Kepler's laws
Kepler's third law
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Kepler's third law.
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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):
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dA/dt = l/(2 m).
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equation 1
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We'll multiply both sides by dt and find
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dA = [l/(2 m)] dt.
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equation 2
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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:
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A = [l/(2 m)] T.
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equation 3
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If we square both sides and solve for T2, we find that
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T2 = 4 (m2/l2) A2.
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equation 4
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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
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T2 = 4 pi2 (m2/l2) a2 b2.
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equation 5
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We know that b is related to a and c, the focus-center distance, by
a2 = b2 + c2, so b2 = a2 - c2:
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T2 = 4 pi2 (m2/l2) a2 (a2 - c2).
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equation 6
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Since c = a e, we find that
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T2 = 4 pi2 (m2/l2) a2 (a2 - a2 e2)
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equation 7
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T2 = 4 pi2 (m2/l2) a2 [a2 (1 - e2)]
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equation 8
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T2 = 4 pi2 (m2/l2) a4 (1 - e2).
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equation 9
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Since we're dealing here with ellipses (and circles), we can use a
property of ellipse geometry which indicates that
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e p = a (1 - e2).
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equation 10
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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
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T2 = 4 pi2 (m2/l2) a3 [a (1 - e2)]
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equation 11
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T2 = 4 pi2 (m2/l2) a3 (e p).
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equation 12
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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
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T2 = 4 pi2 (m2/l2) l2/(G m2 M) a3.
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equation 13
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The m2 and l2 cancel, and we are left with
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T2 = (4 pi2)/(G M) a3.
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equation 14
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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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