Physics 247 October 2001 1
Central Forces
This is an Excel Problem dealing with two-dimensional motion. A lot of the physics
discussed below is ahead of where we are in the lecture, but don't let that bother you. If
you are unfamiliar with parts of it, think of this as a gentle preview. The case explored is
orbital motion about a point that attracts the orbiting object directly toward it. This is
called a “Central Force” problem because the force points directly toward the origin.
An example of such a force is gravity. The radial force between two objects of masses M
and m, separated by a distance r (much greater than their sizes)1 is F = - GMm/r2. The
minus sign indicates that the force is attractive. In the case that M>>m, the large mass
can be thought of as holding still at the origin and the small mass orbiting around it.2
Another example is the orbit of a star in our Galaxy. Except rather close to or very far
from the Galactic center, the distribution of mass in the Galaxy supplies a force that is
approximately proportional to 1/r rather than 1/r2 like the force of a point mass.
A third example is mathematically interesting though I can’t think of an actual physical
situation, maybe an atom in a solid. In it, the force toward the origin is proportional to r,
so in one dimension it is just a harmonic oscillator. It is still called the harmonic
oscillator, but oscillation in each direction is sinusoidal with the same frequency but (in
general) different amplitudes.
A fourth example is the force between quarks, which I am told is attractive and pretty
much independent of the separation, meaning that it is constant.
In all these cases, energy is conserved. The respective potential energies for F = -A/rp are
U = -A/r for p=2 (point mass), U = A*ln(r/ro) for p=1 (Milky Way Galaxy, where ro is the
arbitrary radius at which the potential is zero), U = (1/2) A*r2 for p=-1 (harmonic
oscillator), and U = A*r for p=0 (quarks). In each case, if I got them right, F = -dU/dr.
In addition, for central forces there is no torque supplied by the force (it has zero lever
arm about the origin). As a consequence, angular momentum is also conserved. The
angular momentum of the orbiting mass m about the origin is L = m(x*vy – y*vx)
counterclockwise.
Kepler noticed several things about the orbits of the planets: the orbit of each is in a
plane, is elliptical in shape with the Sun at one focus, has a period “T” and semimajor
axis “a” that are related by T2 proportional to a3, and that the orbits sweep out “equal
areas in equal times.”
Given the origin, the location of m at one time, and its velocity, the orbital plane is fixed,
the plane perpendicular to both r and v that contains the original location. The “equal
areas in equal time” turned out to be the same as angular momentum conservation. The
1 If an object is spherically symmetric, its gravity behaves this way anyway, even when the pulled object is
close (but outside) but “r” is the distance from the center of the object.
2 Otherwise both orbit around their common center of mass.
