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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.

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closed as off-topic by Kyle Kanos, HDE 226868, Gert, ACuriousMind, user36790 Nov 4 '15 at 16:55

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    $\begingroup$ Hi @ms813, welcome to physics.SE. Note that checking line-by-line math is somewhat off-topic... your final, conceptual, question of is this an appriopriate methodology is more on-point. We need more details however. What are you simulating? What's going on? **You're missing a factor of $r$ in the numerator of the second line of the $U = $ integral.** i.e. it should be $\int_0^R r^2 dr$. $\endgroup$ – DilithiumMatrix Nov 3 '15 at 16:30
  • $\begingroup$ Thanks for your reply and for looking at my equations (please feel free to delete and leave the final line if that would be more appropriate). The simulation is for a bunch of 'asteroids' affected by each other's gravity and bouncing around. If one were to be hit with sufficient force it should split into pieces, otherwise it should just bounce off with an appropriate velocity vector. $\endgroup$ – ms813 Nov 3 '15 at 16:35
  • $\begingroup$ That's a very helpful addendum --- I'll add it into the body of your post. $\endgroup$ – DilithiumMatrix Nov 3 '15 at 16:36
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    $\begingroup$ Welcome to Physics! Please note that Physics.StackExchange is not a homework help site. Please read this Meta post on asking homework-like questions and this Meta post for "check my work" problems. $\endgroup$ – Kyle Kanos Nov 4 '15 at 0:03
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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.

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  • $\begingroup$ Is the constant binding energy effectively friction between the pieces of rubble? I'll take a look at working this into my simulation even if it just some sort of magic 'glue' constant. Can a rubble pile be turned into a solid body, e.g. by melting? (theoretically if not actually in practice) Additionally, is the 2D analogy fair? $\endgroup$ – ms813 Nov 3 '15 at 16:58
  • $\begingroup$ @ms813 Added a comment on the 2D aspect. It's not friction --- it's the chemical/crystal bonds in the rock structure. Theoretically yes, you could 'melt' the rubble into a 'monolith' --- but this is very unlikely in practice. It is generally believed that monoliths form only in very large bodies (which generate enough of their own heat to melt things together), and then are later broken up (e.g. by collisions) into smaller fragments. $\endgroup$ – DilithiumMatrix Nov 3 '15 at 17:07
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    $\begingroup$ I think I'll try to model a group of small objects as a single rubble pile type object, that on reaching a certain size can compress into a monolith. The 2D aspect was more for mathematical and visual simplicity (I'm trying to improve my coding at the same time as my astronomy and vector maths :). Thanks for your help! $\endgroup$ – ms813 Nov 3 '15 at 17:12
  • $\begingroup$ @ms813 that makes sense! I might recommend thinking about it still as 3D spheres --- but when you draw that in a 2D plane, it looks (and acts) like circles :) $\endgroup$ – DilithiumMatrix Nov 3 '15 at 17:19
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@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.)

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  • $\begingroup$ Thanks for the detailed explanation. I probably won't be using the actual value of G, but will end up tweaking it so that the simulation runs in a sensible timeframe anyways. From your line about field strength being proportional to $\frac{1}{r}$, does that imply that in 2D the gravitational force changes from $F=G\frac{mM}{r^2} $ to $F=G\frac{mM}{r}$, and the potential becomes $V = Gm\ln{r}$? $\endgroup$ – ms813 Nov 3 '15 at 20:05
  • $\begingroup$ @ms813: Yes, that's it exactly. Note that the natural log in the 2-D potential makes things signficantly harder to deal with, since in 2D you always have an infinite potential difference between a point at finite $r$ and infinity. I'd stick with a pseudo-3D situation the way you were originally proposing. $\endgroup$ – Michael Seifert Nov 3 '15 at 20:22

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