PHY5200 F06

Chapter 8: Two-Body Central Force Problems

Reading

dark matter (today), Final review (Monday), and Rutherford scattering on (Wednesday).

Discovery of Neptune: 1845--1846

The orbit of known planets are perturbed by other planets, especially Jupiter and Saturn. These perturbations were known, and approximations were used to calculate them. The calculations were in good agreement with the observed orbits for all the planets except Uranus, the outermost of the known planets. A number of possible explanations were considered, for instance:

Both options were seriously considered. Two astronomers, Adams (in 1845) and Leverrier (in 1846) used Newtonian mechanics to predict the position of a planet that would account for the deviations of the orbit of Uranus. Both results agreed, and in 1846, astronomers at the Berlin observatory discovered the planet and named it Neptune. This discovery was regarded as a triumph of Newtonian mechanics and the law of Universal Gravitation.

Dark Matter

Newtonian mechanics and the law of Universal Gravitation has been successfully applied to the orbits of more and more distant objects: orbits of Neptune, Pluto, and comets; spacecraft; binary and multiple star systems; and finally within and between galaxies. A strange effect occurred when the motion of stars and gas around the center of the Milky Way and other galaxies was studied. First, we need to determine what we would expect for the motion of a star due to the gravitational force from all other objects in the galaxy.

Gravitational Potential Due to a Mass Distribution

One of the hurdles Newton had to overcome with his law of Universal Gravitation was to prove that the gravitational force due to a spherical distribution of mass was equivalent to the force from a point object of total mass equal to the mass distribution, located at the center of the sphere. He, and many texts, do this with the force vector. We will take advantage of our knowledge that the 1/r² force is conservative and can therefore be computed from a potential function,

Ftot = ∑ Fi = Gm ∑ (Mi/ri²) rihat = ∑ ∇Ui = ∇ ∑Ui = ∇Utot

It is generally much easier to sum (or integrate) the scalar potential energy functions than to sum (or integrate) the vector forces. let's begin by finding the potential for a spherical shell of radius R, total mass M, and uniform mass per unit area ρ, at all points inside and outside of the shell. At a point a distance r from the center of the sphere, call s the distance from an infinitesimal mass to the point. Then the total potential can be written as

Ushell = -G∫(dM/s)

Use spherical coordinates (R (fixed), θ, φ) to integrate over the shell. Choose the polar (z) axis to run through the center of the sphere and the point. Then for fixed R and θ all points on the shell are the same distance s from the point r, independent of φ. So the integral for U becomes

Ushell = -G∫(dM/s) = -2πGρR² ∫ (sinθ/s)dθ

The total mass of the shell is density times area, M = 4πρR². From the geometry of the problem we have r² + R² -2rRcosθ = s². Holding r and R constant, the differentials of both sides are 2rR sinθdθ = 2s ds. Using these, we can substitute for sinθdθ to get

Ushell = -(GM/2rR) ∫ ds = -(GM/2rR) s

If the point is outside the sphere, the integral over ds runs from r-R to r+R and the potential energy is

Ushell, outisde = -GM/r.

If the point is within the shell, the integral over ds runs from R-r to R+r and the potential energy is

Ushell, inisde = -GM/R.

This result says that the potential due to a uniform shell of material is equivalent to that from a point object of the same total mass, located at the center of the shell for points outside the shell, and constant for points inside the shell. At the shell, the potential function is continuous, but the derivative (force) is discontinuous, going to zero inside the shell. It is straightforward to now integrate over a stack of such shells to find the potential due to a solid sphere (with density that can vary with r, but not θ or φ) and get the result that

Usphere = -GM/r

for points outside the sphere. We have been using this result all along. If the point lies within the sphere, only those with radii less than r contribute to the force, those with larger radii contribute a constant to the potential but nothing to the force.

Galaxy Rotation Curves

We will now use this result to understand the speed of stars rotating around the center of our galaxy, or any other galaxy. Our galaxy is not shaped like a sphere, it is more like a disk with a spherical bulge in the center and spiral arms, but this result has the basic qualities we need. It shows the result, generally true, that an object rotating around the center of a mass distribution feels a force that depends on the distance from the center of the distribution, and the amount of mass in a sphere (or disk) of the same radius centered on the distribution. Mass outside this sphere doesn't effect the object.

As an example, let's assume that the mass distribution is uniform so that M(R) = (4/3)πρR³ and the potential energy varies like U(R) = -(4/3)πρkmR². We can use the virial theorem to relate the kinetic energy of a object at distance R from the center to the potential energy and find T = -½U, so that the tangential velocity of the object is v = (2/√3)(πρk)½ R. Within a spherical region where the mass density is approximately constant, the tangential velocity (also called the rotational speed) increases like the distance from the center.

At a point where there is little additional material, like near the edge of a galaxy, the mass within the sphere is approximately constant, so that U(R) = -kMm/R. Again using the virial theorem, we can find the rotation speed v = sqrt(KM/R). Far from the center, where the mass density is much smaller than near the center, the rotational speed decreases like the inverse square root of the distance from the center.

These results are approximate, but give a rough idea of the behavior we expect to see if we measure the rotation speed of objects as a function of their distance from the center of a galaxy. The objects measured are stars and clouds of gas. The expectation is that the rotation speeds will increase linearly near the center of the galaxy, and then turn over and decrease as we move further from the center. In constrast, the measured rotational speeds remain approximately constant to a great distance from the center of the galaxy. [sketch figure from Fowles & Cassidy.]


© Robert Harr 2006