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Pedal Equation and Kepler Kinematics

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Canadian Journal of Physics Pedal equation and Kepler kinematics Journal: Canadian Journal of Physics Manuscript ID cjp-2019-0347.R2 Manuscript Type: Tutorial Date Submitted by the For11-Aug-2020 Review Only Author: Complete List of Authors: Nathan, Joseph Amal; Bhabha Atomic Research Centre, Reactor Physics Design Division Pin and string construction of conics, Pedal equation, Central force field, Keyword: Conservation laws, Trajectories and Kepler laws Is the invited manuscript for consideration in a Special Not applicable (regular submission) Issue? : © The Author(s) or their Institution(s) Page 1 of 9 Canadian Journal of Physics Pedal equation and Kepler kinematics Joseph Amal Nathan Reactor Physics Design Division Bhabha Atomic Research Center, Mumbai-400 085, India August 12, 2020 For Review Only Abstract: Kepler's laws is an appropriate topic which brings out the signif- icance of pedal equation in Physics. There are several articles which obtain the Kepler's laws as a consequence of the conservation and gravitation laws. This can be shown more easily and ingeniously if one uses the pedal equation of an Ellipse. In fact the complete kinematics of a particle in a attractive central force field can be derived from one single pedal form. Though many articles use the pedal equation, only in few the classical procedure (without proof) for obtaining the pedal equation is mentioned. The reason being the classical derivations can sometimes be lengthier and also not simple. In this paper using elementary physics we derive the pedal equation for all conic sections in an unique, short and pedagogical way. Later from the dynamics of a particle in the attractive central force field we deduce the single pedal form, which elegantly describes all the possible trajectories. Also for the purpose of completion we derive the Kepler's laws. Keywords: Pin & string construction of conics, Pedal equation, Central force field, Conservation laws, Trajectories & Kepler laws. 1.Introduction: The accurate observations of Tycho Brahe and the dis- covery of the three laws by Johannes Kepler using it, can be considered an epitome of Observational Science. Kepler after presenting the first two laws in 1609, required 10 years of tediousness, tenacity and pertinence to formulate the third law. This showed what a herculean task it must have been to make such fine and precise sense from the mammoth data. For this extra-ordinary phenomenal work Kepler is considered a central figure of Sci- entific Revolution. Even today he is seen as a magician who pulled tricks out of his hat and his laws like magic which reveals the geometrical plan of the universe. But the secrets of this magic could be understood if the pillars of physics-the laws of conservation, are combined with the Newton's law of gravitation. Because this combination describes the dynamics of the system, the laws of Kepler which are the kinematics of the system become a consequence. We will also see that this combination brings out much more. 1 © The Author(s) or their Institution(s) Canadian Journal of Physics Page 2 of 9 Among several papers on Kepler's laws, we have listed the relevant ones from 1941 to 2017. A innovative derivation of the elliptical orbits for the Kepler problem is presented in Feynman's lost lecture [1]. All the three laws as a consequence of the conservation & gravitation laws are discussed in [2],[3] and in [4] based on geometry & gravitational law. Though pa- pers [5],[6],[7],[8],[9],[10],[11],[12],[13] mention to have a simplified derivation, they discuss only about one or two laws and few among them about their relation to the gravitational law. But none derive the pedal equation and also do not show all the possible trajectories. The classical derivation of the pedal equation for ellipse and hyperbola is shown in [2], along with obtain- ing the RutherfordFor scattering Review law from the Only hyperbolic trajectory. Though in [14] all trajectories are deduced from the pedal form, the pedal equations and Kepler's laws are not derived. So to sum up, no article has covered all aspects of this problem. Hence the motivation to present the comprehensive picture using elementary methods to make it easier to follow and appreciate. 2.Classical derivation of pedal Y-axis equation: Refer to Fig-1. For a curve (Y-y1)=m1(X-x1) g(x; y)=0 and a fixed point Pf (u; v), the pedal equation is the relation between r (x1, y1) and β, where r is the distance from Pf to a point P on g and β is the perpendic- g(x,y)=0=G(r, β) r ular distance from Pf to the tangent at 1 β1 (x , y ) P . The fixed point Pf is called the pedal 2 2 β2 point. The quantities r and β called the r 2 (Y-y )=m (X-x ) pedal coordinates are associated with ev- 2 2 2 P (u,v) ery point on g, similar to the Polar co- f ordinates or any other Curvilinear coor- (0,0) X-axis dinate pair. Then the curve g(x; y) = 0 Fig-1 Pedal Coordinates can be represented by a pedal equation G(r; β) = 0. The locus of the foot of the perpendicular from Pf to the tan- gent at g(x; y) = 0 is called the `Pedal Curve'. The pedal curve is usually different from g(x; y) = 0 unless for all points on g, β = r (for example: Circle). See [15] for an introduction to pedal equations and [16] for pedal curves. For this article pedal curve is of no relevance, we require only pedal equation. To find the pedal equation for any curve we use the standard re- sult from coordinate geometry that the perpendicular distancep from a point (k; l) to a line AX +BY +C =0 is given by jAk +Bl +Cj= k2 + l2. For the function g(x; y)=0 the slope of the tangent at (x; y) is dy=dx=−gx=gy =m and its equation will be (Y − y) = m(X − x), where gx; gy are partial derivatives. Using the above result from coordinate geometry, the perpen- dicular distance from Pf (u; v) to the tangent (Y − y) − m(X − x) = 0, 2 2 1=2 β =[(x − u)gx + (y − v)gy]=[gx + gy] . Then the pedal equation G(r; β)=0 2 © The Author(s) or their Institution(s) Page 3 of 9 Canadian Journal of Physics is found by eliminating x; y between equations r =[(x−u)2 +(y −v)2]1=2 and β. Here we do not use the above procedure to derive the pedal equations, instead derive them using simple concepts from physics. 3.Description of the problem: The exercise is to find the possible tra- jectories of a mass m moving under the influence of the gravitational field due to a mass M m. There is no loss of generality in assuming that all trajectories of m are in XY -plane and their points of closest approach to M(impact parameter) lie on X-axis. We first derive the pedal equations of ellipse, parabola and hyperbola with the focus as the pedal(fixed) point. Though for an individual plot of a trajectory in the cartesian coordinates the origin could beFor arbitrarily Review anywhere, conventionally Only the origin for the ellipse and hyperbola is at the midpoint of their respective foci and for parabola it is at the vertex. But for the purpose of plotting all trajectories together in one axis (explained in Section-8 ), except the ellipse the cartesian forms of the parabola and hyperbola will be written with the origin shifted along X-axis. Next while describing the dynamics of m and deriving the Kepler's laws, we show the conic sections are the only possible trajectories and that M is at the focus of each trajectory, which is also the pedal point. Finally we plot the trajectories together for a simple case. 4.Ellipse: Refer to Fig-2. Y-axis T Let ae and be be the semi- major and semi-minor axis S(0,be) P β=rcosθ of the ellipse respectively, P’ β be the perpendicular dis- b r e 2θ tance from the focus F to e 2ae-r the tangent at the point P θ X-axis R 푭′ (-c ,0) Q(a ,0) on the ellipse, r the radial 풆 e Oe(0,0) 푭풆(ce,0) e distance from F to P and e 풄 = 풂ퟐ − 풃ퟐ 0 풆 풆 풆 \FePFe =2θ. When a string 0 with constant length=FeP + 0 PFe > FeFe and ends tied to ae 0 Fe and Fe is stretched by the Fig-2 Ellipse pencil tip and moved the el- lipse is traced. Let P be a point anywhere on the ellipse. To keep the string stretched, the force applied should be along the normal at P . Let τ be the tension in the string and the normal divide the angle 2θ into θ1 and θ2. To mark the point P the tension in the string resolved perpendicular to the nor- mal should balance, then τ sin θ1 = τ sin θ2 ) θ1 = θ2 = θ, hence the normal 0 at P bisects \FePFe. Since the normal and FeT are parallel \PFeT = θ. 0 0 0 Now consider point Q, since RFe = FeQ, FeP + PFe = FeFe + 2FeQ = 0 0 0 RFe+FeFe+FeQ=2ae. Then for any point P on the ellipse, FeP +PFe =2ae. 0 0 p 2 2 Consider point S, FeS = SFe = ae, OS = be, so FeFe = 2 ae − be = 2ce. Us- 3 © The Author(s) or their Institution(s) Canadian Journal of Physics Page 4 of 9 0 2 2 2 2 ing law of cosines in 4FePFe (2ae − r) + r − 2(2ae − r)r cos 2θ =4(ae − be) 2 2 2 gives 4ae − 2r(2ae − r)(1 + cos 2θ) = 4ae − 4be.
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