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.

2. Suppose a point mass is located at the origin (0, 0) and a point massis located at . If we let r denote the vector and r denote the length of this vector (i.e., ), then the gravitation force that exerts on is given by

where G is a universal constant, independent of , , and r.

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 forceacting on (i.e., a force always in the direction of r =. Note ris the position of massm relative to M. By Newton’s law of universal gravitation, gravity is such a force. Let be the velocity vector of.

Define the angular momentumof to be . Then

Proof.

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 express the area swept out in a given amount of time. The most convincing way I know is to express it in polar coordinates based at the origin.

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

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

So

Hence

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 rand , i.e.,

Then

In the limit,

Differentiate the unit radius vector

One way to derive the first law is to differentiate the vector .

Hence,

Note that is in the plane of motion; the plane of motion is the plane whose normal is and .

Taking the dot product with r,

or

where is the angle between and rnow.

Kepler’s First Law

Case 1. e = 0.

Then

The orbit is a circle.

Now let

Then

Convert to Cartesian coordinates:

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

Also if e > 1, then we get Case 3: If e > 1, the orbit is a hyperbola.

Now assume. Complete the square to get

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

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

Therefore

But or

So

Or

Newton’s Correction

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

Or

Subtract the first equation from the second:

Or

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 m + M.

The corrected analysis is essentially identical as the one above. M is replaced by

M + m. Kepler’s Third Law becomes

Further analysis (which is essentially the same analysis as above) shows for an outside observer both M and m orbit the center of mass of the system in ellipses with the same period.

See excellent animations at

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