2
$\begingroup$

In my CIE A level course, the gravitational potential energy of a mass in a gravitational field is defined as the work done in bringing the mass from infinity to that point without changing it’s k.e. energy.

I thought about the gravitational potential energy if a system of more than 2 masses; it would obviously be lower but I cannot compute it with this definition. What would be the proper definition of the g.p.e of an $n$-body system and how would it be calculated?

P.S: I know that it is nonsensical to define g.p.e of a mass, but that’s just how they define it for some reason, and it works for a 2-body system.

$\endgroup$

1 Answer 1

3
$\begingroup$

The gravitational potential energy of a system of $n$ bodies is calculated by the formula, similar to the formula of potential energy of the two bodies. You just need to apply the two-bodies-formula for all pairs of bodies and sum up the results: $$ E = - \sum_{i=0}^n{\sum_{j=i+1}^n}Gm_im_j/r_{ij}$$

Why is it so? Suppose the formula is correct for $n-1$ bodies. Now you bring $m_n$ from the infinity. At any moment the total gravitational force acting on $m_n$ is equal to the sum of gravitational forces produces by individual masses $m_i$, so the total work done by gravitational forces will be the sum of of works done by these individual forces, so the total work would be $E = \sum_{i=0}^{n-1}Gm_nm_i/r_{ni}$.

$\endgroup$
7
  • $\begingroup$ Is it possible to calculate potential energy of one body in such system (so sum of kinetic and potential is constant for one body)? Like $mgh$ formula was for one body only. I suspect that no, probably because energy is split between all bodies and is no longer constant for each of them, but just wanted to clarify. $\endgroup$
    – Somnium
    Commented Mar 13, 2023 at 10:10
  • 1
    $\begingroup$ @Somnium Yes, it is possible. Let's take some path starting from the position of our body and leading to infinity. We can calculate the work one needs to do to slowly move the body along to this path to infinity.Split the path into many-many small parts, energy required to move the body along the small part would be $\vec{F}*d\vec{s} = ( \sum \vec{F_i}) * d \vec{s} = \sum (\vec{F_i} * d\vec{s}) $. This product is linear, so we can first sum only force from first body - the result will be equal to potential energy of our body in the field of $m_1$, then forces from second body, etc. $\endgroup$
    – lesnik
    Commented Mar 13, 2023 at 23:03
  • 1
    $\begingroup$ @Somnium So, to remove body number $k$ from this system it is necessary to do a work $ E = - \sum_{i=0; i\ne{k}}^n{ G m_i m_k /r_{ik}}$ $\endgroup$
    – lesnik
    Commented Mar 13, 2023 at 23:08
  • $\begingroup$ But if you sum all $E_k$ then we count each term twice and totall energy appears to be two times bigger than in your answers. What I am missing here? $\endgroup$
    – Somnium
    Commented Mar 14, 2023 at 7:40
  • $\begingroup$ @Somnium Well, let's start with 2 bodies. Potential energy is $E = - G * m_1m_2/r^2$. That much works is required to move one of the bodies from initial position to infinity. Any one of them. But to calculate energy required to move both bodies to infinity you can't just sum up works for both bodies - the answer will be incorrect by factor of 2. After the first body is moved there is no gravitational field and energy required to move out second body is 0. Situation is very similar with $n$ bodies. $\endgroup$
    – lesnik
    Commented Mar 14, 2023 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.