Why is the center-of-mass of 2 bodies at the focus of their elliptical orbits?

Why is the center-of-mass of 2 bodies (which interact only via Newtonian gravity) located at a focus of each of the elliptical orbits?

I know that when there are no external forces, the center of mass moves at a constant speed, but that doesn't explain it.

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"they orbit around their center of mass" That's a bit of an oversimplified statement. You do know that orbits are, in general, not circular with a well-defined center point? – leftaroundabout Jun 11 '12 at 13:53
The important thing here is that the center of mass is an invariant in a frame where it is initially at rest. This actually follows from the definition of the thing and Newton's laws, but I've become convinced that the point is subtle and important enough (if basic) to merit a careful answer. – dmckee♦ Jun 11 '12 at 14:00
If the orbit is elliptic, the center of mass will be at one of the ellipse's focuses. But why? – Omer Jun 11 '12 at 14:55
Wait. Is the question about the invariance of the CoM or about the shape of two body orbits? – dmckee♦ Jun 11 '12 at 16:22

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The question is about the shape of two body orbits - why the CoM is one of the two ellipses' focuses. – Omer Jun 11 '12 at 17:13

2 Answers

I'm going to assume that Omer is specifically asking why the centre of mass is at the focus (well, one of the foci) of the orbits. Omer, if this isn't what you meant please ignore what follows because it's completely irrelevant!

If you have a body moving in a central field (i.e. the force is always pointing towards the centre), and the field is inversely proportional to the square of the distance from body to the centre, then the orbit is an ellipse with the centre at one of the foci. For now let's just assume this and we can come back to prove it later.

So if we can show that both of the bodies feel a central inverse square force, with the COM at the centre, this guarantees the orbits will be ellipses with a focus at the COM. Given that the force is due to the two bodies attracting each other, and that both bodies are orbiting around, it may seem a bit odd that each body just feels a central inverse square force, but actually this is easy to show.

The picture shows the two masses and the COM. I haven't shown the velocities because it doesn't matter what they are. For now let's just consider $m_1$ and calculate the force on it. By Newton's law this is simply:

$$F_1 = \frac{Gm_1m_2}{(r_1 + r_2)^2}$$

First is this force central? We know the centre of mass doesn't move. For two bodies this seems obvious to me, but in any case dmckee proved it in his answer. If the COM doesn't move it must lie on the line joining the two mases, otherwise there'd be a net force on it. So the force $F_1$ must always point towards the COM i.e. the force is central.

Second is this an inverse square law force i.e. is $F_1 \propto 1/r_1^2$? Well the definition of the centre of mass is that:

$$m_1r_1 = m_2r_2$$

or

$$r_2 = r_1 \frac{m_1}{m_2}$$

If we substitute for $r_2$ in the expression for $F_1$ we get:

$$F_1 = \frac{Gm_1m_2}{(r_1 + r_1(m_1/m_2))^2}$$

or with a quick rearrangement:

$$F_1 = \frac{1}{(1 + m_1/m_2)^2} \frac{Gm_1m_2}{r_1^2}$$

and this shows that $F_1$ is inversely proportional to $r_1^2$. I won't work through it, but it should be fairly obvious that exactly the same reasoning applies to $F_2$ so:

$$F_2 = \frac{1}{(1 + m_2/m_1)^2} \frac{Gm_1m_2}{r_2^2}$$

This is the key result. Even though the two bodies are whizzing around each other, each body just behaves as if it were in a static gravity field, but the strength of the field is reduced by a factor of $(1 + m_1/m_2)^2$ for $m_1$ or $(1 + m_2/m_1)^2$ for $m_2$. This applies to all two body systems, even such unequal ones as the Sun and the Earth (ignoring perturbations from Jupiter etc).

I did start by assuming that a body in a central gravity field orbits in an ellipse with the foci at the centre, but I'm going to wimp out of proving this since it would double the length of this answer and you'd all go to sleep. The proof is easily Googled.

NB this only applies to two body systems. For three or more body systems the orbits are generally not ellipses with the centre of mass at the focus.

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Thanks! It was exactly what I searched for. – Omer Jun 11 '12 at 18:25

This actually follows from Newton's laws (and it only holds true for an isolated system).

For simplicity we'll consider an isolated system of two bodies on a line. Call their masses $m_1$ and $m_2$ and put them at $x_1$ and $x_2$. Now computer their center of mass: $$X = \frac{\sum_i m_i x_i}{\sum_i m_i} = \frac{1}{M} \sum_i m_i x_i$$

Assume that there is some force, $F$, between them. Newton's second law tells us that the force on body one due to body two $F_1$ is equal and opposite that on body two due to body one $F_2 = -F_1$.

This is enough information to compute the acceleration of the CoM: $$A = X'' = \frac{1}{M} \sum_i m_i a_i = \frac{1}{M} \sum_i m_i \frac{F_i}{m_i},$$ cancel the masses inside the sum and take note of the relation between the forces and you get $$A = 0 .$$

For more bodies and more dimensions the math gets more complicated but the result is the same.

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I think Omer is specifically asking why the centre of mass is at the focus of the orbits – John Rennie Jun 11 '12 at 15:11
@JohnRennie: This is the reason they move "around" the CoM. Any movement away on the mart of mass one must be mirrored by a mass-ratio weighted movement away on the part of mass two (as long as only internal forces are acting). And likewise for movements toward, and in all relevant dimensions. I know it seems like a different question, but it's not. That is the subtle point that makes this question interesting and pedagogically important. – dmckee♦ Jun 11 '12 at 16:19
Ah...seeing the OP's comment above maybe I have answered the wrong question. – dmckee♦ Jun 11 '12 at 16:21
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