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At $r = \infty$, the gravitational potential energy of an object due to Earth’s gravitational field is at a maximum value of $0$. I understood this to mean that an object will have maximum potential energy at a distance infinity away from the Earth’s core. This seems to make sense, as the higher you lift an object, the more potential energy it has.

But we also know that $ F= \frac{GMm}{r^2}$, so when $r=\infty$, $F=0 $. So how can an object have maximum potential energy when the Earth is exerting $0$ attractive force on it?

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    $\begingroup$ Absolute values of potentials never mean anything in physics, only differences of potentials i.e. their derivatives. $\endgroup$ Commented May 22, 2021 at 16:58

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You are comparing apples and oranges here. Force is not the same as energy. The fact that at some point in space the force vanishes does not mean that you must have zero potential. It might be useful to recall the relation between the two: $$ \vec{F} = - \boldsymbol{\nabla} U(\vec{x})$$ As you can see from this relation we understand potential energy in a sense as an integral of force over a distance $$ U(\vec{b})-U(\vec{a}) = -\int_a^b \vec{F}(\vec{x}) \cdot {\rm d}\vec{x}$$ which is nothing else but Work. So a way to understand the situation that the OP describes is by realizing that the potential energy "stored" comes from the work needed to place the object there. In our case by working against the force of gravity until you reach $r=\infty$.

As you can see, absolute values of potential make no sense alone but just differences, for the case of gravitational potential as has already been pointed out, the reference is a free particle at infinity. The fact that the potential is negative and increases with distance is only telling you that gravity is attractive (see the minus sign in the first equation), if the potential were positive and decreasing, the derivative would be negative and thus the force positive and repelling.

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I'm +1 on the answer by @ohneVal, but to add a different perspective, I think part of your conceptual problem is reflected by writing "$r=\infty$", whereas the rigorous statements are of the form "$r \rightarrow \infty$". The way you wrote it and described in your question suggests that you believe the particle can actually "reach" infinity. That's not true. It can get farther and farther away, but it never gets "all the way" to infinity. Hence it never reaches a maximum on the potential energy curve. At any finite distance there will always be some force pointing back to the central body, although that force gets very, very small when the particle gets very, very far away.

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The fact that at $r\rightarrow\infty$ the gravitational potential energy $U_\mathrm{grav} = 0$ is only an arbitrary definition. The potential energy is only defined up to an additive constant, because all observable phenomena either take the derivative (like the gravitational force) or differences of it; both these operations cancel out any constant we would add to $U_\mathrm{grav}$.

We could also define $U_\mathrm{grav} = 0$ at some arbitrary other distance $r$. Then anything closer would have negative potential energy, anything further away would go positive. Defining it to be zero at $r \rightarrow \infty$ just makes some formulas nicer.


Regarding how something can have maximum potential energy when no force is acting on it; imagine a (useless) constant potential everywhere. It also has its maximum everywhere, and there is no force from it (because the derivative is zero everywhere). Maximum energy does not mean a lot of force. The force is always related to how quickly the potential changes, i.e. its derivative. If the potential is really high but flat, the resultant force is zero.

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  • $\begingroup$ Yes, I understand your point about how our choice of reference point makes things easier. I’m still confused though. At a distance infinity away, an object’s GPE has a numerical maximum value of 0. But it doesn’t make sense to me how an object can have ‘maximum’ GPE when it has completely escaped Earth’s gravitational field? $\endgroup$
    – Freddie
    Commented May 21, 2021 at 15:05
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But we also know that $ F= \frac{GMm}{r^2}$, so when $r=\infty$, $F=0 > $. So how can an object have maximum potential energy when the Earth is exerting $0$ attractive force on it?

Rather than rehash what others have already said, I will focus on the above statement.

Yes, as $r\rightarrow\infty$, $F\rightarrow 0$, and therefore the work $W\rightarrow 0$ and increase in potential energy $GPE\rightarrow 0$ as the object moves further away. But the total potential energy is the sum of all the work done to move the object from the surface of the earth to that point far away.

We usually don't think about the fact the gravitational force on an object that we lift on the earth is less after we lift it because the change is infinitesimally small at the surface of the earth, so we assume $g$ to be constant. But technically each time we lift it further the force decreases and so does the subsequent increase in GPE.

Hope this helps.

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So how can an object have maximum potential energy when the Earth is exerting 0 attractive force on it?

Actually, you can only have a maximum or a minimum of the potential energy where the force is zero. That is, in fact, one of the defining characteristics of a minimum or a maximum potential. It doesn't matter if it is at infinity or somewhere else, a minimum or maximum of the energy must always coincide with 0 force.

For example, for a spring the force is 0 at the relaxed position which is also a minimum in the elastic potential energy. A hypothetical anti-spring which pushes things away from the center $F= k x$ (instead of $F=-kx$) would also have 0 force at the relaxed position and it would be a maximum.

This makes sense because the force is the slope of the potential energy and the slope is always 0 at a min or a max.

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