The radius of a planet is $R$. A satellite revolves around it in a circle of radius $r$ with angular velocity $\omega _0.$ The acceleration due to the gravity on planet’s surface is
$\frac{r^3\omega _0}{ R}$
$\frac{r^3\omega _0^3}{ R}$
$\frac{r^3\omega _0^2}{ R}$
$\frac{r^3\omega _0^2}{ R^2}$
In order to shift a body of mass $m$ from a circular orbit of radius $3R$ to a higher radius $5R$ around the earth, the work done is
The angular speed of earth in $rad/s$, so that bodies on equator may appear weightless is : [Use $g = 10\, m/s^2$ and the radius of earth $= 6.4 \times 10^3\, km$]
A particle of mass $M$ is situated at the centre of a spherical shell of same mass and radius $a$. The gravitational potential at a point situated at $\frac{a}{2}$ distance from the centre, will be
The additional kinetic energy to be provided to a satellite of mass $m$ revolving around a planet of mass $M$, to transfer it from a circular orbit of radius $R_1$ to another of radius $R_2\,(R_2 > R_1)$ is
A mass $m$ , travelling at speed $V_0$ in a straight line from far away is deflected when it passes near a black hole of mass $M$ which is at a perpendicular distance $R$ from the original line of flight. $a$ , the distance of closest approach between the mass and the black hole is given by the relation