A projectile is projected with velocity $k{v_e}$ in vertically upward direction from the ground into the space. ($v_e$ is escape velocity and $k < 1$). If air resistance is considered to be negligible then the maximum height from the centre of earth to whichit can go, will be : ($R =$ radius of earth)
$\frac{R}{{{k^2} + 1}}$
$\frac{R}{{{k^2} - 1}}$
$\frac{R}{{1 - {k^2}}}$
$\frac{R}{{k + 1}}$
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
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
A satellite moving with velocity $v$ in a force free space collects stationary interplanetary dust at a rate of $\frac{{dM}}{{dt}} = \alpha v$ where $M$ is the mass (of satellite + dust) at that instant . The instantaneous acceleration of the satellite is
A body weighs $72\, N$ on the surface of the earth. What is the gravitational force(in $N$) on it, at a helght equal to half the radius of the earth$?$
The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be a sphere of radius $R$ of uniform density is as shown figure below
The correct figure is