Reference Point and Change in Potential Energy

Question

Okay so I am VERY confused. Everything online is telling me that I can choose any reference point for potential being zero and still get a consistent result for potential difference HOWEVER I have only been able to see this when force is a constant (like with mg) and doesn’t depend on position then yes work done which is minus the change in potential is a constant as F remains a constant no matter what 2 points we choose to measure distance from so long as d (the separation) remains a constant and consequently as we can choose any reference point for $x=0$ or $r=0$ and only the difference in distance matters, we can choose any point for $U=0$. Now here is the problem. The cases for where the force is not a constant like outside of space and the Earth (where it depends on $1/r_2$) doesn’t seem to be as lenient with me. Let us assume (ignoring the sign and $GMm$) that the change in potential energy is $1/B - 1/A$ where $r=0$ at Earth. I can’t just say $U$ is zero at $A$ and get the same result as saying $U$ is zero at Earth’s center! This is because the equation for potential energy at a point a distance r away from the object producing the force field is $U(r)=\dfrac{GMm}{r}$ and I can’t just say $\dfrac{GMm}{A} = 0$ by choice! Someone help :’(

Answer 1

Sure you can. You can only measure changes in the potential energy, so you can add any constant to a potential energy function and it will describe the same physics.

Anytime you write down a potential energy function $U$, you are making a choice of reference point and the potential energy at that point, whether explicitly or not. In the case of the gravitational potential energy function \begin{align} U(r) = - \frac{GmM}{r}, \end{align} we're choosing $U(r) = 0$ for $r\rightarrow \infty$. Nothing about the physics changes if we choose instead $U(a) = 0$, so that \begin{align} U(r) = - GmM\left(\frac{1}{r} - \frac{1}{a}\right). \end{align}