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
$12\,km/s$
$1\, km/s$
$\sqrt 2\,km/s$
$2\sqrt 2\, km / s$
At what altitude will the acceleration due to gravity be $25\% $ of that at the earth’s surface (given radius of earth is $R$) ?
A geo-stationary satellite is orbiting the earth at a height of $5R$ above surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of earth is
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)
The magnitudes of gravitational field at distance $r_1$ and $r_2$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively. Then
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 $g$)