Kinetic energy constant, but net Work done is not $0$

Question

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 :

$$F_{us}+F_g=F_{net}=0$$

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.

$$W_{net}=W_{ext}+W_{int}=\frac{GMM}{r}+W_{int}$$

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 ?

Answer 1

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.

Answer 2

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.