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The work done on an object by an external agent in bringing it closer to a planet is given by mΔV. So, when the object is brought closer to a planet, the work done by the external agent is negative. But why is it so? What does it mean? The kinetic energy of the object increases, does that have anything to do with the explanation?

When an object is moved away from a planet, the work done by the gravitational force is negative, but of the external agent it is positive. I am confused on how we can determine when the work done by the gravitational force and external agent is negative and when it is positive.

Great thanks for your help!

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So, when the object is brought closer to a planet, the work done by the external agent is negative

This may be a source of confusion. As stated, it is not necessarily true. An external agent is not constrained to provide either positive or negative work in such a scenario.

The easiest to consider is that there is no external agent, so that there is no change in total energy. In that case as the object free-falls closer it will gain KE. Because the total energy is conserved the PE must decrease.

To determine the work of an external agent you need to specify what the agent does. Often the external agent acts to keep the KE constant. In that case, the PE decreases, but the KE does not change. This means that the earth-object system loses energy. Hence, the external agent does negative work on the system.

I am confused on how we can determine when the work done by the gravitational force and external agent is negative and when it is positive.

Always look at the energy. Positive work done on a system increases the energy of the system. Negative work done on a system decreases the energy of the system. Work done by a system has the opposite effect.

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  • $\begingroup$ Hey thank you very much for your explanation!! Definitely makes sense now! :D $\endgroup$
    – Drita Raci
    Commented Mar 31, 2022 at 12:24
  • $\begingroup$ So the main idea here is that the potential and kinetic energy of the object is conserved (conservation of energy law). But now, when we look at the total energy change of the object when moving it, it is not zero. This is because the external agent either took some energy or provided some energy. Therefore, when the total energy change is negative, it means that energy was taken away by the external agent, hence it did negative work. But if the energy change is positive, then the external agent did positive work. So now is my understanding correct? :) $\endgroup$
    – Drita Raci
    Commented Mar 31, 2022 at 12:29
  • $\begingroup$ @NurAhmed yes, that is correct. For Newtonian mechanics energy is conserved for an isolated system, and for a non-isolated system the change in energy is equal to the work done by the external agent. $\endgroup$
    – Dale
    Commented Mar 31, 2022 at 12:33
  • $\begingroup$ Great! Thank you loads! Have a good day! :) $\endgroup$
    – Drita Raci
    Commented Mar 31, 2022 at 12:41
  • $\begingroup$ Sorry! I also wanted to clear something out: When the work done by an external agent is positive, the work done by the gravitational force is equal and negative. Is it so, so that the energy would be conserved? For example, taking a mass m a distance r away from a planet, the work done by the gravitational force is negative, but that of the external agent it is positive. $\endgroup$
    – Drita Raci
    Commented Mar 31, 2022 at 13:20

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