Back to Contents!

Kepler's Laws Proven

Newton's prescription for the gravitational force is universal in that it provides the direction and magnitude of the force for any two point masses separated by a distance. To be consistent with his calculations about the attraction between the earth and the moon, the force must be directed along a line between the centers of mass of the moon and the earth and to be consistent with our observations, we have to make sure the mathematical description always indicates that the force is attractive. Therefore, for two point masses as shown in the figure below, a vector form of the universal law would be as follows:
F₁₂ = -F₂₁ = -Gm₁m₂ r/r³

where the minus sign insures that the force is always attractive under the convention that the direction of r is always away from the particle creating the force and pointing toward the particle upon which the force acts. With this form of the force and his laws of motion in hand, Newton went on to show that all of Kepler's Laws are just consequences of his truly general laws of motion. We will not do Newton's proof of Kepler's 1st law, but you can find it described in Chapter 11 of Stewart's text. The proof of the Kepler's Second Law is not difficult and is instructive because it points out one of the more powerful constraints available for solving problems. First, we need to define some conventions. Assume that one particle (say m₁) is much more massive than the other so that it sits at one of the foci of the elliptical orbit path while the much lighter m₂ orbits around it (see the figure below). This is certainly the case for the description of the motion of the planets around the Sun.

We should note that, as with circular motion, the velocity is tangential to the path taken by m₂. We should also note that the acceleration of m₂ according to Newton is

So the angular momentum is constant since the gravitational force is a central force, namely the acceleration points toward the center of the path defined for the particle by the force and the particle's velocity.

What does this have to do with Kepler's Laws? First, if the vector L = r x p is a constant, then, L is a constant vector pointing to a fixed direction in space (we say, as Newton did, that the direction is fixed with respect to the distant stars, i.e. the stars far enough removed from earth that they show no observable motion with respect to each other and therefore come closest to defining an "absolute" reference frame). Since a vector resulting from a cross-product is always perpendicular to both vectors in the cross-product, r is always perpendicular to L. Since L is fixed in direction, the planet's motion is confined to a plane which is perpendicular to L. With a little more work, we can prove that the orbit must in fact be a conic section: an ellipse, parabola, or hyperbola, but that's more detail than we want to explore right now.

As for Kepler's Second Law, well, we can take another look at our diagram for the orbit and note that the cross-product of two vectors also gives us the area of a parallelogram defined by those two vectors. For the case of Kepler's Second Law, we want to know the area swept out by the radial vector for some infinitesimal time dt so that we can calculate the area swept out for any time period by integrating. Consider the figure below

Here we need to consider the area dA swept out by the radial vector due to the displacement, vdt of the particle m₂. Notice that this is just ½ of the area of a parallelogram defined by r and vdt, so

Since we have already proven that r x v is constant, we know that the area will be a constant times time. In fact, ½ (r x v) = L/(2m₂), so A(t) = [L/2m](t_f - t₀). Therefore, for two equal time periods t_f - t₀, the area swept out during that time will be the same as shown in the figure below.

Functionally, what this means is that the velocity of the particle in orbit must be fast when it's near the center of its orbit and slow when its in the part of its orbit that's further away from the center so that the rate of area swept out is always constant at L/2m.

For those who want another shot at playing with the cross-product, the resource from Syracuse is here.

To get a graphical view about orbits and Kepler's Three Laws of Planetary Motion, look at NASA's Observatorium site.

Kepler's Third Law

Let's consider the figure of an elliptical orbit, as shown below, in which the long axis has a length 2a and the short axis a length 2b.

The period is the time it takes for a particle to make one complete orbit. Let's refer to this time as T. In time T, the radial vector will have swept out an area equal to the area of the entire ellipse (Pi*a*b), so

You may remember that the equation for an ellipse is

From section 9.7 of Stewart, we learn that the equation

describes an ellipse in polar coordinates (i.e. r and theta). In this equation, r is the length of the radial vector, theta indicates the direction of the vector r and e and d can be defined so that

In math-speak e is usually referred to as the eccentricity and d as the directrix. Had we gone through the derivation of Kepler's First Law (you can see it in Stewart, sect. 11.9), we would have found that Newton's Law of Universal Gravitation predicts an elliptical orbit where L²/ (G*m₁²m₂²) = e*d (you'll have to trust me on this). With that, we find that the period is related to the long axis, a, of the elliptical orbit by remembering that we have already shown that T = 2m₂*Pi*a*b/L, so L = 2m₂*Pi*a*b/T

Hence Kepler's Third Law is true. The period squared is proportional to the mean radius cubed. Note that for a circular orbit of radius r (for which we derived Kepler's Third Law in the last lecture), a = b = r and this result reduces exactly to our previous version, T² proportional to r³.