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Suppose there are three mass, $m_1$, $m_2$, $m_3$. In the case of equilateral triangle we bring $m_1$ which has zero potential energy then we bring $m_2$ and consider $m_1$ and $m_2$ as one pair and calculate the potential energy of the 1st pair. After this my confusion starts, when we bring the third mass i.e. $m_3$, we calculate its potential energy considering again that at infinity the PE is zero and then get the total potential energy. So do we have choose the same reference point for each pairs to calculate the gravitational potential energy. can we change the refence point for different pairs, eg for the 1st pair $PE=0$ is at Infinity and then for the second pair $PE=0$ is when seperation is let say $a$: because Each pair is independently added. if not then why we choose same reference i.e, at infinity for each pair?

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The point of the potential energy is that its difference gives you work and thus it is determined up to an additive constant. You can use any constant whatsoever, but for certain problems some choices might be better than others.

For example, having n-point masses that interact gravitationally, it makes sense to ask how much energy there needs to be for these masses to break apart. This leads to zero at infinity choice. Even if you do not wish to break them apart, it makes sense as a relative comparison - it is a binding energy of the system and gives you nice scale for comparison of different energy values.

You don't need to make such a choice though. When we deal with systems near the surface of the Earth, like falling rocks and such, we choose as a reference point to be the surface of the Earth (or some other convenient surface). This is again nice choice that gives you additional information about the system.

You can also make choice for each mass to have different reference point, there is no problem. The question is why would you do it? What would be the use of such a choice?

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  • $\begingroup$ Thank u so much for your help. So it means we can choose different reference point for each pairs but it is not useful for comparing energies. Isn't? $\endgroup$
    – Md Faiyaz
    Jun 6, 2022 at 14:15

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