MSA-The Gravitational 2-Body Problem

__NOTITLE__ Table of Contents = The Gravitational 2-Body Problem =

Absolute Coordinates
A decision was made to let Carol take the controls, for now. Taking the keyboard in front of a large computer screen, she opens a new file in her favorite editor... the exact name of which depends on which of many parallel universes this particular Carol is in, and as such which language the three chose to work in. Expectantly, she looks at Erica, sitting to her left, for instructions, but Dan first raises a hand.

Dan: I'm a big believer in keeping things simple. Why not start by coding up the 2-body problem first, before indulging in more bodies? Also, I seem to remember from an introductory physics class for poets that the 2-body problem was solved, whatever that means.

Erica: Good point. Let's do that. It is after all the simplest case that is nontrivial: a 1-body problem would involve a single particle that is just sitting there, or moving in a straight line with constant velocity, since there would be no other particles to disturb its orbit.

And yes, the 2-body problem can be solved analytically. That means that you can write a mathematical formula for the solution. For higher values of N, whether 3 or 4 or more, no such closed formulas are known, and we have no choice but to do numerical calculations in order to determine the orbits. For N=2, we have the luxury of being able to test the accuracy of our numerical calculations by comparing our results with the formula that Newton discovered for the 2-body problem.

Yet another reason to start with N=2 is that the description can be simplified. Instead of giving the absolute positions and velocities for each of the two particles, with respect to a given coordinate system, it is enough to deal with the relative positions and velocities. In other words, we can transform the two-body problem into a one-body problem, effectively.

Dan: A one-body problem? You mean that one body is fixed in space, and the other moves around it?

Carol: I guess that Erica is talking about a situation where one body is much more massive, so that it doesn't seem to move much, while the other body moves around it. When the Earth moves around the Sun, the Sun can almost be considered to be fixed. Similarly, when you look at the motion of the Moon around the Earth: the Earth has a much larger mass than the Moon, and the mass of the Sun is again much larger than that of the Earth.

Erica: Yes, a large mass ratio makes things simpler. In that case we can make the approximation that the heavy mass sits still in the center, and the lighter mass moves around it. But that is only an approximation, and not completely accurate. In reality, both masses move around each other, even though the heavier mass moves only a little bit.

Even for the case where the masses are comparable, we can still talk about the relative motion between the two particles. And we don't have to make any approximation. Here, let me draw a few pictures, to make it clear.



Since my notepad is two-dimensional, let us start by discussing the two-dimensional two-body problem. Later, we can easily go to 3D. Here, in Figure 5 I show the positions of our two particles by drawing the position vectors r$1$, r 2.

Carol: But in order to define those vectors, you first have to choose a coordinate system, right?

Erica: Yes. We have to choose an x-axis and a y-axis, and they together make up a coordinate system. The point of intersection of the two axes is called the origin of the coordinate system. With respect to the origin, the positions of the two bodies are pointed out by the tips of the arrows that stand for the two vectors r 1, r$2$.

Dan: I see the vectors, but I don't see the bodies.

Erica: You have to imagine the bodies to reside at the very end of each arrow.

Carol: They are point masses, remember, so they are too small to see!

Erica: Well, yes, but of course I could still have drawn them as small points. However, I wanted to keep the figures simple.

Coordinate Systems
Dan: What would happen if you had chosen a different coordinate system?

Erica: In that case, the tips of the arrows would stay at the positions of the particles, since they would not change. However, the arrows themselves would change, because they would be rooted in the new origin.

Dan: But could you really have chosen any coordinate system? And could you then let it change or rotate or let it jump up and down? That seems rather unlikely.

Erica: Ah, no, certainly not. Or more precisely, preferably not. If you choose a coordinate system that moves in a strange way, you then have to correct for those strange motions of the coordinate axes, which would be reflected in the description of the motions of the particles. And you would then have to correct for that, in the equations of motion.

Carol: I'm not sure I follow all that, but I guess the upshot is that you want to keep things simple, and therefore you prefer to use coordinates where the origin stays at rest in space.

Erica: Yes, either at rest, or otherwise the origin should move at a constant speed in one fixed direction. A coordinate system that moves in such a way is called an inertial coordinate system. By the way, we use the term uniform motion for the type of motion that occurs at constant speed in the same direction.

Dan: Ah, that is a term I remember from my physics class. This is related to the fact that Newton discovered that an object which does not feel any forces will keep moving in a straight line with constant speed. Because of the inertia of an object, you have to exert a force in order to change that type of motion.

Erica: Indeed. And this means that any object that feels no force will move in a straight line at constant speed in any inertial coordinate system. But if you start with, for example, a rotating coordinate system, then even an object at rest seems to rotate in the opposite direction, when you describe its motion with respect to the rotating frame.

Dan: And all of this is related to Newton's concept of absolute space, right? And Einstein showed that there is no such thing as absolute space.

Erica: The idea of an absolute space was a brilliant invention, very useful to describe the phenomena on the level of classical mechanics. But when you deal with high velocities, approaching the speed of light, or even with lower velocities and very high accuracies, you have to take into account the fact that special relativity gives a more accurate description of the world.

Carol: And general relativity is even more accurate, I presume?

Erica: Yes, but in general relativity space and time are curved, in a way that is influenced by the distribution of masses and energy in the world. This means that it is extremely difficult to make a computer simulation of the motion of objects such as black holes when they approach each other.

Dan: I'm happy to keep things simple, for now at least, and to stick to Newton's classical mechanics.

Carol: So am I.

A Fourth Point
Dan: Okay, I got the picture. Now where does the one-body representation come in?

Erica: It comes in through a clever coordinate transformation. The trick is to add a fourth point to the picture given in fig. 5, besides the origin and the positions of the two particles. The extra point is what is called the center of mass of the system, often abbreviated to c.o.m. for simplicity.

Dan: You mean the point right in the middle of the two masses?

Erica: No, not if the particles have different masses. Here is the basic idea. Let me draw two masses at rest, in fig. 6. Let particle number 1, at the left, be twice as massive as particle 2, at the right: M$1$=2M$2$. As soon as we let the particles move freely, from rest, they will start falling toward each other.



Carol: Can you write down the equations? Sooner or later we'll have to program them, after all.

Dan: I hope it will be sooner!

Erica: In this very simple example, let us define the x-axis of our coordinate system to go through both particles, with the positive side at the right, as usual.

Carol: Where is the origin?

Erica: You can put the origin anywhere you want. For example, if you like to put it to the left of M$1$, as in fig. 7, then both particles are located on the positive x axis, and therefore both particles have positive values for their positions r$2$ and r$1$ with respect to the origin. Moreover, r$1$&gt;r$2$, so the distance between the two particles is given as r=r$2$-r$1$.



But really, the position of the origin is not important; if you shift the origin to the left or to the right, the value of r remains the same, and that's the only thing that counts.

Dan: Yes, I can see that as long as O stays to the left of both particles. But how about the other cases? Let's check. If we take O to be at the right, then the two particle positions have negative values, with M$2$ further away with a more negative value, say r$1$=-5 and r$1$=-3. In that particular case, r = r$1$-r$2$ = -3-(-5) = 5-3 = 2. Okay, that gives the right answer.

The last possibility is to have O located right in between the two particles. In that case, r$2$ is negative and r$1$ is positive, and clearly r=r$1$-r$2$ is positive as well. Good! I am convinced now that r=r 2 -r$2$ is the right expression for the physical distance between the two particles, with a value that is always greater than zero, if the particles are not at exactly the same place.

Erica: In any of the tree cases that you just explored, the force that particle 1 feels from particle 2 is:

$1$

where r is the distance between the two particles and G' is Newton's universal constant of gravity. Particle 1 is being pulled in the direction of the positive x axis. This means that the velocity changes from zero to a positive value, which in turn means that the acceleration is positive.

In contrast, particle 2 will be pulled to the left, and it will start to move with a negative velocity, so its acceleration is negative. Other than this important minus sign, we can simply reverse subscripts 1 and 2, which gives:

(Equation 2) $$ F_2 ~ = ~ -G\frac{M_1 M_2}{r^2} $$

Dan: So the two forces are equal, but opposite in direction.

Carol: Action equals reaction, that's one of Newton's laws, right?

Erica: Right indeed, that is Newton's third law. Now let us use Newton's second law, to see how the particles will actually start moving. His second law is the famous

(Equation 3) $$ F_1 ~ = ~ M_1 a_1 $$

and of course similarly

(Equation 4) $$ F_2 ~ = ~ M_2 a_2 $$

where a$1$ is the acceleration that particle i undergoes, due to the force F$$.

If we use these relations with Eqs. 1 and 2, we find

(Equation 5) $$ a_1 ~ = ~ G\frac{M_2}{r^2} $$

and

(Equation 6) $$ a_2 ~ = ~ -G\frac{M_1}{r^2} $$

Carol: So even though the forces are equal, the accelerations are not. In the case you mentioned, particle 1 is twice as massive as particle 2. That means that the acceleration a 2 is twice as large as the acceleration of particle 1.

Dan: That makes sense. It takes twice as much force to move a particle that is twice as massive.

Center of Mass


Erica: Let's draw the accelerations, as arrows in the picture, fig. 6 that I sketched earlier. Here they are, in figure 8.

Carol: And this means that the two particles start falling toward each other at different rates, and therefore they will not meet in the middle.

Dan: They will meet closer to particle 1. In fact, because particle 2 will always be accelerated twice as fast as particle 1, I bet they will meet at a point that is one third of the way over, going from 1 to 2.

Carol: Could that be the fourth point that you talked about, Erica, what you called the center of mass?

Erica: Yes, indeed! The center of mass, c.o.m., in the case of the two-body problem, is the point of symmetry, around which each of the two bodies moves in a way that is inversely proportional to their mass. As I remember it, the position of the c.o.m. is given by

(Equation 7) $$ r_{com} ~ = ~ \frac{M_1r_1 + M_2r_2}{M_1 + M_2} $$

Let me show you explicitly how this all works, using the equations of motion, Eqs. 5, 6, where I will remind us that, for each particle, the acceleration is the second derivative of the position:

$i$

and

$i$

We are looking for the point around which everything else turns, a point that does not get pushed, in other words, a point that does not undergo an acceleration. Perhaps you can see how to find a point with zero acceleration, when you add those two equations?

Dan: No, I don't see that. Adding the right hand sides does not give zero.

Carol: Ah, but multiplying the first equation by M$$ and the second equation by M$$ gives

$1$

and

$2$

so when we add the two equations we get zero at the right hand side:

$$

Dan: Okay, that I can see. But what does it mean?

Carol: Erica, correct me if I'm wrong, but I think what it means is that the attractive gravitational forces that the two particles exert on each other are balanced: equal but opposite in direction. For my own peace of mind, let me summarize what I think has happened so far.

Eq. 1 is Newton's gravitational equation, and so is Eq. 2, while 3 and 4 express Newton's law connecting force and acceleration. Combining both of these Newtonian laws allows us to write 5 and 6 as Newton's gravitational equations in terms of acceleration rather than force. After that, 8 and 9 are just rewriting 5 and 6 in a different notation. Then, 12 adds 8 and 9, showing thereby in a more formal way that the attractions between the two stars cancel each other. We could have drawn that conclusion directly from 1 and 2 as well, but there we would not yet have had the information about the second derivatives of the positions.

Erica: Yes, that sums it up nicely. And we can write Eq. 12 as:

$$

Notice that this expression looks a lot like the definition of the center of mass, Eq. 7. Indeed, it implies that:

$$

and with Eq. 7 it can be written as:

$$

This means that the c.o.m. position will either be at rest, or move at a uniform velocity, with zero acceleration.



Carol: Let's check whether that makes sense in our case. We have M$$=2M$$, so:

$1$

Indeed, this is the point that is twice as close to particle 1 as to particle 2.

Dan: Not so fast, how can you see that?

Carol: You can choose the origin of our coordinates where you like. Take particle 1 to be the origin, for example, and you will get r$2$=0 and then Eq. 16 gives you r$$=1/3r$1$. Or take particle 2 as the origin, which means r$com$=0 and Eq. 16 gives you r$2$=2/3r$2$, which means at a distance from particle 2 that is two thirds on the way to particle 1.

Dan: Well, let me try your game too, at a somewhat more complicated point. Let me put the origin of the coordinates exactly half-way between the particles. In that case, r$com$=-r$1$. Now let's see what Eq. 16 gives me. Aha: r$1$ = -2/3r$2$+1/3r$com$ = -1/3r$2$. What do you know! One third of the way from the origin in the direction toward particle 1. Exactly the point that is twice as close to particle 1 as to particle 2. Okay, I'm convinced!

Relative Coordinates


Erica: In the last few pictures we have dealt with a one-dimensional configuration, two mass points on a line. This meant that we only had one coordinate the worry about, the x value of the position. In the more general case of two or three dimensions, we have to go back to vector notation. Eq. 7 in vector form becomes:

$2$

It is also convenient to define r as the relative position of the second particle with respect to the first particle:

$2$

I have drawn them in fig. 10 as they are constructed from r$$, r$$.

Dan: But now you have changed your mind, assuming that particle 2 is heavier than particle 1!

Erica: Why not? These figures should be correct for any choice of masses. I just didn't want to get stuck with the same simple choice for all the figures we will draw.

Carol: So far, you have told us what the center of mass is, but you haven't told us why it is called that way. Looking again at Eq. 17, it seems that the c.o.m. is the mass weighted position of an extended object. We use that notion in computer graphics for computer games all the time. One way to look at it is to ask yourself where you should support a rod with two weights attached at the end. If one of the weights is twice as heavy as the other, it should be twice as close to the support point as the other is, in order to balance the rod.

Dan: I see, that helps. So with `mass weighted' you mean that I multiply the position of each particle with its mass, so that more massive particles have a larger vote, so to speak, in where the center of mass will be.

Carol: Yes, and then you have to divide by the total mass. Already on dimensional grounds it is clear that you have to divide by a mass, as Erica told us, to wind up with a result that has the dimension of length. And in the equal mass case, it is clear that the c.o.m. should be the average of the two positions, the point right in between them.



Dan: Okay, I am happy now with Eq. 17, which leads to fig. 10. But now I'm confused about something else. Let us draw a figure that contains only the two new vectors, R, r that define the alternative coordinate system. Here it is, fig. 11.

Now here is what seems odd. In fig. 5 the two vectors r$1$, r$2$ start at the same point, namely the origin. But in fig. 11 only the c.o.m. vector R starts in the origin, while the relative position vector r connecting the two particles does not.

In other words, the two coordinate systems do not seem to be compatible.

Erica: Good point. It is true that the information contained in the pair of vectors r$1$, r$2$ is the same as the information contained in the vector pair R, r. But in the first case, both vectors appear in the same inertial coordinate system, while in the second case, only R is a vector in an inertial coordinate system, while r points to particle 2 within a coordinate system anchored to particle 1, which is certainly not inertial.





Carol: It is not inertial because particle 1 is not moving in a straight line at constant speed. Is that what you mean?

Erica: Yes. Any coordinate system anchored to a particle that deflected by forces acting upon it cannot be an inertial system.

Dan: So the vectors r$1$, r$2$, R are all defined in the same original inertial coordinate system, while r is defined in a different coordinate system, which is not inertial. Let me make that clear, and draw two new figures. Figure 12 shows the c.o.m. vector in the original coordinate frame, while figure 13 shows the relative separation vector in a different frame, anchored on particle 1, as you just explained.

Erica: Yes, that is a nice way to split it out.