Kepler's Laws - Mathematical Proof.

    Introduction. Johannes Kepler (1571-1630) was contemporaneous with Galileo Galelei (1564-1642) who died the same year as Sir Isaac Newton (1642-1727) was born. Kepler's master, Tycho Brahe (1546-1601), had painstakingly and most likely passionately recorded with his naked eye at the Royal Observatory at Denmark specially built for him, the positions of the 5 visible planets night after night, for decades, and bequeathed the carefully bound volumes of these records to Kepler, his most capable disciple. After years of experimenting with different models of the solar system, Kepler finally got one which the data would fit like a glove. These empirically discovered laws of the planets were sensational, and became a convincing proof of Copernicus' theory that the planets revolve around the Sun, and not around the Earth.
    The first law says each planet's orbit must be an ellipse, of which the Sun is at the focus.
    The second law says the line joining the Sun to a planet sweeps out the same area in a fixed time interval, regardless of how far the planet is from the Sun.
    The third law says the periods of planets are proportional to the orbits' diameters, raised to the three-halves power.
    Now came Newton, whose overwhelming passion was to explain Kepler's laws mathematically. His starting point was the laws of dynamics, especially the 2nd law, which says that force = mass × acceleration. He also guessed the Law of Gravitation, that the force between the Sun and a planet is proportional to the mass of each and inversely proportional to the square of the distance between them. After guessing all these laws, he also invented calculus, as a tool for his mathematical proofs. With calculus, he showed that Kepler's laws have to be so, due to his Gravitation Law and the laws of dynamics.
    In the following pages the sensational, historically almost unprecedented achievements and life-long toil and labor, of 3 brilliant scientists and mathematicians, are condensed down to only 2 hand-written pages, to facilitate memorization by students, and regurgitation at final exams. Their beauty should evoke your sense of awe and praise towards the Creator of the Universe. It is therefore a moral obligation for you not only to memorize the whole proof, but to understand its beauty, then teach it to others.

    Mathematical Preparation
    In the xyz-space, let i,j,k the unit vectors (1,0,0), (0,1,0) and (0,0,1) parallel to the x-axis, y-axis and z-axes. Let a point P(x,y,z) have cylindrical polar coordinates (r,q ,z). Then r² = x² + y². Let r be the vector (x,y). Then the unit vector i_r in the direction of r is (x/r, y/r) = (cos q sin q ) where r = Ö(x² + y²). The perpendicular unit vector to this is i_q = (-sin q , cos q ) .
    Now let the point P moves in time t, so that x, y, r, q , are all functions of t. Even r and i_r and i_q are vector functions of t, i.e., their x- and y- components are functions of t. Let us denote the t-derivative of a function f(t) by f'(t). Then we assert

    Prop. 1. i_r' = i_q q ', and i_q ' = - i_r q '.
    Proof: i_r' = (cos q , sin q )' = (- sin q , cos q )q ' = i_q q '. Similarly, i_q ' = (-sin q , cos q )' = (-cos q , -sin q )q ' = - i_rq ' .

    Prop. 2. i_r × i_q = k. Proof: exercise, using the definition of cross-product.

    Prop. 3. The equation of an ellipse in polar coordinates is r = e a / (1 - e cos q ). Its major and minor radii are a e / (1 - e²) and a e / Ö(1 - e²).
    Proof: In the xy-plane, let O be the focus, and x = - a be the equation of the directrix. The ellipse is then points P(x, y) such that OP = e PM, where PM is the distance of P from the directrix. e < 1 is called the eccentricity.
    \ Ö(x²+y²) = e (a + x), or, in polar coords (r, q ), r = e a + e r cos q
\ r = e a / (1 - e cos q ) .
    The major radius = (½)(r(0) + r(p)) = (½)[e a /(1 - e) + e a /(1 + e)] = e a /(1 - e²).
    The minor radius = (major radius) Ö(1 - e²) = e a /Ö(1 - e²) .

    Kepler's 2nd Law. Let the Sun be at the origin O of a coordinate system. Let r(t) be the position vector of a planet P at time t, = (x(t),y(t),z(t)). Let r = Ö(x²+y²+z²) = |r|. Let v(t) = r' be the velocity vector and a(t) = r(t)" = v(t)' be the acceleration. Let i_r = r / r be the unit vector from the Sun to P, m = mass of P, M = mass of the Sun, G = Gravitational constant.
    Then Newton's Law of Gravitation gives force = m a = ( - G M m / r²) i_r. \ a = ( - G M / r²) i_r .
    By a stroke of genius, Newton examines (r × v)' = r' × v + r × v' = v × v + r × a = 0 × 0 = 0. .
    Integrating, we get r × v = c - - - (1)
where c is some constant vector. c ^ r, so that the planet moves in a plane ^ c. Let us choose the coordinate axes so that c = c k is parallel to z-axis, while x- and y-axes are still arbitrary. Then P is (r, q , 0) in cylindrical coordinates.
Let i_q be (-sin q , cos q ), ^ i_r. Then from Prop. 1, i_r' = i_q , and i_q ' = - i_r . Since r = r i_r, v = r' = r' i_r + r i_r' = r' i_r + r q ' i_q , \ c = r × v = r i_r × (r' i_r + r q ' i_q ) = r² q ' k. \ c = r² q ' - - - (2).
     In time interval dt, the area swept by the line OP joining the Sun to P
= sum of triangles each with height r and base rdq
= ò₀^dt (½)r² dq = ò₀^dt (½)r² q ' dt = ò₀^dt (½)c dt = (½)c dt. This is Kepler's Second Law.

    Kepler's 1st Law. So far we did not make use of the "inverse square" property, and already the planet must move in a plane curve. Now we start with v' = (- G M / r²) i_r. Multiplied by the equation (2), we get
c v' = - G M q ' i_r = G M i_q ' . Integrating, we get
c v = G M i_q + b - - - (3)
for some constant vector b. Since v and i_q are both parallel to the xy-plane, so is b. Now let us re-choose the x- and y-axes so that b is parallel to the y-axis. We want to derive a relationship between r and q , but so far i_r and i_q are still functions of t.
    Here it is: c k = c = r × v = r i_r × ((G M / c)i_q + (1/c) b)
= (G M r / c)k + (1/c)b r sin f k, where b = | b | and f is the angle between i_r and the y-axis. Thus f = p/2 - q
\c = r (G M / c + b cos q / c), or r = a e / (1 + e cos q ) where e = b / G M, and a = c² / b. - - - (4)
Thus the orbit is an ellipse, which is Kepler's First Law.

    Kepler's 3rd Law.
Let T = the planet's year = ò₀^{2 p} (dt/dq ) dq = ò₀^{2 p} (1 /q ') dq = ò₀^{2 p} (r²/c) dq from (2), = 2 A / c where A is the area of the ellipse.
Let R = major radius, then T = 2 A/ c = (1/c) ( 2 p R² Ö(1 - e²)).
Hence T² = (4 p²/c²) (1 - e²) R⁴ = (4 p²/c²) R³ a e;
From (4), a e = c²/G M, so that T² = (4 p² / G M) R³ Notice (4 p²/G M) is independent of c or b, hence it is true for all planents. This is Kepler's 3rd Law.