Is Earth part of the system when writing Gravitational Potential Energy?

Here is a question which stumped me when teaching high school students.

The Work-Energy equation can be written as:

$$W_{ext} + W_{non-conservative} = \Delta{U} + \Delta{K}$$

Here, $\Delta{U}$ refers to the difference in potential energy of the system in consideration. Potential Energy is nothing but the negative of work done by conservative forces.

I would like to talk about one special potential energy, and that is Gravitational Potential Energy.

Now, let's say I have a block of mass $m$. We write the gravitational potential energy for this block as $mgh$. When doing so, we say that this potential energy is the potential energy of the block-earth system. So, we mean that Earth is a part of our system.

Now, if earth is a part of our system, everything on earth is a part of the system. It means if I am standing near this block and apply some force on it, that force will not be external and hence its work done would not be counted in the $W_{ext}$!! This doesn't make any sense.

As a student, I never looked at gravitational potential energy this way. But now, when I look at it, it is mind-boggling to think the whole earth is part of the system.

Please clarify where my logic/reasoning is going wrong.

• Have you considered that with F=mgh, in my opinion, when calculating the velocity thus work of the block the "earth doesn't move for you", for the block? Intricately, if the earth is said not to be moved by the block's gravitational force why should earth be taken as a reference to the "external worker"? It's both internally and externally "out", I think. "...everything on earth..." may just reflect my uneasiness about disregarding earth's potential energy (and not the block's, that lies in its height; earth's depth seems neglected). Cheers. Commented Nov 28, 2022 at 13:34