Suppose I have a book kept on the floor. I pick it up, and keep it on the table, at a constant velocity. This means I've to apply a force $mg\hat{y}$ to counteract the force of gravity. The work that I do to pick up the book is $mgh$. Similarly, gravity is also acting on the book with a force $-mg\hat{y}$. Hence, the work done by gravity is $-mgh$.
Thus, if we consider the book to be our system, the net force and by extension, the net work comes out to be $0$. Moreover, the work done by gravity is defined as the negative of the change in potential energy of the book-Earth system. Hence, we can say that the change in potential energy $\Delta U=mgh$.
Now I want to analyze the same situation by considering the book and the Earth to be our system.
The force that I apply, is now the external force, and the internal forces are the equal and opposite forces between the book and the Earth. Remember, in order to raise the book with a constant velocity, I have to apply force on both the book and the Earth in the opposite direction. This will ensure that the center of mass moves at a constant velocity.
Thus, as you can see, the net force on each of the objects is $0$, and so, the net work done must also be $0$, just like the previous case.
let us now calculate the net work done by each of the four forces $F_{us/book},F_{us/earth},F_{book/earth},F_{earth/book}$
Well, as I know, the work done by us, on the book is simply $F_{us/book}. h=mgh$
Similarly, the work done by the Earth, on the book, is given by $F_{earth/book}=-mgh$
Now, I'm confused about two things here. First of all, what is the work done by us, on the earth ? Since the Earth stays in place, I can say this work done is $0$. However, what is the work done by the book on the Earth ? The answer again, should be $0$, but isn't the Earth moving relative to the book ? From the book's perspective, the earth is moving downward, while the gravity is acting upward, towards the book. So, shouldn't the book do a work of $-F_{book/earth}.h=-mgh$ on the Earth ? I'm told this is not true, and I don't seem to understand why. Is there a special choice of origin that we stick to, throughout the entire analysis ? Else, if the Earth does $-mgh$ work on the book, then from the book's perspective, it is stationary, and the Earth is moving, so shouldn't it do $-mgh$ work on the Earth ?
But this would mean that the total work done by the internal forces is $-mgh-mgh=-2mgh$. However, we have established that we do no work on the Earth, so the total external work done by us is $mgh$. If we add up all of these, it doesn't come out to be $0$. It is clear that we are overcounting the work done by the book on the earth and viceversa, but I don't seem to understand why.
Finally, what exactly is the potential energy here ? Is it the negative of the work done of the earth on the book and the book on the earth ? Is the work done by the book on the Earth $0$ , since the Earth doesn't move ? But the Earth doesn't move with respect to us, who is picking up the book, with respect to the book the Earth does move.
Is the change in potential energy of a system the negative of the sum of work done by all the internal conservative forces in the system ? In this case there are two internal forces, and the work done by the earth on the book is $-mgh$ and the work done by the book on the earth is $0$(?). Then we can safely say that total potential energy of the system changes by $mgh$.
However, since the book moves up from the Earth, and the earth moves down, according to the book, why is the work done by the earth on the book non zero, while the work done by the book on the earth is zero. Is there a special reference line with which we measure the change in distance, and if so why ?