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if we have $m_1$ and $m_2$ as two particles separated by a distance $r$, then gravitational potential energy of this system is $-Gm1m2/r $. why do we not add their gravitational potential energies with respect to each other and we say that the system has (-2Gm1m2/r) just as we do the same thing (i.e adding) in case of kinetic energy of the same system when they (the two particles) start moving.

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We define the potential energy of the system so that $$M.E. = K + U$$ is constant for internal interactions of the masses. Let $m_1$ be fixed at the origin. When $m_2$ moves from $r_0$ to $r_f$, the work done on $m_2$ by $m_1$ is $$W = \int_{r_0}^{r_f} -\frac{Gm_1m_2\hat{r}}{r^2}\cdot\hat{r}\,dr = Gm_1m_2(\frac{1}{r_f}-\frac{1}{r_0}).$$ This works for any path that $m_2$ might follow. (Why? Because all paths can be resolved into radial and circumferential parts, and since gravity acts centrally, the work done along the circumferential segments is zero.) By the work-energy theorem, this work is the change in the kinetic energy of the system, which must be the negative of the change in potential energy: $$\Delta U = -\Delta K = -Gm_1m_2(\frac{1}{r_f}-\frac{1}{r_0}).$$ This suggests that $$U = -\frac{Gm_1m_2}{r}$$ when the masses are separated by a distance $r$. Any scalar constant could be added to $U$, as only the change in energy is important. The formula you suggested doesn't fit with this analysis, so it is wrong.

Edit:

I meant to add that, in calculating the potential energy of a system of masses, it is the pairs that we are concerned with. In the case of three masses (1, 2, and 3), there are three pairs: 12, 13, and 23. We sum the potential energies associated with each pair to get the total potential energy: $$U = U_{12}+U_{13}+U_{23}.$$ This is completely different from the method you discussed, wherein each particle has an associated energy, and the total energy is the sum of those individual energies. The reason that we care about pairs is that gravity, fundamentally, is a mutual interaction. One mass produces a gravitational field, which exerts a force on another mass. But the second mass also produces a $\vec{g}$-field, which pushes on the first. Regardless of which one is the source, there have to be at least two masses. It is nonsensical, then, to talk about the energy possessed by a single mass-- that energy belongs to the system as a whole.

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This is because potential energy is inherently the amount of energy due to a pair of interacting objects, and not just the properties of a single object (like kinetic energy).

Note that the reason this might be confusing is because you could define a new kind of "potential energy" which is associated with each individual object, and has a value of $-G' m_1 m_2/r$, where I've used $G'$ to stress that the gravitational constant in our new definition has a different value. For two gravitationally interacting particles, the total amount of energy would be, by our new definition, $-2G'm_1m_2/r$, which would be equivalent to the regular expression if we set $G'=G/2$.

This definition is not used, however, since this doesn't correspond to the actual amount of energy as soon as you move beyond two objects. If you had the next simplest possibility, with three gravitationally interacting particles of the same mass, arranged in an equilateral triangle of side length $r$, the total gravitational energy would be:

$$U = 3 \times \left(-\frac{Gmm}{r}\right)$$

whereas our new definition would give a value of

$$U' = 3 \times \left(-\frac{G'mm}{r}\right)$$

which we know is different from our original expression since $G' \neq G$.

tl;dr: We could define a new quantity $-G'm_1m_2/r$, but it wouldn't be useful, because it doesn't actually indicate the amount of potential energy once you move beyond two objects.

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  • $\begingroup$ "you could define a new kind of "potential energy" which is associated with each individual object" Isn't that exactly what the potential is? $\endgroup$
    – my2cts
    Oct 13 '21 at 17:13

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