Potential energy of system or particle of the system?

Question

Why in gravitational potential or also electrical potential energy we refer to potential energy of 1 particle of the system and not whole system? for example in system earth and a ball we speak about the potential energy of ball at height $h$. Why we dont include earth? Same happens with electric potential energy where we refer to potential energy of a single electron or charged particle

Answer 1

As you suspect, we should refer to the potential energy of the system.
More specifically, the potential energy is a property of an interaction (not of just one particle).

Here is a quote from a textbook that emphasizes this point

Six Ideas that Shaped Physics - Unit C (Thomas Moore) - 2nd edition, p.105
Exercise C6X.4

In circumstances where an object interacts with the earth, it is tempting to think of the potential energy as belonging to the object instead of the interaction, since the potential energy varies as the object changes position (the earth seems fixed). Many textbooks in fact would say that "the object's potential energy is converted to kinetic energy" as it falls. Why is this kind of language not helpful when the interacting objects have comparable mass?

Answer 2

Potential energy is always associated with a system of two or more interacting objects.

We usually refer to the potential energy associated with just one of the objects but it would be more appropriate to assert 'potential energy associated with the system'.

Actually it would be incorrect to link the potential energy to just one of the objects, ignoring the Earth (for example).

Answer 3

First of all when we usually refer to the gravitational potential energy or electrostatic potential energy in most common cases, we usually refers to that due to earth(grav. Pot.)or a unit positive carge (electro. Pot.). Secondally we usually don't use that term for a system of particle in most common use to make calculations simpler .For example if you consider gravitational potential energy of yourself

• which is $$(-GMm/r)$$ Where r is the diatance of your body from centre of earth ,M mass of earth., m is your mass,G is universal gravitational constant.

• Now if you consider that for a system of particle for your body you should spoose to write

• $$P.E=((-GMm1)/r1)+((-GMm2)/r2)+........((-GMmn)/rn)$$ where $$m1,m2,.... mn$$ are the masses of different particles of your body and $$r1,r2,....rn$$ are distances of different particles of your body from centre of earth .Now your height will be like 2metre which can be ignored if you see radius of earth that is $$6.4×10^6m$$ so all of the distances $$r1,r2.....rn$$ will nearly be equal and add up to give you the same result as calculated by equation for single particle system.

Answer 4

My answer will depart in some respects from the others I've seen. Yes of course,

Potential energy is always associated with a system of two or more interacting objects.

as JD_PM (and all others, myself included) do say. But I ask to anybody: are you sure you (and me) never used potential energy the other way, as a quantity pertaining to a single body in a given environment?

Just an example among thousands possible. An electron is accelerated between two electrodes, with a potential difference 200 volt. If the electron starts from rest, find its final velocity.

There is someone who would attribute the potential energy to the whole system in solving this problem? (And it would be far from trivial: what is exactly the whole system? Electron + electrodes? Or should we include the electric field? And why not the generator?)

In short: there are a lot of situations where to attribute potential energy to a single body is a perfectly legitimate approximation. So good, to be sure, that it would sound ridiculous to worry about it. Surely all problems of motion of a body in Earth's gravitational field are part of the set. The only exception I can see is Moon.

It's not only "to make calculations simpler": It's that in most cases we couldn't observe the difference. We are out of experimental possibilities for several orders of magnitude. It's an important part of a physicist's skill to be able to identify what objects, actions, effects, are relevant for a given problem and for the desired level of accuracy.

A little numbers just to show OP how things go as Earth is concerned. (Physics without numbers is often just chatter.) Consider ISS, whose mass is $4\cdot10^5\,\mathrm{kg}$. Its speed is about $8\,\mathrm{km/s}$. Then kinetic energy is $1.3\cdot10^{13}\,\mathrm J$. Now think of Earth. Its mass is over $10^{19}$ times larger, so in the common c.o.m. frame its speed will be less in the same ratio, and the same for kinetic energy (can you see why? velocity enters squared). So we find for Earth's KE something like $10^{-6}\,\mathrm J$. Can you imagine how small an energy is it?

I reasoned on kinetic energies, but my argument holds for potential energy too. Should ISS' KE be brought to zero, because an irrealistic orbit change, an equivalent gain in potential energy would be produced. Neglecting Earth's contribution we would err by $10^{-6}\,\mathrm J$ in about $10^{13}$.

Now an exercise for OP. Try a similar estimate for the electron's problem. Then examine the Sun-Jupiter system: what can you say about it?