# Time period of a satellite close to earth

The time period $T$, of a satellite revolving around the earth and close to it is given by, $$T = \frac{2\pi r}{v}$$ where $v$ is the linear velocity of the satellite and roughly equal to $8\ \mathrm{ km/s}$. Why are we not taking $\omega$ instead of linear velocity ?

Notice, the linear velocity $v$ is co-related with the angular velocity $\omega$ as follows $$\text{linear velocity}=(\text{radius})\times (\text{angular velocity})$$ $$v=r\omega$$ by setting $v=r\omega$, we get $$T=\frac{2\pi r}{v}=\frac{2\pi r}{r\omega}=\color{red}{\frac{2\pi }{\omega}}$$
Its an application of the formula $$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$. $2\pi r$ is the distance covered in one revolution and $v$ is the speed. In terms of $\omega$ $$T = \frac{2\pi}{\omega}$$. (Also note that $\omega=v/r$)