Gravitational binding energy of 2D circle [closed]

Question

I'm interested in calculating the gravitational binding energy for an object modelled by 2D circle for a small collision simulator.

In the simulation, I'm using a 2D equivalents of 3D properties (e.g density of the circle is calculated as $\frac{mass}{\pi r^2}$ in units of $kg \ m^{-2}$)

By following the derivation on the wikipedia page, but for a circle instead of a sphere, I end up with the following:

$$ m_{shell} = 2 \pi r \rho \ dr \\ m_{interior} = \pi r^2 \rho $$ Integrating over all circular shells $$ U = -\int_0^R G\frac{(2 \pi r \rho)(\pi r^2 \rho)}{r} dr $$ Solving the integral and substituting in $\rho = \frac{m}{\pi R^2}$: $$ = -\frac{2}{3}\frac{Gm^2}{R} $$

Can someone confirm/correct my derivation, and explain if this is a reasonable way to go about calculating the energy needed to 'destroy' one of the balls in the simulation?

Edit: the simulation is a bunch of 2D 'asteroids' that are affected by each other's gravity and can collide with each other. If one is struck with sufficient force, it should split into smaller pieces, else it should just bounce off.

Answer 1

Asteroids are held together by a combination of gravity and cohesive (electromagnetic) forces (the same forces which hold rocks together on earth). For small asteroids (smaller than about 1 km), the gravity is negligible, while for larger bodies (larger than 10s to 100s of km) the cohesive forces become negligible*.

If you're only interested in the larger bodies, then your purely gravitational approximation is good.

If you only include gravity then what your modeling are so called 'rubble piles' --- which isn't a terrible approximation. But if you want to be a bit more realistic for the smaller bodies, you could include a constant binding energy which doesn't depend on mass. The magnitude could be comparable to the gravitational binding energy for a roughly 1 km radius object, but you should choose it to match your desired dynamics.

Regarding the 2D formalism: it's not entirely clear why you're choosing to do this rather than 3D, but it is effectively the same. You're just using 'surface density' instead of volume-density. It's like your asteroids are pucks instead of spheres.

Answer 2

@DilithiumMatrix has talked a bit about the physical meaning behind your answer, and the important distinction behind the gravitational and electromagnetic binding forces. I want to point out that your gravitational calculation is not actually correct (though it might be good enough—see the end of the post.)

Why your answer isn't actually correct:

The problem is that if you have a 2-D asteroid in 3-D space, then it's not true that the potential at the edge of a disc is equal to the mass of the disc divided by its radius (times $G$): $$ \Phi \neq -\frac{ G (\rho \pi r^2)}{r}. $$ It's true that the gravitational force on the surface of a spherical shell depends only on the mass enclosed in that shell; in this case, the force acts just like there was a point mass at the center of the shell. But for any other configuration of mass inside the spherical, the force will vary in direction & magnitude over the surface, and that means that the potential will also vary in magnitude over the surface. It doesn't act like a point mass concentrated at the center any more.

If, on the other hand, you have a 2-D asteroid in a 2-D Universe, then it's not really natural for the gravitational force to fall off proportionally to $1/r^2$, and so the potential probably shouldn't fall off proportionally to $1/r$. This is easiest to see in terms of electric fields, which behave just like gravitational fields in 3-D (i.e., they obey an inverse-square law.) If you think about the field lines of an isolated point charge, they stream out from the point charge in straight lines in all directions; the fact that they get farther apart from each other is tantamount to saying that the field gets weaker as you get farther away. Specifically, the strength of the field at a particular distance is proportional to the number of field lines per area at that distance; thus, since we have the same number of field lines at any particular distance, and the areas of enclosing spheres go proportionally to $r^2$, then it follows that the field strength is inversely proportional to $r^2$.

But if you translate this argument to 2-D, and you want the gravitational/electric field lines to have the same interpretation, then the field lines are going to have to spread out over the circumference of a circle, not the surface of a sphere. Therefore, since the circumference of a circle is proportional to $r$, the field strength in such a Universe will be proportional to $1/r$, not $1/r^2$. The potential, it can be shown, is then proportional to $-\ln r$ rather than $1/r$.

Returning to the case of a 2-D asteroid in a 3-D world: the actual calculations for this are pretty nasty, and I don't think there's an exact numerical answer. The problem is basically equivalent to asking what the electric potential of a uniformly charged thin disc is, and a closed-form solution in terms of "nice" functions probably doesn't exist. However, I've done a related problem in the past, and I was able to fire up the code and get what I think is an approximate numerical answer: $$ U \approx -0.424 \frac{Gm^2}{R}. $$

Why your answer might still be good enough:

In some sense, the numerical factor in front doesn't really matter all that much. If, for example, one used asteroids that were twice as dense, this would double all of their masses and thereby quadruple all of their binding energies. It's not too hard to see, in fact, that changing the numerical factor in front of the formula is equivalent to changing the density of the asteroids. Really, what's important to get is the scaling of $U$ relative to $m$ and $R$ correct, and the only possible answer to this is $$ U = - k \frac{GM^2}{R} $$ for some dimensionless constant $k$. (The combination $G M^2/R$ is the only way to combine $G$, $M$, and $R$ to get something in units of Joules; the only thing that can differ about your final answer is the value of $k$.)

I think that as long as you give your asteroids a binding energy that is proportional to the square of their mass, and inversely proportional to their radius, then the physics will still be pretty realistic. If this were for a game, for example, I'd probably tell you to ship that code as-is. If, on the other hand, this were for an academic monograph, then you'd want to be a lot more careful with the exact value of $k$. (You'd probably be working in 3-D in the first place, though, so that's as may be.)