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In a two particle system, suppose both are moving by their internal gravitational force, and hence the potential energies of the system is changing. Since both are acted upon by the conservative forces i.e. gravitational force and we know that the potential energy is equal to negative of work done by the conservative forces, can't we calculate potential energy of both the masses individually because both conservative forces is doing some work. Then add the values of potential energies to get the net change in potential energies of the system?

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2 Answers 2

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Gravity is a pairwise interaction with an associated potential energy of the form

\begin{equation} E = - G\frac{m_{1}m_{2}}{\left|{\vec r_{1}} - {\vec r_{2}}\right|} \end{equation}

where $m_{1}$ and $m_{2}$ are the two masses involved in the interaction and $r_1$ and $r_{2}$ are their positions. From the form of this potential its obvious that it makes no sense to ask about individual potential energies.

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  • $\begingroup$ It still makes sense to say they have individual potential energies. Just to say that they mutually share this energy instead. $\endgroup$ Jun 6 at 16:12
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The fundamental answer to your question is that the potential energy between 2 charges is mutually shared.

It is true that mass $M_{1}$ has a potential energy of $$U_{1} =-\frac{Gm_{1}m_{2}}{R}$$

It is also true that mass $M_{2}$ has a potential energy of $$U_{2} =-\frac{Gm_{1}m_{2}}{R}$$

It is NOT true however, to state that the potential energy of the system is $U_{1} + U_{2}$

"because both conservative forces is doing some work"

The definition of potential energy, is the amount of work required to bring a charge from infinity to some r,against the force of gravity.

This is work done by ME, not the gravitational fields.

The definition of potential energy of a system, is the amount of work to build up a distribution.

Let's say that we want to make mass $m_{1}$ move to location x, what is the work that is needed to be done to move this mass from infinity to x?

There are currently no external gravitational fields, so zero work is needed.

Now let's build up mass $m_{2}$ in the presence of the gravitational field from $m_{1}$

$\int_{\infty}^{r} -\vec{F} \cdot \vec{dr} = -\frac{Gm_{1}m_{2}}{R}$

"Paradox 1":

You may now think that "hold on a minute, you've said that 0 work is required to move $m_{1}$ in its origin position, yet it HAS gravitational potential energy"

Yes, that is very true, since the calculated potential energy of $m_{1}$ assumed $m_{2}$ "existed" first. This is the amount of work required, given $m_{2}$ Existed at point y, to move $m_{1}$ to its location, in the presence of gravitation forces.

Potential energy of a particle, assumed the other parts of the distribution existed "first". Which is why both particles have some potential energy, That is mutually shared. It is the same energy!

It is the work required to bring them together.

If I fix "m1" in place, "m2" gains that energy if it moves to infinity. If I fix "m2" in place, "m1" gains that energy if it moves to infinity

Paradox 2:

if I move mass 1, its PE changes,corresponding to work being done on mass 1. But if PE is mutually shared, mass 2's PE changes, so work must be so done on mass 2.

No, potential energy required fields are static. Using the change in potential for mass 2, as proof of work being done, when in reality its because mass 1 moves, is invalid as there is a change of "external field"

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