Derivation of Kepler S Laws from Newton S Laws

111Equation Chapter 1 Section 1STEM 699 2014 Derivation of Kepler’s Laws from Newton’s Laws

Kepler’s Laws

1. A planet orbits the sun in an elliptical path with the sun at one of the foci.

2. The area swept out by a planet radially from the sun over a fixed period of time is constant. (Hence the planets vary their speed as orbit the sun; planets travel faster when they are closer to the sun.)

3. The square of the orbital period is proportional to the cube of the semimajor axis of the orbit.

Newton’s Laws

1. (Newton’s Second Law)

2. (Newton’s Law of Universal Gravitation) Suppose a point of mass is located at the origin (0, 0) and a point of massis located at . We assume (for time being) that the point of mass M is fixed and doesn’t move. Let r denote the vector , r denote the length of this vector (i.e., ), be the unit vector in the direction of r (so ). Then the gravitation force that exerts on is given by where G is a universal constant, independent of , , and r.

Putting these two equations together we get

or equivalently Angular Momentum of planet about the origin is conserved Suppose a point mass is located at the origin (0, 0) and a point massis located at . Suppose that at all times, there is a central force acting on (i.e., a force always in the direction of r =. Note r is the position of mass m relative to M. By Newton’s law of universal gravitation, gravity is such a force. Let be the velocity vector of. Define the angular momentum of to be . Then

Proof. where a is the acceleration. The first term is 0, because . As for the second term, a is in the direction of r so because , it is also 0.

A corollary of this result is that the motion of must lie in a single plane determined by the normal vector , .

Areas swept out are constant (Kepler’s Second Law)

The key challenge here is to express the area swept out in a given amount of time.

Define . NOTE: r and are functions of time. is relative to some (arbitrary) fixed ray.

Here is an elementary way to get the formula for the area swept out in a given amount of time.

Another way I know is to express it in polar coordinates based at the origin. Now the area swept out by from time 0 to time t is given by

Taking the derivative with respect to time we get

Now taking the derivative of , we find

where L denotes the magnitude of the constant angular momentum vector.

Another way of approaching this is to think about the incremental area swept out as one half of the area of the parallelogram formed by r and , i.e.,

In the limit, Conservation of Energy

Prop. For an object moving in with a central force whose magnitude varies according the inverse square law, the quantity is constant.

Proof. Note the dynamical variables v and r are scalars in the expression. and . We are going to differentiate, and one needs to be a little careful in doing so.

So the quantity is constant. We will denote this quantity E.

Another Proof using x, y, and z coordinates.

, , and . So the second vector is 0 and the derivative is zero.

A Derivation of Kepler’s First Law

We will use the previously derived result is constant. As usual let use L to denot the length of , i.e., .

We also need an expression for in polar coordinates:

Substituting and into the expression .

Now solve for :

Notice that if we take the square root of the previous expression we get a differential equation with time as independent variable that one could solve. However we are interested in the equation of the orbit, not the equation as a function of time. We would like or . This is an integral one can evaluate actually using simple, if awkward, Calculus. It doesn’t look that way at first, but it really is straightforward. Notice that if you let the integral becomes

for some constants a¸b, c, and d. This is an arccos integral. I have left the details of evaluating in the integral in the appendix. You get

We usually take θ0 = 0 because θ0 is an arbitrary angle from which we start.

Kepler’s First Law

Case 1. e = 0.

The orbit is a circle.

Now we convert to Cartesian coordinates:

We now obtain Case 2: If e = 1, the orbit is a parabola.

Now assume . Complete the square to get

If e > 1, then this last equation is the equation for a hyperbola. If e < 1, the equation is the equation of an ellipse. Let us assume e < 1, the ellipse case, for the rest. Let , . Then distance from the center to the ellipse to the focus is c where .

So . So (0,0) is a focus of the ellipse. Note that the eccentricity of the ellipse is

. Kepler’s Third Law

The derivative of the area is a constant . Thus over an entire closed orbit of time T

Therefore

But or

Or Newton’s Correction

Newton realized that mass m will pull on M as well. A more realistic analysis goes as follows.

Subtract the first equation from the second:

Therefore, taking into account that the larger mass is also affected by the smaller mass, we see that the distance between bodies actually satisfies the same equation as above with M replaced by . This describes the motion of one body relative to the other.

The corrected analysis for the relative motion is therefore essentially identical as the one above with M replaced by , so Kepler’s Third Law becomes

Aside: The equation can be written as

The quantity is known as the “reduced mass” and is typically denoted by μ. So the relative motion can be described a force of magnitude on orbiting object of mass μ, the reduced mass, i.e.,

Further analysis (which follows) shows for an outside observer both and orbit the center of mass of the system in ellipses with equal periods.

The center of mass is defined to be .

By adding the two equations

we see that ″= 0. This means that the center of mass moves in a straight line at a constant speed. It is an inertial reference system. Now let us change coordinates so that the origin is at . So let and . The goal is to find the equations that and satisfy.

First notice that , so satisfies exactly the same equation as , namely .

Next notice that

Also notice

and so

and so So

Notice that the original set of coupled differential equations are decoupled in this reference system.

The conclusion is that the center of mass moves in a straight line at a constant speed and the first object moves with respect to the center of mass as if a fictitious object of mass were located at the center of mass and the second object moves with respect to the center of mass as if a fictitious object of mass were located at the center of mass. Notice that in practice you really only need to solve one equation because .

See excellent animations at http://commons.wikimedia.org/wiki/File:Orbit1.gif Appendix: Details of evaluating the integral

Complete square in the denominator

Finally let

so the integral becomes

Note for future reference, that

Going back to the integral:

Let so

The integral becomes:

Or which is the polar equation for a conic with eccentricity ε It turns out that as mentioned above, so i.e., the total energy of an object in orbit is a function only of its semi-major axis.