Kepler's laws
Kepler's third law
 16Kp6 Kepler

Kepler's third law.
After getting Kepler's first and second laws (though not in that order) out of our way, we're ready to tackle Kepler's third and final law.

Actually, after all of the trouble we've gone through, the third law is easy to prove and seems almost an afterthought. The third law relates the period of a planet's orbit, T, to the length of its semimajor axis, a. It states that the square of the period of the orbit (T2) is proportional to the cube of the semimajor axis (a3), and further that the constant of proportionality is independent of the individual planets; in other words, each and every planet has the same constant of proportionality.

Note that since we're talking about a planet's period, we clearly must be referring to a planet which orbits the Sun in a closed curve -- that is, either a circle or an ellipse: a conic section with e < 1.

We'll start with the expression we derived for the rate of change of the area that the Sun-planet ray is sweeping out (Kepler's second law):

dA/dt = l/(2 m). equation 1
We'll multiply both sides by dt and find
dA = [l/(2 m)] dt. equation 2
We can integrate once around the orbit (from 0 to A and from 0 to T) to get an expression relating the total area of the orbit to the period of the orbit:
A = [l/(2 m)] T. equation 3
If we square both sides and solve for T2, we find that
T2 = 4 (m2/l2) A2. equation 4
The area A of an ellipse is pi a b, where a is the length of the semimajor axis and b the length of the semiminor axis. Thus our expression of T2 becomes
T2 = 4 pi2 (m2/l2) a2 b2. equation 5
We know that b is related to a and c, the focus-center distance, by a2 = b2 + c2, so b2 = a2 - c2:
T2 = 4 pi2 (m2/l2) a2 (a2 - c2). equation 6
Since c = a e, we find that
T2 = 4 pi2 (m2/l2) a2 (a2 - a2 e2) equation 7
T2 = 4 pi2 (m2/l2) a2 [a2 (1 - e2)] equation 8
T2 = 4 pi2 (m2/l2) a4 (1 - e2). equation 9
Since we're dealing here with ellipses (and circles), we can use a property of ellipse geometry which indicates that
e p = a (1 - e2). equation 10
This relates the semimajor axis a and the eccentricity e of an ellipse to its focus-directrix distance p. If we factor out a (1 - e2) from our expression of T2 and replace it with e p we find that
T2 = 4 pi2 (m2/l2) a3 [a (1 - e2)] equation 11
T2 = 4 pi2 (m2/l2) a3 (e p). equation 12
Here we have an expression which indicates that T2 is proportional to a3; but the constant of proportionality doesn't look like it is the same quantity for all planets -- after all, it seems to be a function of m and l, which are certainly different for each planet. Thus, we'll continue on with our proof. We know from Kepler's first law that e p = l2/(G m2 M). We can use this information to substitute into our expression for T2 to find that
T2 = 4 pi2 (m2/l2) l2/(G m2 M) a3. equation 13
The m2 and l2 cancel, and we are left with
T2 = (4 pi2)/(G M) a3. equation 14
The constant of proportionality, (4 pi2)/(G M), is the same quantity for all planets -- it depends only on G, the constant of universal gravitation, and M, the mass of the Sun. Therefore, the square of the period of a planet is proportional to the cube of the length of the semimajor axis, and this proportionality is the same for all planets. This is Kepler's third law.

Quod erat demonstrandum.

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