2
$\begingroup$

Lets define the Gravitational Binding Energy of a set of point masses as the work required to separate all point masses by an infinite amount.
First Question: Is the amount of energy required to do this independent of the path along which each point travels? Will the amount of work required to do this always be constant, as long as all points end up infinitely far apart?
Second Question: Based on the assumption that the Gravitational Binding Energy is independent of how the points are pulled apart, I calculated the amount of energy to separate all points by a finite scale factor, then defined the binding energy of the limit of that as the scale factor goes to infinity, and I came up with the following equation. Given $n$ masses, the binding energy $E_B$ is: $$E_B=\sum_{i=1}^n \vec F_i \cdot \vec p_i$$ Where $\vec F_i$ is the total gravitational force from the other masses on mass $i$, and $\vec p_i$ is the position of mass $i$ in space. Can someone confirm the correctness of this equation?
Third Question: Imagine you have a cloud of gas and dust that's undergoing gravitational collapse. As it collapses, the Gravitational Binding Energy increases (since everything is getting closer together), and the kinetic energy of particles within the cloud increases (since they're accelerating inward). Ignoring the effects of friction, etc, if the kinetic energy increases a set amount, will the gravitational binding energy increase a corresponding amount? That is to say, will the difference between the gravitational binding energy and the kinetic energy of the system remain constant?
Fourth Question: If all of the above are true, given that $$K_E + U = (constant)$$ and $$E_B - K_E = (constant),$$ can I use $$U=E_B - 2 K_E$$ As a metric for the potential energy of a gravitational system? In the above equations, $U$ is potential energy. Should I use this as a metric for potential energy? If not, how else should I ensure that total energy remains constant within a gravitational simulation?

$\endgroup$
1
$\begingroup$

Your reasoning is correct.

1st question: Yes, gravitational binding energy is a state quantity depending only on the system's configuration. This follows from the fact that gravitational force is conservative.

2nd question: The gravitational binding energy should be $$E_{\mathrm B} = \sum_{i,j\\i<j} F_{ij} \cdot (r_j-r_i)\,,$$ where $F_{ij}$ is the gravitational force on mass $i$ due to mass $j$ and where $r_i$ is the position of the mass $j$.

3rd question: Yes, the increase in kinetic energy is equal to the work done (which, by definition, is equal to the increase of the gravitational binding energy). This would be called the work-energy principle for a many-body system.

4th question: Yes, you can define the potential energy like that, though it is more common to define it as $$U = -E_{\rm B}$$

Justification of 1st (and partly 2nd) answer:

(Based on The work-energy theorem in a many-body system.)

$$W = \sum_i \int \sum_{j \neq i}-F_{ij} \cdot {\mathrm d}r_i = -\sum_{i,j\\i<j} \left(\int F_{ij} \cdot {\mathrm d}r_i + \int F_{ji} \cdot {\mathrm d}r_j\right)$$ The gravitational force is conservative: $$F_{ij} = -\nabla_i U(|r_i - r_j|)$$ and thus: $$W = \sum_{i,j\\i<j} \left(\int \nabla_i U(|r_i - r_j|) \cdot {\mathrm d}r_i + \int \nabla_j U(|r_i - r_j|) \cdot {\mathrm d}r_j\right)$$ In other words: $$W = \sum_{i,j\\i<j} \int {\mathrm d}U_{ij} = \sum_{i,j\\i<j} \Delta U_{ij}\,,$$ where $U_{ij} = U(|r_i - r_j|)$. Now if we define $U = \sum_{i,j\\i<j} U_{ij}$, we get $$W = \Delta U$$ Observe that $\Delta U$ depends only on the initial and final configurations.

$\endgroup$
  • 1
    $\begingroup$ Gravitataional binding energy should definitely be an integral of force over a path. $\endgroup$ – Jerry Schirmer Nov 2 '16 at 2:14
  • $\begingroup$ The integral has a closed-form solution $\endgroup$ – J. Antonio Perez Nov 2 '16 at 2:46
0
$\begingroup$

trosos has answered all your questions but I think the relation under question 4 might be an error depending on what you meant.

If you are trying to relate known/calculated values $K_E$ and $E_B$ with the unknown $U$, then you are kind of right. You put forward the relations $$K_E + U = c_1$$$$E_B - K_E = c_2$$ and proceeded to solve them to give $$U=E_B - 2 K_E$$ but this is only true if the constants in the above two equations are the same. If you have not defined $U$ yet, then your relation is valid as potential energy is not an absolute value. The potential energy of masses at infinite separation can be chosen to make the constants of the equations to be equal.

In case you are using the conventional or textbook definition of $U$, then your relation is not correct as the constants don't match.

This answer may be completely unnecessary if I misunderstood you so I apologise beforehand.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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