$ at the equator is $1/2$ of $g$ at the poles. Find the escape velocity of a particle on the planet in terms of $V$.