PHYS 211 Lecture 17 - Orbital Stability 17 - 1

Lecture 17 - Orbital Stability

Text: same material as Fowles and Cassiday, Chap. 6, but with different method

There is more to understanding the motion of an object than just solving a set of force equations. For example, consider a sphere set on top of a hemisphere:

Although we can use Newton’s laws to find a configuration in which the upper sphere is “balanced” in the sense that all forces and torques to which it is subject are balanced, nevertheless we know intuitively that the position is unstable. That is, the upper sphere cannot recover from small perturbations in its position.

Effective potential

Do the same phenomena occur for orbits as well? Can an orbit be bound, for example, yet unstable? To answer this question, we introduce the idea of the effective potential. We start with our expression for the total energy E: (m /2)[(dr /dt)2 + r 2(d / dt)2] + V(r) = E

The rotational motion is contained in the term d /dt = u 2 = /r 2, so (m /2)[(dr /dt)2 + 2 /r 2] + V(r) = E

We can regroup this as (m /2)(dr /dt)2 + m 2 /2r 2 + V(r) = E or (m /2)(dr / dt)2 + U(r) = E where the effective potential U(r) is U(r) = V(r) + m 2 /2r 2 (1)

The second term reflects the rotational energy, as can be seen from 2 2 2 2 2 2 2 2 Krot = I /2 = (I ) / 2I = L / 2mr = m(L/m) / 2r = m /2r

Eq. (1) is a one-dimensional equation in r and allows the minimum and maximum values of r to be determined just by solving U(r) = E, since the extrema are at dr/dt = 0. For the gravitational potential with V(r) = -GMm/r, large r U(r) -> -GMm/r physical potential dominates small r U(r) -> m 2 / 2r 2 angular momentum dominates.

U(r) angular momentum ~ 1/r 2

r gravitational potential ~ 1/r

The behaviour of the Keplerian orbits easily can be seen from the effective potential:

U(r)

hyperbola

parabola

ellipse

circle

The lines of constant energy intersect the effective potential when (dr /dt)2 = 0. The circular orbit has only one value of r (obviously) corresponding to dU/dr = 0. An elliptical orbit, on the other hand, covers a range in r (from the perigee to the apogee, to use a terrestrial example).

Orbital stability

Whether an orbit is stable depends upon the behavior of the effective potential. Consider the following two situations, each of which involves an attractive potential:

angular term ~ 1/r 2

physical potential ~ 1/r physical potential ~ 1/r 3

A circular orbit can be found for each of these potentials by finding the radius at which dU/dr = 0. But in one case (V(r) ~ -C/r), the orbit is a global minimum in U(r), while in the other case (V(r) ~ -C/r 3), it is a local maximum. Just like with the balancing spheres, a small perturbation in r at the local maximum will result in a decrease in the effective potential, and the orbit will decay.

What’s the mathematical difference between the two orbits? For the global minimum, d 2U / dr 2 > 0, while the local maximum, d 2U / dr 2 < 0. Hence, the requirement for orbital stability is d 2U / dr 2 > 0. (2)

Let’s see what this equation says about forces. First, the radius R is determined from dU / dr = 0 or d é m 2 ù dV m 2 m 2 0 = V(r ) + l = +(- 2) l = - f - l dr ëê 2r 2 ûú dr 2r 3 r 3

Evaluating this at r = R, -f(R) = m 2 / R 3 (3)

Next, we evaluate the second derivative to get the stability condition: d 2U / dr 2 > 0 or d æ m 2 ö ç - f - l > 0 dr è r 3 ø 2 df 3ml - + 4 > 0 (4) dr r Eq. (4) is to be evaluated at r = R, from Eq. (3): df 3 m 2 - l < 0 dr R R 3 r = R or R df · + f (R) < 0 (5) 3 dr r =R

Example Consider power law forces like f(r) = -cr n . Then Eq. (5) reads R - n cR n - 1 - cRn < 0 3 or 1 + n / 3 > 0 hence n > -3.

If the force must obey f ~ r > -3, then the potential must obey V ~ r > -2, as we saw in the diagrams earlier in the lecture.