# Gravitational potential energy of a two-body system

We say the gravitational PE of a system is $$-GMm/r$$. This is for a constant gravitational field. But, when we try to calculate PE for a two-body system, the distance the body moves is not the same as in the first case, since the second body is also moving. This creates a changing gravity field. In this case, how do we calculate potential energy? Doesn't the second body have PE because of the gravity produced by the first body (which we ignore while calculating PE, I dont know why)?

I think your intuition about potential energy comes from the gravitational potential energy of objects relative to the ground/earth. In this case, it makes some sense to say that the object possesses PE by virtue of its position relative to the ground.

A more general and useful way to thing about potential energy will be to instead consider the total energy needed to assemble the system of masses. In this case, the PE is not a measure of how much energy there is "contained within each object", but rather a measure of the total energy used to put the system together.

Regarding the question, consider the following experiment: Say we have 2 equal masses, m1 and m2 in empty space separated by a distance x that attract each other.

Case 1: Fix the position of m1 and allow m2 to fall towards m1

Case 2: Allow m1 and m2 to freely attract each other

Now plot graphs of force against displacement for m1 and m2. Clearly, the sum of the areas under the graph is the increase in KE of m1 and m2, which is also the change in PE of the system.

Now for case 1, the graph for m1 has 0 area since it does not move. The graph for m2 goes from X to 0 (or some finite value if m1 and m2 has some radius) and the force, F, goes from -F0 to -F1.

For case 2, the graph for m1 goes from 0 to X/2, and F goes from F0 to F1. The graph for m2 goes from X to X/2, and F goes from -F0 to -F1. If you compare the sum of the areas under the graphs, they are equal. That means that the change in PE of both systems are the same. The general result for gravitation is that the PE of the system does not depend on how the system got to that configuration, it only depends on the configuration of the system.

If the bodies start with a separation of $$r_0$$ and end with a separation $$r_1>r_0$$ (and are initially and finally at rest with respect to one another) then the work done to separate the bodes (assuming we can neglect the work done by all other forces apart from their mutual gravitational attraction) is:

$$\displaystyle \int_{r_0}^{r_1} \frac{GMm}{r^2} dr = \left[ -\frac{GMm}{r}\right]_{r_0}^{r_1} = \left( \frac {GMm}{r_0}-\frac{GMm}{r_1}\right)$$

This work is the change in potential energy of the system. It is not specific to one body or the other. And because gravity is a conservative force, the change in potential energy is the same however we separate the bodies. Whether we hold one body or the other body still, or move both bodies at the same time, we do the same work because the mutual gravitational attraction between them depends only on their separation $$r$$ and not on their absolute positions or velocities.

I am going to answer on the basis of newton mechanics.

while reading your question I feel that you are misunderstanding the potential energy. You asked, " Doesn't the second body have PE because of the gravity produced by the first body?". The answer to this question is "yes, but the second body doesn't have energy. It's the energy of the system of both body".

when we have two body interactions due to gravity. The political energy we talk about is the system of both the body. concept of gravitation field of one object on other objects is to make our calculation easy (choosing one as the reference point).

[...]A single object does not have the potential energy.[...]

(see comment below)

Potential energy function can be derived as the negative work done by a conservative force.

In this case the only force acting on the body (1) is gravitational force due to another body (2). It means that the system must consist of at least two bodies in order to derive such function.

The usual way to calculate gravitational potential energy between two bodies is to define a potential function $$U(R)$$: $$U(R)=-W=-\int^R_{\infty}-G\frac{m_1m_2}{r^2}\hat r \cdot d\vec r=\int^R_{\infty}G\frac{m_1m_2}{r^2}dr=Gm_1m_2\bigr[-\frac{1}{r}\bigl]^R_{\infty}=-\frac{Gm_1m_2}{R}$$

As for movement of bodies, it is the distance between them that matters. We can always chose a reference point to coincide with one of the bodies, making it stationary.

• The Wikipedia statement is, at best, misleading. I would say that it is wrong. A single object does not have the potential energy. As other answers clearly state, the potential energy is due to the system. In fact, the whole first few paragraphs are poor. In a gravitational system of two 3 kg balls, according to Wikip's definition, which object has the potential energy, the ball on the left or the ball on the right? Neither. The system has the energy. Commented Oct 29, 2020 at 14:53
• After rereading the citing, it certainly is misleading. It does the opposite I wanted to stress out, that potential energy is due to the system, not the single body. Thanks for pointing this out. Commented Oct 29, 2020 at 15:23
• Citing my comment. LOL! I think I'll add that to my CV. :) Commented Oct 29, 2020 at 22:42