Suppose I have two objects of equal mass and volume, in space, in contact with one another.

The two objects exert equal and opposite gravitational force on each other. Let us apply a force $F$ on one of the objects, to separate it from the other. This is similar to lifting a mass on the surface of the Earth.

The external force is such that, the velocity remains constant. Hence, the change in Kinetic energy is essentially $0$. Hence, the net Work done in this scenario must also be $0$.

Let us look at the moving object and consider that to be our system.

There are two external forces : The force due to us, and the force of attraction due to the other mass. Hence :


Thus, $$W_{net}=0\,\,\,and\,\,\,\,\,F_{us}=-F_g=\frac{GMM}{r^2}$$

Now let us consider the two objects to be our system.

Now there is only one external force, and two equal and opposite internal forces. The net work done must still be $0$. However, I run into trouble, when I try to expand and write this.


However, there are two internal forces here, of the same magnitude. According to the first mass, the second mass is moving away, and hence, work done is $-GMM/r$. Similarly, according to the second mass, the first mass is moving away, opposite to the force, so the work done must again be equal to $-GMM/r$.

Hence, total internal work $W_{int}=\frac{-2GMM}{r}$

If I plug this back in, I'd get $W_{net}=\frac{-GMM}{r}\ne 0$

What am I missing here, and how come the two situations don't agree with one another. What would be the correct way to think about this ?

What is the correct definition of potential energy here. Is it $\Delta P=-W_{int}$ ? Since there are two internal forces, shouldn't there be two internal works ?

  • $\begingroup$ I think you are being too cordial with you use of the $r$ variable. In this dynamical system, the distance between the two bodies is a function $r(t)$ that follows from the equations of motion for the two bodies. If you apply an external force $F$ to one of the bodies, such that the velocity of that body is constant, that force cannot be constant but depends on $r(t)$. In that case you get a complicated $r(t)$-dependant situation, and the work done does not simplify so neatly as you describe. $\endgroup$ Dec 6, 2021 at 9:31
  • $\begingroup$ @MariusLadegårdMeyer yes, it is not a constant force. However, I'm just trying to see that the work-done by me, in raising the object is equal to the potential energy increase of the system. $\endgroup$
    – RayPalmer
    Dec 6, 2021 at 9:56

2 Answers 2


The net work done must still be 0.

Here is the key to your problem. It is not correct that the "net work" must still be zero. You have changed your system definition, and there is no guarantee that different systems will have the same "net work".

In the first system, both $F_{us}$ and $F_g$ are external forces, so the net force is their sum, which is zero by construction. With a zero net force the "net work" is also zero.

In the second system, only $F_{us}$ is an external force, so the net force is not zero. Additionally, the center of mass of the system is accelerating. There is a zero net force on the first object, but the second object has only the unbalanced gravitational force. Thus the second object is accelerating and hence the center of mass of the system is moving.

Because the net force is not zero and the center of mass of the system is not stationary, the "net work" in the second case is also not zero. The kinetic energy is also not constant as the second object is accelerating.

  • $\begingroup$ Suppose I have a book kept on the ground. I pick it up and put it on the table. Suppose I've done this at constant velocity. The force on the book must be $0$, since my force acts against gravity. Hence the net work is also $0$. How do I analyze this same scenario, but taking the Earth-book system as a whole ? $\endgroup$
    – RayPalmer
    Dec 6, 2021 at 13:47
  • $\begingroup$ According to what you said, the earth would accelerate when I lift up the book, and the center of mass of the earth book system would also move upward. Does this mean that the net work is no longer zero ? $\endgroup$
    – RayPalmer
    Dec 6, 2021 at 13:51
  • $\begingroup$ @RayPalmer you lifting a book is not exactly equivalent to the scenario in the question. When you lift a book with your hand, you are also exerting a force on the earth with your feet. So in that case $F_{us}$ is applied to both objects (in opposite directions). In that case the center of mass does not move and the net force is zero, so the net work is again zero. $\endgroup$
    – Dale
    Dec 6, 2021 at 13:57
  • $\begingroup$ in that case how do I analyze the scenario ? If I consider the book as the system, then nothing changes. However, I guess if i consider the book+earth to be the system, then I should write something like : $$F_{us/book}+F_{us/earth}+F_{book/earth}+F_{earth/book}=0$$ $\endgroup$
    – RayPalmer
    Dec 6, 2021 at 14:01
  • 1
    $\begingroup$ @RayPalmer unfortunately, you are being deceived by terminology. The "net" in "net work" refers to the "net force". It does not imply that it is the total work obtained by all forces acting on the system. Unfortunately, the terminology is very standard even though it is confusing. The "net work" never tells you any thing about the total work done by all forces on the system. There is no shortcut for that. $\endgroup$
    – Dale
    Dec 6, 2021 at 14:18

Let us apply a force F on one of the objects ...

So the external force acts on one body but not on the other. This external force is keeping one body stationary (relative to some inertial frame of reference) but not the second body. The second body is accelerating towards the first body due to gravitational attraction, so the kinetic energy of the second body (and hence also the KE of the system of two bodies as a whole) is not constant. And so $W_{net}$ is not zero.

  • $\begingroup$ but I'm giving it a force that exactly counteracts this gravitational attraction. Ofcourse it is not a constant force that I'm providing. So, shouldn't the kinetic energy remain constant ? Think of it as raising a book from the ground by giving an external force to counteract gravity, only now, the book is as large as the earth. $\endgroup$
    – RayPalmer
    Dec 6, 2021 at 9:50
  • 1
    $\begingroup$ @RayPalmer When you lift a book from the ground you are usually applying a force to the book and an equal and opposite force to the earth. If you lifted the book by somehow applying a force to the book only then the earth would accelerate (very slowly !) towards the book so the KE of the book/earth system would not be constant. $\endgroup$
    – gandalf61
    Dec 6, 2021 at 10:45

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