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The gravitational potential energy of a mass at a point in a field is defined as the work done by an external agent in bringing that mass from infinity to that point, without a change in kinetic energy.

I don't understand why it's "the work done by an external agent in bringing a mass from infinity to the point", because work done from infinity to the point is in the direction of the gravitational field, so why isn't the work done by the conservative force, aka gravitational force? Shouldnt the work done by an external agent separate the masses further and further, which is from a point towards infinity?

To answer my own question, I thought that I may be misinterpreting the statement,"work done by an external agent in bringing that mass from infinity to that point, without a change in kinetic energy". Instead of placing emphasis on "work done by an external agent in bringing that mass from infinity to that point", which I think happens due to the gravitational force, the "work done by external agent" is not responsible for bringing the test mass from "infinity to a point", but is instead responsible for ensuring the velocity doesn't change during the process (KE kept constant).

Based on the reasoning above, the force exerted by the external agent must be equal in magnitude to gravitational force to ensure velocity is constant and thus, increases as the test mass approaches the central mass since r decreases and $F=GMm/r^2$. Thus, the work done by the external agent, which is GPE is given F×displacement. Displacement is away from the central mass,since external agent has to act opposite to gravitational force to ensure net force is zero, so it is $-r$, where $r$ is towards the central mass. Thus $gpe = - GMm/r$.

This implies that when radius decreases, gpe because more negative and since gpe is a scalar, gpe decreases.

So to summarise my reasoning, I would like to confirm if the following is true:

  1. The work done by the external agent does not bring the test mass from infinity to a central mass. Instead it acts in the direction away from the central mass, towards infinity.

  2. The test mass is brought from infinity to the point due to work done by gravitational force, but although potential energy will be negative when work is done by conservative force, this is not the reason why gpe is negative since the definition of gpe says nothing about the work done by the conservative force and is instead about the work done by an external agent.

  3. The reason why gpe is negative is because the external agent acts opposite to r, and using formula for work done by external agent displacement is negative so gpe is a negative. Alternatively, we can integrate F with respect to r to get -GMm/r.

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  • $\begingroup$ Related:physics.stackexchange.com/questions/754638/… $\endgroup$
    – khaxan
    Jun 8, 2023 at 10:49
  • $\begingroup$ Basically, yes, yes and (partly) yes. Why GPE is negative in this definition is because GPE zero point can be chosen freely and in this declaration zero point is chosen to be at infinite distance from target. However you can choose zero point differently, for example it can be at the surface of planet, then GPE would be positive and defined as $mgh$. So, GPE sign depends how(where) you choose zero point and how you integrate elementary work. $\endgroup$ Jun 8, 2023 at 14:57

2 Answers 2

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You got it. Often the potential energies are defined by the formula $\Delta U=-L_{cons.}$, but if you consider an external force opposing to gravity (so that $F_{ext}=-F_g$) then you have: \begin{equation} \Delta U=U(r)-U(\infty)=L_{ext} \end{equation} but you can arbitrarily put $U(\infty)=0$, so you obtain the definition given by your book.

Clearly, the assumption $F_{ext}=-F_g$ implies that the net force is zero and the velocity does not change, in agreement with the definition.

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Gravitational potential energy is all to do with a system of at least two masses.
So if a mass $m$ approaches a fixed mass $M$ it is the system of two masses which has the decreasing gravitational potential energy.

If the separation of the two masses is decreasing with the mass $m$ moving at constant velocity then the external force on that mass is in the opposite direction to the displacement of the external force and so negative work is done by external force and the gravitational potential energy of the system of two masses decrease.
At the same time if the internal gravitational force on the $m$ due to mass $M$ has an equal magnitude but opposite direction to the external force, that gravitational force does positive work (force and displacement in the same direction) and again the gravitational potential energy decreases as another way of defining gravitational potential energy is via minus the work done by the gravitational field.
So the net work done on mass $m$ is zero and hence its kinetic energy does not change.

If you want to use the concept of gravitational potential energy then you must have a system of at least two mass which would be $M$ and $m$ in your case.
Then whenever you are referring to gravitational potential energy it is the gravitational potential energy of the system of two masses, not just one of the masses.

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  • $\begingroup$ The external agent does negative work, ie the gravitational field does positive work...seems to conflict with "we do positive work against gravity and gravity too does work but negative". I am familiar with second $\endgroup$
    – khaxan
    Jun 8, 2023 at 10:45
  • $\begingroup$ @khaxan Although I have streamlined my answer, and I still stand by the statement The external agent does negative work, ie the gravitational field does positive work and I never made the contradictory statement. If the second statement were true the masses would be separating and the gravitational potential energy would be increasing. $\endgroup$
    – Farcher
    Jun 8, 2023 at 12:16
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    $\begingroup$ So where is the problem here. If distance between bodies increases, then potential energy increases. This idea seems perfectly reasonable to me $\endgroup$
    – khaxan
    Jun 8, 2023 at 12:25
  • $\begingroup$ Why in your comment have you written that there is a conflict? $\endgroup$
    – Farcher
    Jun 8, 2023 at 12:30
  • $\begingroup$ I said "we do positive work against gravity". That is in conflict with ur proposition that "we do negative work". $\endgroup$
    – khaxan
    Jun 8, 2023 at 13:43

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