A body of mass $m$ falls from a height $R$ above the surface of the earth, where $R$ is the radius of the earth. What is the velocity attained by the body on reaching the ground? (Acceleration due to gravity on the surface of the earth is $g$)
$gR$
$\sqrt {gR} $
$\sqrt {g/R} $
$g/R$
The kinetic energy needed to project a body of mass $m$ from the earth's surface (radius $R$ ) to infinity is
Masses and radii of earth and moon are $M_1,\, M_2$ and $R_1,\, R_2$ respectively. The distance between their centre is $'d'$ . The minimum velocity given to mass $'M'$ from the mid point of line joining their centre so that it will escape
A particle of mass $M$ is at a distance $'a'$ from surface of a thin spherical shell of uniform equal mass and having radius $a$
A small ball of mass $'m'$ is released at a height $'R'$ above the Earth surface, as shown in the figure. If the maximum depth of the ball to which it goes is $R/2$ inside the Earth through a narrow grove before coming to rest momentarily. The grove, contain an ideal spring of spring constant $K$ and natural length $R,$ the value of $K$ is ( $R$ is radius of Earth and $M$ mass of Earth)
A body tied to a string of length $L$ is revolved in a vertical circle with minimum velocity, when the body reaches the upper most point the string breaks and the body moves under the influence of the gravitational field of earth along a parabolic path. The horizontal range $AC$ of the body will be