Kepler, Newton and Planetary OrbitsAndrew SmithApril 24, 2009AbstractIn 1609 Johannes Kepler (1571-1630) published the first two of his three laws of planetarymotion. After analyzing the data Tycho Brahe (1546-1601) had collected from observing themotion of Mars over a period of more than 20 years, he reached the conclusion that the orbitof a planet about the sun is an ellipse with the sun at one focus (his first law ). His work wasa scientific triumph because it created a model for the solar system that was not only moreaccurate than any before, but simpler as well. Though brilliant, Kepler’s result amounted tofitting a curve to a set of data without discovering any fundamental principles. In 1687 Newtonprovided the missing principles. For example, Kepler’s first law, combined with Newton’s lawof universal gravitation, led to Newton’s law of ellipses: “objects attracted to a center by a forceinversely proportional to the square of the distance travel in conic sections”. Hence the behaviorof the planets could be explained by the same laws which govern the path of an apple as it fallsfrom a tree to the ground; for the first time it became clear that the so-called heavenly bodiesbehaved no differently than the seemingly more substantial bodies of our everyday experience.The solution of this problem is one of the greatest triumphs of the human intellect in generaland of calculus in particular. Here, I present you with a more modern derivation; one thatstarts out in the complex plane, a concept that took a couple hundred years after Newton tofully develop, yet the basic concept is true to Newton and Kepler.1 The Search for the Equation of Planetary MotionKepler’s second law of planetary motion states that the line joining the Sun with any planet sweepsout equal areas in equal times.This tells us that as a planet approaches the Sun, the line joiningthe two becomes shorter, so the planet must travel faster. This law then led Kepler to his thirdlaw: ”The ratio of the cube of the semi-major axis of an elliptical orbit to the square of the orbitalperiod is the same for all planets”. Newton deduced both the inverse square law for gravity, andthe direction of that force, from Kepler’s second and third laws. Then, using that deduction, heshowed how Kepler’s first law is not an independent statement, but rather a necessary consequenceof these laws. Newton’s derivation was completely geometrical and extremely complicated.12 Newton’s Law of GravitationSuppose we have two bodies, one of mass m, and a larger one of mass M. Newton’s laws of motionhold in a coordinate system with the origin located at the center of mass of the two bodies. If welet M be the Sun and m a planet (or, an asteroid or comet), then the center of mass is inside of M.Therefore we choose polar coordinates in the complex plane so that M is at the origin, and we letz(t) represent the position of m with respect to M at time t. Thus,z = reiθWhere r and θ are real-valued functions of t.Using Newton’s “Law of Universal Gravitation”, the gravitational force betweenM and m is given byF =GMmr2θi sin θcos θmrimaginary axisreal axisM(reiθ)tiFigure 1: The Complex Plane and the orbit of m about M. The position of m at time t is givenby reiθG is the gravitational constant 6.67 × 10−11Nm2/kg2, and r is the distance between M and mat time t . (force is in Newtons, distance in meters, and mass in kilograms). Since gravity is anattractive force and we are assuming M to be at rest at the origin, F is directed from m towardM, giving usF = −GMmr2eiθFor sake of simplicity (and without loss of generality), we’ll assume that this is the only forceacting on the two bodies. By Newton’s second law of motion, F = ma, we have2ma = −GMmr2eiθSubstituting k = GM and dividing through by m , this becomesa = −kr2eiθ(1)2.1 Velocity and Acceleration at time tVelocity is the first derivative of the position function z(t). Thus,v(t) =dzdt=ddtreiθ= ireiθdθdt+ eiθdrdtAcceleration is the first derivative of velocity. Thus,a(t) =dvdt=ddtireiθdθdt+ eiθdrdt= ireiθddtdθdt+dθdtddtireiθ+ eiθddtdrdt+drdtddteiθ= ireiθd2θdt2+dθdti2reiθdθdt+ ieiθdrdt+ eiθd2rdt2+drdtieiθdθdt= ireiθd2θdt2− reiθdθdt2+ ieiθdθdtdrdt+ eiθd2rdt2+ ieiθdθdtdrdt= −reiθdθdt2+ eiθd2rdt2+ ireiθd2θdt2+ 2eiθdθdtdrdt(2)2.2 Deriving the Second Order O.D.E for Determining The OrbitPutting 1 and 2 together gives−kr2eiθ= −reiθdθdt2+ eiθd2rdt2+ ireiθd2θdt2+ 2eiθdθdtdrdt,3After dividing through by eiθwe have−kr2= −rdθdt2+d2rdt2+ ird2θdt2+ 2dθdtdrdt. (3)Equating the real and imaginary parts of 3, we have−kr2= −rdθdt2+d2rdt2. (4)And0 = rd2θdt2+ 2dθdtdrdt. (5)Multiplying both sides of 5 by r gives us0 = r2d2θdt2+ 2rdθdtdrdt. (6)However,r2d2θdt2+ 2rdθdtdrdt=ddtr2dθdt,so 6 implies thatddtr2dθdt= 0 .Since a function with 0 for its derivative must be a constant function, it follows thatr2dθdt= c , ordθdt=cr2(7)We will now use the substitution s = 1/r to get 4 into a simpler form. First,drdt=ddt1s= −1s2dsdt= −1s2dsdθdθdt. (8)Putting s into 7,dθdt=cr2= cs2. (9)Substituting 9 into 8,drdt= −1s2dsdθ(cs2) = −cdsdθ.Differentiating again,4d2rdt2= − cddtdsdθ= − cddθdsdθdθdt= −cdθdtd2sdθ2. (10)Hence, using 9 on the previous page in 10d2rdt2= −cdθdtd2sdθ2= −cs2d2sdθ2. (11)Finally, substituting 11,and s into 4 on the preceding page gives us−ks2= −1scs22− c2s2d2sdθ2= −c2s3− c2s2d2sdθ2, (12)Dividing both sides of 12 by −c2s2, we haved2sdθ2+ s =kc2.(13)This is the differential equation that we’ve been seeking, the solution of which will be anexpression for s(θ) (ultimately r(θ)) and allow us to determine the path of motion of m.To solve 13, we first note that if y(θ) is a solution to the equationd2ydθ2+ y = 0 ,then the functionx(θ) = y(θ) +kc2satisfies the equationd2xdθ2+ x =kc2.This is true becaused2dθ2y +kc2+y +kc2=d2ydθ2+ y +kc2= 0 +kc2=kc2.Hence to solve 13, we only need to solve the equationd2sdθ2+ s = 0 .That is, we need only find a function s(θ) such thatd2sdθ2= −s . (14)Now 14 says that s is a function with the property that its second derivative is the negative ofitself. We already know of two such functions; sin θ and cos θ, and, for any constants A and B, the5function A sin θ + B cos θ also has
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