# Clarification on universal gravitational potential energy re: planet orbiting a star?

## I know that ${U}_{g} = \frac{\text{-} G M m}{r}$, but I also know that gravitational potential energy increases as distance increases because, for example, if a planet is orbiting a star it is faster (has more kinetic energy) when it is closer to the star and slower (has less kinetic energy) when it is farther from the star and since mechanical energy is conserved, more kinetic energy means less potential energy and vice versa. Conceptually, a small $r$ means a large kinetic energy and thus a small potential energy and vice versa, but the formula does not verify this. What am I doing wrong?

The minus sign is there precisely because PE should be greater as r increases. We achieve that by saying that $U \left(\infty\right) = 0$ (!!) and that $U < 0$ everywhere else. So we use negative energy as an accounting convention that matters not.
You may have noticed that the actual value of potential energy is neither here nor there. What matters in physics is the difference between energy levels. So in the simple linearised version of gravitational PE, $U \left(h\right) = m g h$, we are treating the earth's surface as being at zero potential because $U \left(h = 0\right) = 0$.