Chapter 10. Relationship between orbital period and radius for a circular orbit (10-12)

Review

Another useful way to describe how rapidly a satellite moves around its circular orbit is in terms of the \(\textbf{orbital period}\), which is the time required to complete one orbit. In uniform circular motion the speed \(v\) is constant, so the orbital period \(T\) is just the circumference \(2\pi{r}\) of the orbit (the distance around the circular orbit of radius \(r\)) divided by the speed: \(T = 2\pi{r}/v\). Using Equation 10-11 for \(v\), this becomes

\(T = \frac{2\pi{r}}{v} =2\pi{r}\sqrt{\frac{r}{Gm_{\mathrm{Earth}}}}\)

It's convenient to square both sides of this equation so as to eliminate the square root. Note that the square of \(r\) is \(r^2\) and the square of \(\sqrt{r}\) is \(r\), so we end up with a factor of \(r^2 \times r = r^3\) on the right-hand side of the equation: