$A$ satellite of mass $m$ is revolving around the Earth of mass $M$ in an orbit of radius $r$ with constant angular velocity $\omega$. The angular momentum of the satellite is ($G$ = gravitational constant).

The correct formula for the height $h$ of a satellite from the Earth's surface is:

$A$ satellite of mass $m$ is revolving close to the surface of a planet of density $d$ with time period $T$. The value of the universal gravitational constant $G$ in terms of $d$ and $T$ is given by:

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

Suppose the gravitational force varies inversely as the $n^{th}$ power of the distance. Then,the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

The angular momentum of a planet of mass $M$ moving around the sun in an elliptical orbit is $\overrightarrow{L}$. The magnitude of the areal velocity of the planet is: