An artificial Earth satellite moves in a circular orbit.…

An artificial Earth satellite moves in a circular orbit. What is the relationship between its kinetic and potential energy?

Step 1: Idea

The speed of a satellite on a circular orbit is not arbitrary: gravity must provide exactly the centripetal force. This fixes the kinetic energy in terms of the radius, and so ties it to the potential energy.

Step 2: Kinetic energy

\[ m\frac{v^2}{r} = \gamma\frac{Mm}{r^2}, \]

\[ mv^2 = \frac{\gamma Mm}{r}, \]

\[ E_k = \tfrac12mv^2 = \frac{\gamma Mm}{2r}. \]

Step 3: Potential energy

With the zero of potential energy at infinite distance (Problem 14): $E_p = -\frac{\gamma Mm}{r}$.

Step 4: Relationship

Comparing the two expressions, the potential energy is exactly $-2$ times the kinetic energy, whatever the orbit radius: \[\begin{gathered} E_p = -2E_k \\ E = E_k + E_p = -E_k \end{gathered}\]

Consequences: a higher orbit (larger $r$) has a smaller kinetic energy (the satellite is slower), but a larger (less negative) potential energy and a larger total energy. To raise a satellite we must add energy, even though it ends up moving more slowly. Conversely, a satellite losing energy to air drag sinks to a lower orbit and speeds up.

Check

This is the virial theorem for a $1/r$ potential: $2E_k = -E_p\ \checkmark$

Problem 7: $E_2 - E_1 = \gamma Mm/12R_Z > 0$, energy must be supplied to go higher $\checkmark$