Energy of an object or a system?

Question

We know that the gravitational potential energy of a system consisting of at least two bodies is given by $U=-\frac{Gm_1m_2}{r}$ where masses $m_1$ and $m_1$ are at a distance $r$ from each other. My question is since this energy is defined for a system, what will be the individual potential energies of them?

For example, take the example of potential energy of an object $m$ at a height $h$(which is pretty small compared to the radius of the earth)from the ground. Here we say that the potential energy of the object is $mgh$ which is derived from $\frac{mgh}{1+\frac{h}{R}}$ considering that fact that $\frac{h}{R}=0$. But this energy is actually the energy of the configuration which consists of earth and the object. Then why do we say that the potential energy of the object is $mgh$? Does that mean the potential energy of the earth is also $mgh$?

Answer 1

Saying the potential energy of the object is $mgh$ is technically incorrect, but nothing bad happens if you think of it like this, at least for simple systems.

The reason is that we can just think of the object as the system. Then the constant gravitational field is external to the system. If we want to look at changes in mechanical energy of the object moving under the force of gravity only, then we have

$$\Delta E=\Delta K=W_\text{ext}=-mg\Delta h$$

or we can just say that

$$\Delta K+mg\Delta h=0$$

This $mg\Delta h$ is usually stated as the change in potential energy of the object. But it doesn't really matter what you call it as long as you recognize the validity of the above equation.

In general you do need to be careful with it. For example, if we have two planets approaching each other, if we were to add up the (incorrect) changes in "individual potential energy", then we would be double-counting the change in potential energy.