Find the speed required to maintain a satellite in a near-Earth circular orbit. (The gravitational attraction of other bodies is ignored.)

Solution Let $R$ be the distance of a satellite from the center of Earth. Then from (2), the magnitude of the acceleration of the satellite whose motion is circular is $$\begin{equation*} \left\Vert \mathbf{a}(t)\right\Vert\;=\;\dfrac{v_{0}^{2}}{R} \end{equation*}$$

For the satellite to remain in orbit, the magnitude of the acceleration $\left\Vert \mathbf{a}(t)\right\Vert$ of the satellite at any time $t$ must equal $g$, the acceleration due to gravity for Earth. As a result, $$\begin{eqnarray*} \frac{v_{0}^{2}}{R}&\;=\;&g \\[3pt] v_{0}&\;=\;&\sqrt{gR} \end{eqnarray*}$$

The speed $v_{0}$ required to maintain a near-Earth circular orbit is $$\begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {v_{0}=\sqrt{gR}} \end{equation*}$$

where $R$ is the distance of the satellite from the center of Earth and $g$ is the acceleration due to gravity.