0
$\begingroup$

We define two classes of forces, namely conservative forces and non-conservative forces.

When we do work against non-conservative forces , then our work is dissipated in the form of heat, sound or light. When we do work against conservative forces, our work gets stored in the form of potential energy. What my doubt is that how does it get stored as potential energy

For example, consider the block and earth system , When I make the block move up slowly to a height $h$ , I do $mgh$ work on it, and the earth's gravitational field does $-mgh$ work on it. As a result, the $mgh$ energy goes from me to block and $mgh$ from block to the earth's gravitational field. There's no change in the kinetic energy of the block, but the energy of the gravitational field of the earth increases, and my chemical energy decreases.

The doubt that arises here is that the energy of the earth's gravitational field increases, then why do we say that the energy of the earth+block system increases.

Even if we say that energy goes to total field, then we can also say that block's gravitational field has done work on block which is false. What's the flaw in my reasoning here?

$\endgroup$
6
  • 1
    $\begingroup$ When you move a ball to a height $h$ from the surface of the Earth, you need to provide an energy $mgh$ to do so, not $mgh+(-mgh)$ nor $2mgh$, just $mgh$. $\endgroup$
    – Mauricio
    Sep 5, 2021 at 14:23
  • $\begingroup$ @Mauricio What else I wrote in the question? $\endgroup$ Sep 5, 2021 at 14:24
  • $\begingroup$ I was under the impression, that you suggested that $mgh$ was given to the ball and you took something or added something to Earth. What you really do is to separate Earth and the ball a distance $h$ at the same time. The total energy needed is $mgh$. $\endgroup$
    – Mauricio
    Sep 5, 2021 at 14:26
  • $\begingroup$ @Mauricio Yes $mgh$ was added to ball by me and taken from ball by gravitaional field of earth $\endgroup$ Sep 5, 2021 at 14:27
  • $\begingroup$ The energy of the system is the internal energy of the ball, the internal energy of earth and the gravitational energy between the two (and kinetic energies if there is some movement). If you add some external energy, you are increasing the energy of the total system ball+Earth. $\endgroup$
    – Mauricio
    Sep 5, 2021 at 14:31

2 Answers 2

1
$\begingroup$

The doubt that arises here is that the energy of the earth's gravitational field increases, then why do we say that the energy of the earth+block system increases.

It is because gravitational potential energy, like all forms of potential energy, is a system property. Potential energy is the energy of the position of something relative to the position of something else. In the case of gravitational potential energy, is not the energy of the earth's gravitational field alone or the energy of the block alone, but the energy of the combination of the earth+block system.

Hope this helps.

$\endgroup$
10
  • $\begingroup$ _ In the case of gravitational potential energy, is not the energy of the earth's gravitational field alone or the energy of the block alone, but the energy of the combination of the earth+block system_ But how ? when the energy is being taken away by gravitational field of earth only, then how it is stored as energy of block+earth $\endgroup$ Sep 5, 2021 at 15:02
  • $\begingroup$ The potential energy would not exist unless both the object and the earth are present. Although we commonly refer to the gravitational potential energy of an object, technically the object alone would have no gravitational potential energy in the absence of the earth (or other mass) and the gravitational field alone has no gravitational potential energy in the absence of the object. Gravitational potential energy is a system property. See physics.stackexchange.com/questions/440099/… $\endgroup$
    – Bob D
    Sep 5, 2021 at 15:15
  • $\begingroup$ I know it is a system property but then where's flaw in my reasoning. I only want to have correct understanding of the fact that how does work done against conservative forces gets stored as potential energy of a system $\endgroup$ Sep 5, 2021 at 15:19
  • $\begingroup$ @LalitTolani OK, then I'm not sure exactly what your reasoning is in the last paragraph of your post. Gravity does negative work taking energy away from the object (that would otherwise been kinetic energy per the work energy theorem) and storing it as gravitational potential energy of the earth-block system. Precisely what is it about that that you don't understand? You know the block has energy if it were to fall on your head. But it wouldn't "fall" in the first place without gravity doing positive work on it. $\endgroup$
    – Bob D
    Sep 5, 2021 at 15:37
  • $\begingroup$ @LalitTolani Potential energy can be tricky to grasp because it is not clear how or where the energy is stored. Nevertheless it is stored in the system and not within either object. Think of the block and the Earth as were they either end of a spring. You can't "remove" one end. We wouldn't know what that even means. That would mean that the spring doesn't exist at all to store energy in. Similarly, you can't remove the block or the Earth - they are the two "ends" of the gravitational system. Removing one "endW would mean no system to store gravitational energy in at all. $\endgroup$
    – Steeven
    Sep 5, 2021 at 16:27
0
$\begingroup$

If you write the gravitational potential energy of the total system it is written as $$E_{\rm p}= -\frac{GM_{\oplus}m}{R_{\oplus}+h}$$ where $G$ is Newton's constant, $M_{\oplus}$ is the mass of Earth, $m$ is the mass of the block, $R_{\oplus}$ is the radius of Earth and $h$ is the distance between the surface of Earth and the block.

In the limit where $r\ll R_{\oplus}$, it is written as $$E_{\rm p}\approx mgh+\mathrm{constants}$$ where $g={GM_{\oplus}}/{R_{\oplus}}$. If you add one gram to the mass of the Earth, the potential energy increases, if you add a gram to the mass of the block, the potential energy increases. As both the Earth and the block variables appear in $E_{\rm p}$ you cannot say it is the gravitational potential energy of either Earth or the block, it is the potential energy of the system Earth+block.

Imagine that instead of a block, we discuss the attraction of 2 planets. Where does the energy go? It does not go to the field of any of the two, it goes to the total gravitational potential energy of the system of two planets.

With respect to work, yes the gravitational field does work, but on the Earth-block system not the block alone. When we consider Earth as fixed ($M_{\oplus}\gg m$), we can effectively describe the gravitational field as some background field and discuss its effect on the block as if the field was some kind external and directly related to the block (in that approximation, we can say that the field does work on the block).

Edit: sorry for mixing up ball and block in my answer and comments.

$\endgroup$
8
  • $\begingroup$ Yes but then what's the fault in my reasoning? $\endgroup$ Sep 5, 2021 at 14:50
  • $\begingroup$ What I get from your question, is that you are assuming that some external energy can be transferred to the gravitational field of Earth. That is partially true. The energy goes to the total gravitational field, which is the sum of both Earth+ball. $\endgroup$
    – Mauricio
    Sep 5, 2021 at 14:52
  • $\begingroup$ If energy goes to total field, then we can say that ball's field has done work on the ball which is false $\endgroup$ Sep 5, 2021 at 14:54
  • $\begingroup$ From what I understand, you are the one adding the external energy from some chemical source, why would the field do work on the ball? $\endgroup$
    – Mauricio
    Sep 5, 2021 at 15:01
  • $\begingroup$ Because field is applying force and there's displacement of point of application of force $\endgroup$ Sep 5, 2021 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.