An earth satellite of mass $m$ revolves in a circular orbit at a height $h$ from the surface of the earth. $R$ is the radius of the earth and $g$ is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by

$A$ planet revolves around the Sun in an elliptical orbit. When it is closest to the Sun,its distance from the Sun is $1.6 \times 10^{12} \ m$ and its velocity is $60 \ m/s$. What is its velocity in $m/s$ when it is farthest from the Sun,given its distance is $8 \times 10^{12} \ m$?

The height of a geostationary satellite is approximately ........ $km$.

$A$ 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 $G = 6.67 \times 10^{-11} \, N \cdot m^2/kg^2$]

The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is $v$. For a satellite orbiting at an altitude of half the earth's radius above the surface,the orbital velocity is:

$A$ geostationary satellite is orbiting the earth at a height of $5R$ above the surface of the earth,where $R$ is the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of the earth is: