The acceleration due to gravity near the surface of a planet of radius $R$ and density $d$ is proportional to
$\frac{d}{{{R^2}}}$
$d{R^2}$
$dR$
$\frac{d}{R}$
If a planet consists of a satellite whose mass and radius were both half that of the earth, the acceleration due to gravity at its surface would be ......... $m/{\sec ^2}$ ($ g$ on earth $= 9.8\, m/sec^2$ )
If earth has a mass nine times and radius twice to the of a planet $P$. Then $\frac{v_e}{3} \sqrt{x}\; ms ^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_e$ is escape velocity on earth. The value of $x$ is
The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$, the radius of the earth, is
$T$ is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be:
The ratio of inertial mass and gravitational mass has been found to be $1$ for all material bodies. Were this ratio different for different bodies, the two bodies having same gravitational mass but different inertial mass would have?