## Physics Friday 42

Last week, we considered a thin ring of radius a and mass M, and the energy of a mass m on the perpendicular axis at a distance z from the center of the ring. We noted that the equilibrium position of the mass m is the center of the ring, and that, at least with regards to motion perpendicular to the plane of the ring, the mass will be attracted toward this point.
Now, let us consider motion within the plane of the ring; in particular, is the equilibrium point at the ring center stable or unstable?

As before, we calculate the gravitational potential energy. Here, due to symmetry, the only element of the position of mass m relative to the ring that will affect the energy is the distance ρ from the center of the ring. Note that we are considering ρ<a.

Let us choose our origin to be the center of the ring, and let us choose an angular coordinate θ so that the mass is on the ray θ=0. Note that a small segment of the ring of arc would have mass . For such a portion at a coordinate angle θ, the distance between that piece and the mass is given by the law of cosines:
.

Thus the gravitational potential energy due to this portion of the ring is
.

And we integrate over all angles to find the energy:

This, however, is an elliptic integral, and cannot be done analytically.
Remember, however, we are interested in stability of the origin, and thus are considering ρa. In particular, to obtain stability, we need to know , which means we need only consider to second order in .
Our integrand is
.

 to second order in x is
.
And so our integrand is, to second order,
.
Integrating over 0≤θ≤2π, the terms independent of theta are simply multiplied by 2π, while the  and  terms integrate to zero; thus, we have:


.
This gives  at the center, as expected, and shows us that the potential energy decreases as the mass moves from the center of the ring, and thus the center is an unstable equilibrum. (The Ringworld is unstable!)