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)
$\frac{R}{{{k^2} + 1}}$
$\frac{R}{{{k^2} - 1}}$
$\frac{R}{{1 - {k^2}}}$
$\frac{R}{{k + 1}}$
An object is taken to height $2 R$ above the surface of earth, the increase in potential energy is $[R$ is radius of earth]
Figure shows the variation of the gravitatioal acceleration $a_g$ of four planets with the radial distance $r$ from the centre ofthe planet for $r \ge $ radius of the planet. Plots $1$ and $2$ coincide for $r \ge {R_2}$ and plots $3$ and $4$ coincide for $r \ge {R_4}$ . The sequence of the planets in the descending order of their densities is
Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to
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 in figure below
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)