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My current concept about potential energy is that 'If work is done on a body when it is at a point (taken as zero P.E), it covers some distance. Afterward, when it is allowed to move freely it does the same work. So, I get my work back as if I stored it in that body. But if the body doesn't come back to my assigned value of zero or in other words starts to move away from it then potential energy would be negative'. Refer to the following fig.:

enter image description here

According to my understanding, if the object is taken from point A to B then it had positive P.E.
But if it started to move towards O than it's P.E would be negative. Correct me if I am wrong.

Now here's the Question:

Absolute Gravitational P.E is defined as 'Work done by Gravitational force for moving a body from a certain position to infinity'

Ok, agreed. So it's obvious that if Earth's center (O) is taken as zero P.E then Absolute P.E would be negative.

Now my book says:

enter image description here

Here U is absolute P.E. r is the distance from the center of Earth.

Consider the following figure: enter image description here It can be clearly seen that Workdone'y' is greater than Workdone'x', which is contrary to my textbook.

Or in other words, if Workdone is -50 and -5, which one would be greater?

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  • $\begingroup$ Your set up confuses me. In both cases, the work done is negative, so the potential energy increases. It increases more in case"$y$" than in case "$x$". Can you explain why this bothers you? I point out that you have some language that sounds a little sloppy. In your paragraph starting "According to my understanding", you conclude that the PE is positive going away from $0$ and negative going towards $0$. In fact PE is negative in both cases, but the change in PE has different sign. Are you confused about that? $\endgroup$
    – garyp
    Commented Oct 29, 2017 at 17:46
  • $\begingroup$ @garyp Yes P.E increases, but the reason for this in my book is "if r increases U will also increase" And that bothers me. PE is positive when obj. is going away from 0 (When allowed to move freely) and negative when obj. is going towards 0 (When allowed to move freely). Although obj. cannot move away from o when allowed to move freely but IF it moves then PE at that point where it was initially, should be positive. HOw you can say it's neg. in both cases? And my major question is "if Workdone is -50 and -5, which one would be greater?" $\endgroup$
    – M.Ahmad
    Commented Oct 29, 2017 at 19:03
  • $\begingroup$ "but IF it moves then PE at that point where it was initially, should be positive." This is not correct. If you can explain why you think it is, we might be able to untangle things. It might also explain why you want to know the answer to "if Workdone is -50 and -5, which one would be greater?" As written, that question is ambiguous. I don't see the connection between that question and what you have written before it. $\endgroup$
    – garyp
    Commented Oct 29, 2017 at 20:00
  • $\begingroup$ I think that when I've done any work on a body which stores energy in it, and when it's allowed to move that body will do the same Work that I did on it (PE will be +). But If it did not do the same work I've done on it's PE will be neg. @garyp $\endgroup$
    – M.Ahmad
    Commented Oct 29, 2017 at 20:42
  • $\begingroup$ Your understanding of potential energy is incorrect. I will try to find some time to give a complete answer. For now, maybe these points will help. 1.) objects do not have potential energy, systems do 2.) potential energy is defined this way: $\Delta U = -W_\mathrm{internal}$, the work done by forces internal to the system. If I push an object away from the Earth, the source of work is external (me), and the work I do is positive. The Earth applies an internal force (gravity), and does negative work. The change in potential energy, the negative of the internal work, is positive. $\endgroup$
    – garyp
    Commented Oct 30, 2017 at 1:16

3 Answers 3

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The point where you assign a value of zero to gravitational potential energy is completely arbitrary. Over the years, physicists have settled on defining zero gravitational energy for an object as being at an infinite distance and stationary. This assignment has some advantages when considering the total mechanical energy of an orbiting body (e.g., a planet), which consists of the gravitational potential energy plus the kinetic energy of that body:

  • An object with negative total mechanical energy is "bound", meaning that you know for a fact that such an object is orbiting a planet or star in an elliptical orbit
  • An object with zero total mechanical energy is either already at infinity and not moving, or it is passing a planet or star along a parabolic path, and it will eventually end up infinitely far away and not moving
  • An object with positive total mechanical energy is either infinitely far away and moving away, or it is passing a planet or star on a hyperbolic path and will eventually end up infinitely far away and moving away
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In my opinion, not much good in defining absolute gravitational potential energy. Gravitational potential energy is defined as: $$ U = -G \frac{m_{1}m_{2}}{r} $$ So when $r \to \infty $, then $U \to 0$. Thus every physical object at infinite distance from earth will have zero potential energy. See also Gravitational energy and Gravitational potential.

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In case $x$ the system starts with Gravitational Potential Energy (GPE) of $-5$ and ends with GPE of $0$. So the GPE has been increased by $+5$ because $-5+5=0$. The amount of work done is $+5$.

In case $y$ the system starts with GPE of $-50$ and ends with GPE of $0$. So the GPE has been increased by $+50$ because $-50+50=0$. The work done is $+50$.

More work has been done in case $y$ than in case $x$ because $+50 > +5$.

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