This investigation will attempt to establish a mathematical proof of Kepler’s Laws in order to assess the validity of them. His three laws consisted of the ideas that: planets move in elliptical orbits, planets sweep equal areas in equal times, and the square of the period is proportional to the cube of the semi-major axis. These laws proved to be a major step in planetary science and helped Newton later formulate his laws based on the planets’ motions. By using Newton’s three laws combined with the law of gravitation we will mathematically manipulate equations to show that Kepler’s Laws can be proved. Some of the methods that were used are representing hypothetical situations using polar coordinates, creating cylindrical coordinated by extending a xy-plane into a 3-dimensional polar plane, products of vectors, integration and derivatives. In doing so the equations represent what the laws state and allow the modeling of other real life situations with similar conditions. Therefore proving mathematically valid.      Word Count:161Table of Contents?Abstract 2Table of Contents 3Introduction 4Investigation 5Kepler’s 1st Law 5Kepler’s 2nd Law 12Kepler’s 3rd Law 13Conclusion 15Bibliography 16?IntroductionThe creation of planetary science was very revolutionary for the field of Astronomy. Johannes Kepler was born December of 1571 in Germany, where he attended the University of Tübingen. He studied the work of Nicolaus Copernicus, who declared that the planets orbited the sun instead of planets orbiting the Earth. Copernicus could not base these claims in observational evidence. In 1596, Kepler wrote a public defense of Copernicus. This led to huge criticism of his Heliocentric claims because it questioned the traditional and widely viewed idea of Geocentrism. Disagreeing with these traditional ideas was very dangerous considering that a century and a half later Galileo Galilei was put on “house arrest for his publication on the subject” by the Catholic Church. Tycho Brahe, a Danish astronomer, built an observatory in Prague and invited Kepler to join him as an assistant in his research in 1600. They studied the planets’ paths and took detailed notes on the motions of them in the solar system. Brahe was suspicious of Kepler and hid his notes from him. So he decided to pass the issue of Mars’ strange orbit to Kepler. This is where he finally begins his discovery of Kepler’s laws.This essay will attempt to derive and make sense of Kepler’s laws in a mathematical manner. His laws describe how the motion of planets around a central mass that is the center of mass within the given system. Which proved essential to Sir Isaac Newton for his law of gravitation in 1684. Kepler’s laws also stretch to identify the motions of both natural and artificial satellites. These laws do not take into account the gravitational interactions between other planets within the system, for example how Earth might interfere with Mars’ orbit. Therefore his laws best apply when disrupting forces are insignificant. Although his laws seem to have their limits, they apply to all forces that include an inverse-square effect. These include both gravitational and electromagnetic forces, even applying on the scale of atoms.   InvestigationKepler’s 1st Law After Tycho Brahe’s death in 1601, Kepler finally managed to acquire Brahe’s observations and begin his solution to the Mars issue. The heart of this issue was that Mars appeared to start moving backwards in the sky, referred to as retrograde motion. Many astronomers at the time tried to work around this issue by creating what they call epicycles-a circular orbit along the pre-existing orbit around the center mass(like how the moon rotates around Earth). Even Copernicus had to use epicycles in his Heliocentric models, in order to account for these motions and maintain the idea that orbits were circular. That is when Kepler had the idea that both Earth and Mars had an elliptical orbit that created the illusion of retrograde motion.   Kepler’s 1st Law: The Law of Orbits states that all planets move in elliptical orbits, with the sun at one focus. To begin with I will define the terms/viables related to ellipses. The center is the intersection of the semi-major (a) and semi-minor axis (b). The aphelion is the position farthest from the sun or center of rotation. The perihelion is the position closest to the sun or center of rotation. The foci are two points that lay on the semi-major axis and are equidistant from the center. Another property the foci hold is that the sum of their distances from any given point on the circumference is a constant independent of the point.  The last term is the eccentricity which determines how elliptical the shape is by defining how far the foci are from each other. If e=00.999 then the shape can be defined as an ellipse. By representing a particle P(r,) traveling along a curve in terms of the coordinate plane the resulting units of position, velocity, and acceleration also must be put into unit vectors. This motion can be represented by the equations. ur=(cos)i+(sin)j u=-(sin)i+(cos)j       Vector uis orthogonal to urand point in a direction of increasing . Taking the derivative.           durd=-(sin)i+(cos)j=u dud=-(cos)i-(sin)j=-urDifferentiating urand uwith respect of time and using the chain rule results in. ur’=durd’=’u u’=dud’=-‘ur Since ris position, velocity can be expressed as v=r’v=ddtrur=r’ur+rur’=r’ur+r’uThe same process can be applied to acquire that a=v’ a=(r”ur+r’u)+(r”u+r”u+r’u’)    =(r”-r’2)ur+(r”+2r”)uNow the process of cylindrical coordinates allows for the third axis(z)to create a 3-dimensional coordinate system. That means vector kis added, making the position coordinates now r=rur+zk. We can use this to find vand a. v=r’ur+r’u+z’k a=(r”-r’2)ur+(r”+2r”)u+z”kVectors ur, u, and k make a right hand coordinate system.  uru=k uk=ur kur=uNewton’s Law of Gravitation F=-GmMr2rrwhere r is position, m is a mass, M is another mass located at the origin, F is a force, and G is the universal gravitational constant (6.6710-11 Nm2kg-2). Newton’s Second Law of Motion states F=ma. This can be rewritten in terms of our other symbols as F=mr”. Then we can combine.mr”=-GmMr2rr r”=-GMr2rrSince r”is parallel to r we can imply that.  ddtrr’=r’r’+rr”=0+rr”=rr”=0 rr’must be a constant vector, because if not rand r’become parallel. This means that mass m will fall directly towards mass M, therefore is it not orbital motion. So, C=rr’0 Vector rwill always be orthogonal to vector C, therefore will always be in the same given plane with C.Imagine a situation in which a mass M is located at the origin of a system. Mass m will move under the influence of gravity. Then m travels in a conic section with M at a focus of that conic.Let r(t)=?(t) be the position vector of mass m and let r=?    ddt??=ddt?r=r?’-r’?r2 =r2?’-rr’?r3=(??)?’-(??’)?r3Since, ddtr2=2rr’by the Chain rule and ddtr2=ddt??=2??’, rr’=??’=(??’)?r3 Since, (uv)w=(uw)v -(vw)u    ddt?r=(??’)?r3=C?r3, then multiply both sides by -GM               -GMddt?r=C(-GMr3r)From Newton’s Law of Gravitation and Newton’s Second Law of Motion, ?”=-GM?2??=(-GMr3?) GMddt?r=C(-?”)=?”CBy integrating both sides and adding the constant vector e gets GM(?r+e)=(?’C)Now adding ? to both sides gets                 GM(??r+?e)=(?’C)?     or   GM(??r+?e)=(??’)CBy using the triple scalar product and substituting (??’)=Cresults in GM(r+?e)=CC=C2By manipulating the equation it ends up as r+?e=C2GMJust like before when we proved (??’)=C if C=0then the motion of the mass m will go directly towards mass M. So assume C0. Now if , then r=C2GMwhich is a constant meaning that the orbit around the mass M would be circular. When e0then the equation before we integrated GM(?r+e)=?’Cand the knowledge that C=??’can be used by cross product to identify that both ?r+eand ?are orthogonal to C. Meaning that ?C=0 (?r+e)C=1r(?C)+eC=0+eC=eC=0Proving that e must be orthogonal to C and since C is orthogonal to the plane of motion then e must lie on that plane of motion. Let’s make e a vector on the xy-plane with being the angle that is created between the x-axis and e. Let ?(t) be the position and the arrowhead of it as P(r,)in polar coordinates. Also make the value of =-.  Based off this we can conclude that ?e=recos. With the previous equation we derived r+?e=C2GM we can substitute the new equation in resulting in r+recos=C2GM. By manipulating the equation we get r=C2/(GM)1+ecos.This resultant is a conic section of eccentricity ein polar coordinates (r,). The variable r is at a minimum when the denominator is at its greatest value. That is when cos=1or =0. Using this for the equation gives us r0=C2/(GM)1+e. Rewriting this in terms of C2the outcome becomes C2=GM(1+e)r0. By using the last equation in the previous paragraph we can conclude that r=(1+e)r01+ecos. The equation represents the motion of mass m as a conic section of some eccentricity e in the polar coordinate form (r,). Other interesting information about the equation is that for noncircular orbits r0is referred to as the perihelion distance. If earth is the central mass r0=pedigree distance.  Kepler’s 2nd Law Kepler’s 2nd Law: The Law of Areas states that a line that connects a planet to the sun sweeps out equal areas in equal times. This law was based on the idea of conservation of angular momentum. Since the planet moves faster the closer it is to the sun, then it will sweep a longer path. While the father it is the slower it moves, therefore sweeping less of a path. This is due to the fact that the imaginary triangles that can be drawn for the area swept have a larger base, but smaller height when moving at the perihelion while the opposite will happen at the aphelion. Using Kepler’s 1st Law to find the orbit of an orbiting mass m combined with Kepler’s 2nd Law, the location of that mass at any given time can be calculated only knowing its initial position. Expressing position as a vector in polar coordinates we get that r(t)=r=(rcos)i+(rsin)j. Taking the derivative will result in r'(t)=(r’cos-r’sin)i +(r’sin+r’cos)jSince we know that r(t)r'(t)=Cis a constant and C=?k. So the equation r(t)r'(t)=Cyields                      i                        j                 kr(t)r'(t)=          rcos                  rsin            0        r’cos-r’sin    r’sin+r’sin   0={(rcos(r’sin+r’cos))-rsin(r’cos-r’sin)}k=(rr’cossin+r2’cos2-rrincos+r2sin2)k =(r2′)k=?kIn terms of polar coordinates the area is calculated by the formula A=12abr2()d. By taking the derivative of this by chain rule it becomes dAdt=dAdddt=12r2()’=12r2′. Therefore dAdt=12r2’=12Cand C is a constant. Meaning that the rate of change of time is constant and the radius vector sweeps equal area in equal times.  Kepler’s 3rd Law    Kepler’s 3rd Law: The Law of Periods states that the square of the period of any planet is proportional to the cube of the semi-major axis of orbit. The interesting observation of this law proves that no matter the eccentricity(0.999) of an orbit the period will always depend only on the semi-major axis. Therefore a circular and elliptical orbit will have the same length period. The area of the ellipse that describes the orbit is A=oT(dAdt)dt=0T12Cdt=12CTSince we proved that dAdt=12Cfrom the proof done in Kepler’s 2nd Law. The area of an ellipse with a semi-major axis of length of aand semi-minor of length bis ab. Using this it results in 12CT=ab=a21-e2We can conclude this because a2=c2+b2. This can be rewritten in terms of b. Since c=ea therefore a2=e2a2+b2   or   b2=a2-e2a2  or  b2=a2(1-e2)   b=a1-e2Therefore we can square both sides of the equation to get C2T24=2a4(1-e2) this can be rewritten as T2a3=42(1-e2)aC2The maximum value that rcan achieve is when =rmax=(1+e)r01+ecos()=r0(1+e)1-e Since 2a=r0+rmax a=r0+r0(1+e)}/(1-e)2Using common denominator it can be simplified to a=r0(1-e)+r0(1+e)2(1-e)=r01-eSubstituting this value of ainto the previous equation T2a3=42(1-e2)aC2 we get T2a3=42(1-e2)C2ro1-e=42(1-e2)GM(1+e)r0r01-eSince we already showed that C2=GM(1+e)r0in the proof of Kepler’s 1st Law. The final result becomes T2=42GMa3Therefore proving that the square of the period of any planet is proportional to the cube of the semi-major axis of orbit. Conclusion Kepler made huge discoveries that revolutionized his scientific field. He made these laws without the mathematical tools that future scientists could wield. It was not until Sir Isaac Newton invented calculus that these laws could finally be proven mathematically. The fact that Kepler’s work was completely accurate even without his knowledge of the mechanics of gravity’s interaction. These proofs showed that Kepler’s Laws can be back by evidence and can be applied in many other fields than Astronomy.

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