The reason for weightlessness in a satellite is

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

$A$ satellite is revolving very close to a planet of density $\rho$. The period of revolution of the satellite is

$A$ satellite of mass $m$ is circulating around the Earth with constant angular velocity. If the radius of the orbit is $R_0$ and the mass of the Earth is $M$,the angular momentum about the centre of the Earth is:

Two satellites of mass $m$ and $9m$ are orbiting a planet in orbits of radius $R$. Their periods of revolution will be in the ratio of

If a satellite has to orbit the earth in a circular path every $6 \text{ hrs}$, at what distance from the surface of the earth should the satellite be placed (in $\text{ km}$)? (Radius of earth $R_e = 6400 \text{ km}$)
(Assume $\frac{GM}{4\pi^2} = 8 \times 10^{12} \text{ m}^3\text{s}^{-2}$, where $G$ and $M$ are the gravitational constant and mass of earth, and $10^{1/3} = 2.1$)