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 earth surface is - ( $R =$ Radius of Earth)
$\frac {R}{2}$
$\frac {16R}{9}$
$\frac {R}{3}$
$\frac {R}{8}$
If the radius of the earth were shrink by $1\%$ and its mass remaining the same, the acceleration due to gravity on the earth's surface would
A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $\left(R_e\right)$, the increase in potential energy will be : ( $g =$ acceleration due to gravity on the surface of Earth)
A planet is revolving ground the sun in an elliptical orbit. Its closest distance from the sun is $r_{min}$, the farthest distance from the sun is $r_{max}$. If the orbital angular velocity of the planet when it is the nearest to the sun is $\omega $, then the orbital angular velocity at the point when it is at the farthest distance from the sun is
Imagine a light planet revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $R^{-5/2}$, then,
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 ..........