# 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?

##### 1 Answer
May 2, 2017

What you're fretting about here is well worth fretting about but I would leave KE aside and just think about PE.

The further apart you pull 2 objects, the more gravitational PE you are storing in that system. You are putting work in to create the PE, if you release the objects, they will, all things being equal rush toward each other. The gravitational force between the objects will do work on the objects and increase their KE.

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 $U \left(h = 0\right) = 0$.

This issue also arises about half the time (!?!?) in electrostatics. If the charges are attractive, then you have the same issue. That little sign really matters and so to use thought experiments as a reality check is a good habit to be in.