Originally Posted by

**grapes**
I think I see what you're asking for, good question.

Y'all are better at this than I am, but it is much easier than I thought. [And I have been limited to an iphone lately.]

F = mA

A is acceleration; we will use “a” for the semi-major axis of the system later

1) s1 =1/2 * A1 * t^{2}

2) A1 = 2*s1/ t^{2} = 2*r1/ t^{2}…. Because r1 = s1

F1 = F2….. from the inverse square law (i.e. gravity)

From, F = mA

m1*A1 = m2*A2

Substituting from eq. 2…

m1*(2*r1/ t^{2}) = m2*(2*r2/ t^{2})

Resolves to…

3) m1*r1 = m2 * r2……. This is the seesaw equation; summation of moments = 0 stuff.

To get it to the common barycenter equation….

4) a = r1 + r2………… where “a” is the semi-major axis.

From eq. 3…

r2 = m1*r1/m2

a = r1 + r2 = r1+ (m1*r1/m2)

a = r1*(1 + m1/m2)

thus…

**r1 = a/(1+m1/m2)**

As **Hornblower** has stated, no orbital dynamics are needed to give us a c.g. or barycenter, though as a layperson I suggest the use of "barycenter" be associated with orbital masses. [When I was very little, I still recall seeing a picture (perhaps my first) of an airplane in the air and it was just.... resting there in the air. I don't see two masses in space just sitting there worth determining their barycenter; if they aren't going to fly, let 'em fall.]

The seesaw analogy is helpful but it assigns the force, F, perpendicular to the board (moment arm), whereas here we use F (from gravity) aligned radially with the "board". [The perpendicular F for seesaws is really the same because the F used there is radially with the c.g. of the Earth (gravity). We will ignore the case for a vertical seesaw if for safety only. ]

I had thought that something more than F=ma, presented by **Grant**, would be needed because I couldn't get the orbital motions out of my head. But they aren't needed to derive the barycenter, after all.

Last edited by George; 2017-Mar-26 at 06:43 PM.

We know time flies, we just can't see its wings.