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I have studied about Gravitational potential energy ($GPE$) and Gravitational potential ($GP$). ..while everywhere it is said that $GPE$ is always of a system as it depends on the configuration of the particles with respect to each other ...but nothing as such is said about it in my books..so my question is...

1)is $GP$ also defined for a system like say two particles...??

A)I think it is true because ..$GP$ and $GPE$ are very much related..

B)But then I also think it is not true because of the definition I have that Gravitational potential at a point is also defined as..The work done per unit mass by an external agent in bringing a particle slowly from the reference point to the given point.

2)But nothing is being said about another particle(system) in this definition..

3)I am also really confused..with these terms $GP$ and $GPE$ because there are different definition of these in different places..{I took The standard definition of $GPE$ from (Resnick,Halliday) but the book doesn't talk about $GP$}...

Can someone please clarify these??it will be of great help.

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    $\begingroup$ $GPE$ is gravitational potential energy. $GP$ is gravitational potential energy per mass. $\endgroup$
    – Steeven
    Commented Nov 11, 2015 at 7:28

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What is gravitational potential?

Usually a potential is defined as the potential energy per mass or per charge or similar. This is most often seen in relation to electricity or chemistry and less often to gravity.

$GPE$ is gravitational potential energy. $GP$ is gravitational potential energy per mass:

$$GP =\frac{GPE}{m} $$

Is it defined for the system or the object?

$GPE$ is the amount of energy that can potentially be used as work for pulling an object towards the source. It depends on both the source's mass $m$, on the mass of the other ("target") object $m_0$ as well as on the distance $r$ between them ($G$ being a universal constant):

$$GPE=-G\frac{m m_0}{r}$$

$GP$ is the amount of energy that can potentially be used as work for pulling one kilogram of mass towards the source. It gives an easier basis for comparison as it does not depend on the specific target object's properties anymore:

$$GP=\frac{GPE}{m_0}=-G\frac{m}{r}$$

BUT it still depends on the distance $r$ between them. $GPE$ is system dependent as it depends on both the object properties and on the system configuration, but even though $GP$ does not depend on the object properties, and thus is defined for the source object alone, it still depends on the system configuration. If no other object was present in the near neighbourhood (if $r\rightarrow \infty$), then $GP\approx 0$.

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  • $\begingroup$ .so does that means GP is also of a system(say ..two particles)..(by this equation) $\endgroup$
    – Freelancer
    Commented Nov 11, 2015 at 9:00
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    $\begingroup$ Hi @Freelancer, I have now found the time to elaborate and answer the full question. See the edited answer. Hope this solves it. $\endgroup$
    – Steeven
    Commented Nov 12, 2015 at 7:38
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    $\begingroup$ Yes, it's OK. If you have a 1 kg mass. You might consider an object as made up of many 1 kg masses. The $GP$ is the potential energy associated with each of these small masses. The $GPE$ is then just the total (the sum) for the whole object. $\endgroup$
    – Steeven
    Commented Nov 12, 2015 at 11:11
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    $\begingroup$ Well, yes. It is right for a 1 kg mass. $\endgroup$
    – Steeven
    Commented Nov 12, 2015 at 12:52
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Steeven
    Commented Nov 12, 2015 at 16:04
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is GP also defined for a system like say two particles

Yes.

You can admire in the attached picture the Contours of Equal Gravitational Potential for the system Earth - Moon.

Contours of Equal Gravitational Potential Contours of Equal Gravitational Potential

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  • $\begingroup$ The figure you're shown here is not exactly the gravitational potential. It is the effective potential in the rotating frame (that is it has a centrifugal pseudoforce contribution computed in). $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 12, 2015 at 15:14
  • $\begingroup$ There is no need of any centrifugal force. Because the moon moves around earth, in any point of the orbital plane there will be a time changing potential $V(x, y, t)$ or $V(\omega, r, t)$. Such a time varying potential is quite ugly and not quite suggestive but once you have its mathematical expression you can find, by pure math (the laws of mechanics being no longer involved), a convenient rotating reference system where variable time, $t$ disappears and you get what is called the Contours of Equal Gravitational Potential. $\endgroup$
    – Energizer777
    Commented Nov 12, 2015 at 20:14
  • $\begingroup$ I'm not suggesting adding one, it is already in the figure you are showing. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 12, 2015 at 20:21
  • $\begingroup$ The map "Contours of Equal Gravitational Potential" was not computed using the "centrifugal pseudo-force" concept. It is also not $V(x, y, t_0)$, it is not a picture of $V(x,y)$ at moment $t_0$. The map is exactly what I have already explained in my previous post. $\endgroup$
    – Energizer777
    Commented Nov 12, 2015 at 21:12

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