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Assume an isolated system consisting only of the earth and a basketball interacting through the gravitational force: No air, no other planets/stars etc. Are the following statements accurate?

  1. The total kinetic energy for the system depends on the speed at which the earth and the ball are moving toward or away from one another
  2. The potential energy depends on the distance between the earth and the ball
  3. The internal system-work of earth on ball depends only on the change in speed of the ball (and the current size of the gravitational force of earth on ball)
  4. The internal system-work of ball on earth depends only on the change in speed of the earth (and the current size of the gravitational force of ball on earth... which would be the equal/opposite of earth on ball).
  5. Potential energy changes never come into the bookkeeping for one of the work calculations from #3 because work is defined to depend on the change in speed of a mass (and the force on the mass).
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    $\begingroup$ Points 3 and 4 are a bit odd. Only a force can do work so by the phrase "system-work" done on an object I guess you mean the work done by the gravitational force on that object. But that work doesn't really depend on the speed, rather it causes it. If somebody slowed down the falling ball, then the work done by the gravitational force on it would still be the same in the end. This doesn't happen in your isolated scenario, so all work done just happens to be converted into kinetic energy, giving speed. But it is odd to say that it depends on the speed. $\endgroup$
    – Steeven
    Commented Nov 22, 2018 at 8:38
  • $\begingroup$ @Steeven By "system-work" I mean "internal work" (I say "internal work" because there is nothing outside of the system doing work on anything in system, so we only have parts of the system doing work on other parts of the system). I would say the gravitational force is the source of both work considerations (speed changes) and potential energy considerations (position changes). I have in mind $0=\Delta K_{tot} + \Delta U = (W_{earth} + W_{ball}) + \Delta U$. I'm just clarifying that work by definition reflects the impact on speed while potential energy by definition reflects the impact on pos. $\endgroup$
    – okcapp
    Commented Nov 22, 2018 at 8:54
  • $\begingroup$ point1: The total kinetic energy depends on the speed of their center of mass with regard to our observers' framework.point2 seems fine.For point3 and 4 I agree with Steeven and I could not understand point 5 so I could not say anything. $\endgroup$
    – Looser
    Commented Nov 22, 2018 at 8:57
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    $\begingroup$ I do not agree with the term "internal work" for this use. All internal work cancels out (work done by the Earth-ball force cancels out work done by the ball-Earth force), (which is why only external forces can do any net work), so if you want to specifically talk about only one of these works, I'd stick to saying "work done by a force", which is always how work is done. $\endgroup$
    – Steeven
    Commented Nov 22, 2018 at 9:36
  • $\begingroup$ "that work by definition reflects the impact on speed" As already explained, I do not think that this is true. Not in general at least. Maybe in your specific case. But not as a general definition of work. Work already has a clear definition: force-times-distance. The work done can then be converted into many other types of energy, for example kinetic energy in your scenario. But it could also be for example elastic potential energy in a spring, when I do work by pressing a spring. $\endgroup$
    – Steeven
    Commented Nov 22, 2018 at 9:39

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You can determine the net work done on an object two ways.

1) Calculate it's change in kinetic energy: $$W_{net}=\Delta K$$

2) Directly add up the work done by all forces on the object. $$W_{net}=\sum_i\int \mathbf F_i\cdot\text d \mathbf x$$

In your case, we only have one force acting on each of our objects, and it is conservative. Therefore, we can also use the fact that the work done on an object is equal to the negative of its change in potential energy $$W_{cons}=-\Delta U$$

It seems like your points 3-5 is trying to get you to say that only one of these three is the "right way" to find the work done on an object, when in reality all three are equivalent and equally valid.

Another point to be made. The work done does not depend on the speed of the object alone. It depends on the kinetic energy, which depends on the mass of the object in addition to its speed. A more massive object moving slowly can have the same kinetic energy as a less massive object moving fast.

Similarly, the work done by a force does not depend on the magnitude of the force alone. It depends on the distance over which the force acts as well. A large force acting over a small distance can perform the same amount of work as a small force acting over a large distance.

With all of that being said, your question seems to have some areas where it could be more specific.

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