The periodic time of a satellite revolving above Earth's surface at a height equal to $R$ (where $R$ is the radius of the Earth) is given by (where $g$ is the acceleration due to gravity at Earth's surface):

If the horizontal velocity given to a satellite is greater than critical velocity but less than the escape velocity at the height,then the satellite will

$A$ light planet is revolving around a massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the force of attraction between the planet and the star is proportional to $R^{-3/2}$,then choose the correct option:

$A$ ball is dropped from a spacecraft revolving around the earth at a height of $120 \, km$. What will happen to the ball?

$A$ spaceship orbits around a planet at a height of $20 \, km$ from its surface. Assuming that only the gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in $24 \, hours$ around the planet? [Given: Mass of planet $= 8 \times 10^{22} \, kg$, Radius of planet $= 2 \times 10^6 \, m$, Gravitational constant $G = 6.67 \times 10^{-11} \, N \cdot m^2/kg^2$]

Two satellites of equal mass are launched in circular orbits at heights $R$ and $2R$ above the surface of the Earth. The ratio of their kinetic energies is ($R =$ radius of the Earth).