CELESTIAL MECHANICS. Advisor: Steve Surace Assistant: Margaret Senese

Similar documents
Radial geodesics in Schwarzschild spacetime

GEOMETRY Properties of lines

On the Eötvös effect

Algebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016

1 Using Integration to Find Arc Lengths and Surface Areas

π,π is the angle FROM a! TO b

MATHEMATICS IV 2 MARKS. 5 2 = e 3, 4

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface

Optimization. x = 22 corresponds to local maximum by second derivative test

Answers to test yourself questions

RELATIVE KINEMATICS. q 2 R 12. u 1 O 2 S 2 S 1. r 1 O 1. Figure 1

10 m, so the distance from the Sun to the Moon during a solar eclipse is. The mass of the Sun, Earth, and Moon are = =

1. Viscosities: μ = ρν. 2. Newton s viscosity law: 3. Infinitesimal surface force df. 4. Moment about the point o, dm

6. Gravitation. 6.1 Newton's law of Gravitation

This immediately suggests an inverse-square law for a "piece" of current along the line.

EECE 260 Electrical Circuits Prof. Mark Fowler

Friedmannien equations

Mark Scheme (Results) January 2008

Physics 1502: Lecture 2 Today s Agenda

9.4 The response of equilibrium to temperature (continued)

Two dimensional polar coordinate system in airy stress functions

10.3 The Quadratic Formula

SPA7010U/SPA7010P: THE GALAXY. Solutions for Coursework 1. Questions distributed on: 25 January 2018.

Comparative Studies of Law of Gravity and General Relativity. No.1 of Comparative Physics Series Papers

Section 35 SHM and Circular Motion

r a + r b a + ( r b + r c)

A Cornucopia of Pythagorean triangles

Chapter 7. Kleene s Theorem. 7.1 Kleene s Theorem. The following theorem is the most important and fundamental result in the theory of FA s:

Energy Dissipation Gravitational Potential Energy Power

PLEASE DO NOT TURN THIS PAGE UNTIL INSTRUCTED TO DO SO THEN ENSURE THAT YOU HAVE THE CORRECT EXAM PAPER

Physics 111. Uniform circular motion. Ch 6. v = constant. v constant. Wednesday, 8-9 pm in NSC 128/119 Sunday, 6:30-8 pm in CCLIR 468

Satellite Orbits. Orbital Mechanics. Circular Satellite Orbits

Topics for Review for Final Exam in Calculus 16A

Chapter 21: Electric Charge and Electric Field

Properties and Formulas

B.A. (PROGRAMME) 1 YEAR MATHEMATICS

Physics 11b Lecture #11

Physical Security Countermeasures. This entire sheet. I m going to put a heptadecagon into game.

NS-IBTS indices calculation procedure

Electric Field F E. q Q R Q. ˆ 4 r r - - Electric field intensity depends on the medium! origin

Chapter 2: Electric Field

Electricity & Magnetism Lecture 6: Electric Potential

Physics Courseware Electromagnetism

Previously. Extensions to backstepping controller designs. Tracking using backstepping Suppose we consider the general system

Important design issues and engineering applications of SDOF system Frequency response Functions

Homework: Study 6.2 #1, 3, 5, 7, 11, 15, 55, 57

CHAPTER 18: ELECTRIC CHARGE AND ELECTRIC FIELD

CELESTIAL MECHANICS. Advisor: Dr. Steve Surace Assistant: Margaret Senese

DYNAMICS. Kinetics of Particles: Newton s Second Law VECTOR MECHANICS FOR ENGINEERS: Ninth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

Fluids & Bernoulli s Equation. Group Problems 9

Get Solution of These Packages & Learn by Video Tutorials on EXERCISE-1

(a) Counter-Clockwise (b) Clockwise ()N (c) No rotation (d) Not enough information

Chapter 4 Kinematics in Two Dimensions

13.5. Torsion of a curve Tangential and Normal Components of Acceleration

Week 8. Topic 2 Properties of Logarithms

The Area of a Triangle

SOLUTIONS TO CONCEPTS CHAPTER 11

Solutions to Midterm Physics 201

The Formulas of Vector Calculus John Cullinan

STD: XI MATHEMATICS Total Marks: 90. I Choose the correct answer: ( 20 x 1 = 20 ) a) x = 1 b) x =2 c) x = 3 d) x = 0

MAGNETIC EFFECT OF CURRENT & MAGNETISM

Electric Potential. and Equipotentials

Area. Ⅱ Rectangles. Ⅲ Parallelograms A. Ⅳ Triangles. ABCD=a 2 The area of a square of side a is a 2

Homework 3 MAE 118C Problems 2, 5, 7, 10, 14, 15, 18, 23, 30, 31 from Chapter 5, Lamarsh & Baratta. The flux for a point source is:

Class Summary. be functions and f( D) , we define the composition of f with g, denoted g f by

U>, and is negative. Electric Potential Energy

Chapter 4 Two-Dimensional Motion

AQA Maths M2. Topic Questions from Papers. Circular Motion. Answers

Problem Set 3 SOLUTIONS

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

dx was area under f ( x ) if ( ) 0

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3

Chapter 28 Sources of Magnetic Field

1.3 Using Formulas to Solve Problems

3.1 Review of Sine, Cosine and Tangent for Right Angles

10 Statistical Distributions Solutions

1. The sphere P travels in a straight line with speed

Fourier-Bessel Expansions with Arbitrary Radial Boundaries

Equations from the Millennium Theory of Inertia and Gravity. Copyright 2004 Joseph A. Rybczyk

Chapter 2. Review of Newton's Laws, Units and Dimensions, and Basic Physics

FI 2201 Electromagnetism

Continuous Charge Distributions

A P P E N D I X POWERS OF TEN AND SCIENTIFIC NOTATION A P P E N D I X SIGNIFICANT FIGURES

SSC TIER II (MATHS) MOCK TEST - 31 (SOLUTION)

SURFACE TENSION. e-edge Education Classes 1 of 7 website: , ,

Michael Rotkowitz 1,2

sec set D (sp 2014): PaPer 1

Chapter 6 Thermoelasticity

Tutorial on Strehl ratio, wavefront power series expansion, Zernike polynomials expansion in small aberrated optical systems By Sheng Yuan

KEPLER S LAWS OF PLANETARY MOTION

Central Forces: Circular Motion and Gravitation

Work, Potential Energy, Conservation of Energy. the electric forces are conservative: ur r

Physics 505 Fall 2005 Midterm Solutions. This midterm is a two hour open book, open notes exam. Do all three problems.

7.5-Determinants in Two Variables

Solutions to Problems Integration in IR 2 and IR 3

CHAPTER 2 ELECTROSTATIC POTENTIAL

Appendix D: Formulas, Properties and Measurements

Lecture 10. Solution of Nonlinear Equations - II

cos kd kd 2 cosθ = π 2 ± nπ d λ cosθ = 1 2 ± n N db

Transcription:

CELESTIAL MECHANICS Andew Dvis, My Gemino, Etn Govemn, Semmie Kim, Dniel Mogn, Kte Sfin, Jke Snell, Alexnde Stepn, Denys Voytenko, Roslie Yn, Dniel Yoo, Eileen Zung Adviso: Steve Suce Assistnt: Mget Senese ABSTRACT Te pupose of tis tem poject ws to investigte te motion of celestil bodies Fist, we deived Keple s Lws We wee ten ble to clculte plnet s ngle in obit in eltion to its peielion by tnslting n ellipse, expessed in pol coodintes, nd constucting n uxiliy cicle By intoducing speicl tigonomety, we wee ble to tck body s movement on te celestil spee In ddition, we used tigonomety to clculte te ngles of ottion of sdow t diffeent times of dy, nd constucted sundil Finlly, we successfully unified ou studies of Keple s Lws, te geomety of ellipses, nd speicl tigonomety to pinpoint te position of te Sun nd plnets on te celestil spee t ny given time Ou findings could be fute pplied to te movement of ote celestil bodies INTRODUCTION In 1618, fte yes of obsevtion nd ceful fomultion, te cowning cievement of Jonnes Keple ws complete Te tee lws of plnety motion, evolutiony in tei new ellipticl view of te sol system, suggested diclly new yet unpoven view of plnety motion Te ppence of Newton s gvity lws nd system of clculus mked down in teoy wt ws once postulted yes befoe Te cosmos, eld togete by te univesl Newtonin gvity lws, ws conqueed by new pysics Te im of tis tem poject ws to follow nd esset wt ws nd is n incedibly impotnt conjunction of obsevtion nd teoy Fist, it ws necessy to deive Keple s Lws Te study of ellipticl geomety ten llowed us to mke pcticl use of te equtions we d deived Toug te detemintion of lw of cosines nd lw of sines fo speicl tingles, we wee ble to model plnet s motion on te celestil spee We lso mde sundil PART I: KEPLER S LAWS Keple s Fist Lw Keple s Fist Lw llows one to clculte te distnce between plnet nd te st it obits given θ, te ngle between te mjo xis of te plnet s obit nd te line connecting te plnet nd te st We begin wit digm of te plnet nd st (Fig 1) [7-1]

A plnet of mss m expeiences gvittionl foce F towds st of mss M Te x- xis is defined to be te mjo xis of te plnet s obit Te cente of te st is locted t te oigin Let te x-component of F be Fx nd te y-component of F be Fy Using Newton s Lw of Univesl Gvittion: m F (wee G is te gvittionl constnt), nd Newton s Second Lw of Motion, F m, we cn expess F nd F in two diffeent wys: x y m d x F x cosθ m, Figue 1: A gp of te m d y F sin m foce vecto y θ d x d y (Note tt nd e te cceletions of te plnet in te x nd y diections) Fom te digm, we see tt x cosθ nd y sinθ Now we substitute tese vlues in to get m d F x cosθ m cosθ, (1) m d F y sinθ m sinθ () Finding te second deivtive of cosθ, you obtin d d θ d dθ d dθ cosθ sinθ cosθ cosθ sinθ + (3) Likewise, te second deivtive of sinθ is d d θ d d dθ dθ sinθ cosθ sinθ cosθ + + sinθ (4) Inset te second deivtives (3, 4) into equtions (1) nd () to obtin d θ d dθ d dθ cosθ sinθ cosθ cosθ sinθ +, (5) d θ d d dθ dθ sinθ cosθ sinθ cosθ sinθ + + (6) Multiply (5) bysin θ, multiply (6) by cosθ nd dd te esulting equtions to obtin: d θ d dθ 0 + (7) Multiply (5) by cos θ, multiply (6) by sin θ nd dd te esulting equtions to obtin [7-]

dθ d + (8) dθ To mke clcultions esie, we let p nd substitute tis into (7) Afte lgebic mnipultion, we get dp d dp d p p We integte ec side wee te constnt of integtion is to get d p θ (9) Substituting tis bck into (8) nd simplifying, we obtin (10) A B sinθ + cosθ + 1 Teefoe, to obtin te smllest vlue of, wic we ve defined to occu wen θ 0, we mximize A sinθ + B cosθ + To mximize A sinθ + B cosθ +, we tke te deivtive wit espect to θ to obtin ny citicl points nd get Acosθ Bsinθ Now we equte tis to 0 nd obtin te eqution A cosθ Bsinθ Substituting θ 0, we find tt A 0 Now, ou eqution (10) becomes, B cosθ + 1 te eqution of n ellipse in pol fom Keple s Second Lw A line joining plnet nd its st sweeps out equl es duing equl intevls of time If plnet is close to te sun, it will obit fste, nd if it is fute fom te sun, it will obit slowe As you cn see in Fig, Ae 1 nd Ae e equl in size Fig : Keple s Second Lw A1A In ode to deive tis lw, we must sow tt te cnge in e fo some cnge in time is not dependent on ny ote vibles We stt wit te cicul eqution fo e, ppoximting te ellipse s cicle [7-3]

Afte mnipultion, it becomes dθ da π, π dθ da Fom ou elie clcultions, we cn ecll te following substitution (9) tt cn be used to fute simplify te eqution d θ to da (11) da Since is constnt, is constnt, nd is constnt, te e swept out by te et in its obit ound te sun is equl duing equl intevls of time Keple s Tid Lw Keple s Tid Lw descibes te eltionsip between te peiod (T) of plnet nd te semimjo xis of its obit () Deiving it poduces n eqution elting T nd 3 da We stt wit, wic is constnt ccoding to Keple s Second Lw (11): da da We integte fom 0 to T to epesent te e of te ellipticl obit of plnet: A da T 0, 0 A T We lso know tt te e of n ellipse cn be expessed s A π b, π b T We clculte n equivlent expession fo b: + b c Becuse c e, b b c (1 ( c ) (1 e ), ) [7-4]

b 1 e Ten, becuse (1 e ), (1) Substituting in fo, we get b (1 e ) b We ten substitute tis in fo b nd get π 1 e T Substituting fo, we get (1 e ) π 1 e T, π T 4π 4π 4 3 T T Teefoe, T nd 3 e elted by fcto of 4π / PART II: STUDYING ELLIPTICAL GEOMETRY (13) Fo tis pt of te tem poject, we needed to study te pttens nd equtions ssocited wit n ellipse, to bette undestnd te pocesses of plnety motion Tis knowledge eventully elps to unlock te pcticl pplictions of Keple s Lws fo te gete gol of tying to pinpoint te exct loctions of te plnets t ny time Ellipses s Functions of Pol Coodintes Befoe we begin to uncove ny pttens involved wit te motion of te plnets, it sould fist be mentioned tt fo ll of te plnets obits, te sun lies on one focus of tei ellipticl obits It is esiest to continue discussing plnety motion by plcing te sun t te oigin on stndd xy-coodinte plne In ode to do so, we need to sift te ellipse (Fig 3) to te left by c, te lengt of one focus fom te oigin: Fig 3: An Ellipse Fig 4: An Ellipse centeed t focus [7-5]

( x + b ) y 1 (14) b Now, te focus t te oigin (Fig 4) epesents te position of te sun, wit te pt of plnet s obit ound it Tis sift llows us to simplify te eqution fo te ellipse by conveting it into pol-coodintes (Fig 5) will epesent te distnce fom te sun to te plnet nd θ will epesent te ngle tt te plnet mkes wit te sun nd te x-xis, espectively (14) becomes Fig 5: An ellipse defined in pol fom b x ( x + b ) y + 1 4 + y b b x b, Now we efe to te pol-coodinte definitions ( x cosθ; y sinθ ) nd substitute fo x nd y: b 4 cos θ + sin θ b b c cosθ (15) In ode to simplify fute, we cn cnge (15) to include only cosine by using te tigonometic identity ( sin θ 1 cos θ ) 4 b cos θ + cos θ b b c cosθ 4 cos θ b + b b c cos b x + ( )( ) θ 4 c cos θ b c cosθ + b ( c cosθ b ) ( ) c cosθ b y b 4 b b Note in tis lst step, te negtive oot of ws tken If te positive oot ws tken, te finl nswe would be negtive fo ll vlues of θ, but, te distnce fom te sun to plnet, cn only be positive Tis leds us to b + c cosθ Te following e stndd equtions tt elte te diffeent mesuements of n ellipse to ec ote e denotes te eccenticity of n ellipse, wic in oug tems mesues ow close te ellipse is to cicle Vlues of e nge fom 0 to 1, o pefectly cicul to igly elongted c e; b c b e e + ( 1 e ) (16) cx, [7-6]

Simplifying Iegulities in Ellipticl Motion Afte detemining n eqution tt defines plnet loction long tei obit, we ve to tckle te poblem of motion long tis pt Wen plnet evolves ound te sun, its speed is lwys diffeent depending on its distnce fom te sun Rte tn tying to find n eqution tt will detemine Fig 7: On te uxiliy cicle, tee exists point P te velocity of te plnets t given time, it is esie to elte ellipticl motion to some vlue tt cnges t constnt nd definble te Fo tis pocess, ou ellipse is moved bck to its Fig 6: An uxiliy cicle wit dius oiginl position, centeed on te oigin, wit cicle dwn ound it wit dius, (Fig 6) Fom ny point on te ellipse, P, tee exists point, P, on te uxiliy cicle tt lies on te sotest line pependicul to te x-xis toug point P (Fig 7) Te ngle ceted by P nd te x-xis toug te oigin is temed E Te ngle ceted by P nd te x-xis toug te focus is still θ Wile θ cnges t n iegul te, E cnges muc moe egully Tus, we cn estblis eltionsip between θ nd E wic will eventully be used to define te motion of te plnets s constnt: cos E x cosθ x c cos E e cosθ Remembe now tt by sifting te ellipse nd conveting its fomul to pol coodintes, we estblised eltionsip (17) between nd θ: ( 1 e ) e cosθ (17) Fig 8: An ellipse wit constituent pts, E, c,θ, point P, nd point P Tus, e + cosθ cos E 1 + e cosθ (18) [7-7]

Unifying Ellipticl Geomety wit Keple s Lws eltes Becuse M E esin E cnges unifomly, it is convenient to find n eqution tt to ote constnts Since is te te, must be constnt Diffeentite To get M E e sin E (19) de ( 1 ecos E) (0) Fute diffeentite To get e + cosθ cos E [( sinθ )( ) ( esinθ )( e + cosθ )] ( ) de sin E (1) Substitute (1) into (0) to get Distibuting e, we get We know fom (17) tt de e + cosθ [1 e( )] de ( 1 e ) () ( 1 e ) Substitute (17) into () to get: de, sinθ 1 sin E, [7-8]

sinθ 1 sin (3) E Using tigonometic substitution on eqution (1), we know tt e + cosθ ± sin E 1 ± sinθ 1 cos It is possible to inset tese equtions into te (3) nd get θ 1 cos θ e cosθ e 1 + cos θ cos θ ( ) Toug fute expnsion, we get 1 cos θ e cosθ + cos ( ) θ e cos θ Recll (1): 1 e Simplify using tis eqution nd get 1 1 e Peviously, we d te eqution (13): 4 T π Wit tis eqution (13), solving fo T, we get 3 T π, [7-9]

π T Plugging tis into te min eqution (7), we get π π T T So te finl esult is tt π T Integting, nd finding te constnt of integtion to equl zeo, we conclude: M π t T To demonstte usge of te deived equtions, te following poblem ws dwn up: given dte of te ye, find te efeence ngle θ of te Et in eltion to its position wen it is te closest to te sun In tis cse, we cose te dte August 4, 005 Te point t wic te Et is closest to te sun, te peielion, is found t t 0, nd is ound Jnuy 4 ec ye August 4 is 1 dys fte Jnuy 4; tus fo tis sitution, t 1 Ten, plugging in te vlues T 3655 (te peiod of te Et) nd t 1, we get M 36469 dins Becuse M E e sin E (), E e sin E 36469 dins Te vlue of te constnt e, te eccenticity of te obit of te Et, is 00167 E (00167) sin E 36469 dins Using gping clculto, o using mtemticl mnipultion, we find tt E 36389 dins Using te peviously detemined eqution (1) cos E (36389 dins), nd e + cosθ, we find cos E cos e cosθ [7-10]

e + cosθ 08788 e cosθ Let α -08788 We find Using lgebic mnipultion, we get α e + cosθ e cosθ cos θ α - e 1-α *e Substituting bck in α nd e, we get cos θ -0886, θ 651 dins 080430 degees Tus, te ngle of te Et in eltion to its point t its peielion is ppoximtely 80430 degees pst 180 degees, o bout 080430 degees PART III: SPHERICAL TRIGONOMETRY To detemine te loction of te plnets in efeence to te nigt sky, it is necessy to deive functions tt wok fo speicl tingles, since we ssume fo simplicity tt te plnets tvel on celestil spee Lw of Cosines Fig 9: Te points of speicl tingle pojected on plne plce te spee tngent to plne t C Tee intesecting cs of distinct get cicles fom speicl tingle Sides e given by ngles te tn lengts Ec ngle of speicl tingle is defined s te ngle between te lines tngent to te c t te vetex of te speicl tingle (Fig 9) To tnslte tese spes to Eucliden system, we poject te speicl tingles to plne We stt wit tingle ABC on spee of dius R (Fig 9) To poject te speicl tingle, we [7-11]

We dw dius fom point O to point C Te dii toug points A nd B e ten extended onto te plne on wic te spee ests to points A nd B, espectively Points A, B, nd C fom pojected tingle on te plne Te pln tingle is te bse of tetedon A B CO (Fig 3) Since B C nd A C e te tngent lines to cs b nd, espectively, te ngle BCA on te speicl tingle is te sme s ngle B CA on te pln tingle Fo fce B OC, we pply te Lw of Sines, Fig 10: Te lw of cosines is pplied to te new pol vibles R OB Rsec( ) OB sin( π / ) sin( π / ) Fom tis, we get B C tn( ) B C R tn( ) R We ten cy out n nlogous pocess on tingle A OC nd get OA R sec(b) nd A C R tn(b) To find side B A, we use te Lw of Cosines fist on tingle B CA, nd ten on tingle B OA : ( B' A') ( R tn( )) + ( R tn( b)) ( R tn( )) ( R tn( b)) (cos( C)) ( B' A') ( Rsec( )) + ( Rsec( b)) ( Rsec( )) ( Rsec( b)) (cos( c)) Tis implies: sec ( ) + sec ( b) sec( ) sec( b) cos( c) tn ( ) + tn ( b) tn( ) tn( b) cos( C) Using lgeb, we get sec( ) sec( b) cos( c) tn( ) tn( b) cos( c) And finlly cos( c) cos( ) cos( b) + sin( ) sin( b) cos( C) (4) Lw of Sines Stting wit te Lw of Cosines, cos( c) cos( ) cos( b) + sin( ) sin( b) cos( C), We isolte te ngle tt we e looking fo (C) cos( c) cos( ) cos( b) cos( C) sin( ) ( b) We ten substitute in fo cosine: [7-1]

cos( c) cos( ) cos( b) 1 sin ( C) sin( ) ( b) Ten we pply some lgeb nd get sin ( ) ( b) cos ( c) + cos( ) cos( b) cos( c) cos sin ( C) sin ( ) ( b) Substituting (1-sin (x)) fo ll cos (x) we get sin ( C) ( ) ( b) sin ( c) + sin ( ) + sin ( ) cos ( b) ( b) + cos( ) cos( b) cos( c) We ecognize tt te igt side of te eqution emins constnt egdless of wete we pefom tis pocess fo ngle A, B, o C Teefoe, we equte te following nd get sin ( C) ( ) ( b) sin ( B) ( ) ( c) sin ( A) Fom tis we obtin te following s Lw of Sines fo speicl tingles: ( c) ( b) sin( C) sin( c) sin( B) sin( A) sin( b) sin( ) Finding te position of objects in te celestil spee Using te Lw of Sines nd te Lw of Cosines we deived fo speicl tigonomety, we e ble to find te position of cetin object, suc s te sun, s seen fom te Et In ou model, we visulize celestil spee wit te Et t its cente s te sun evolves ound te plnet (Fig 33) Te celestil spee s defined noten nd souten poles, s well s celestil equto Tis equto is get cicle, one wose dimete is te sme s tt of te spee nd is fomed wen plne intesects te spee toug its cente Te pt on wic te sun tvels, te ecliptic, is lso depicted in te figue Objects seen on te celestil spee fom Et e locted on coodinte system tt utilizes igt scension (ltitude, epesented by α,) nd declintion (longitude, symbolized by δ) Fig 11: Te sun is defined by (α, δ) on te celestil spee α 1 δ 0 0 α 0 δ 0 θ λ (α, δ) α 6 δ 35 ε 35 Te nge of α is fom 0 to 4 ous, nd δ [7-13]

π π is between nd dins Using tese units, one ou cn be expessed s 15 (360 /4) We fute define ε to be te tilt of te et s xis fom te equto A line connecting te Not Celestil Pole nd te Sout Celestil Pole is dwn on te celestil spee, indicting wee time equls 0 ous Tis meidin is nlogous to te imginy line unning toug Geenwic, Englnd on te globe If te sun is positioned wee tis meidin intesects wit te ecliptic, its igt scension nd declintion e bot 0 At te time of te summe solstice, te sun s tveled one qute of its pt ound te Et nd δ ε 35 Te distnce on te ecliptic fom te oigin (α 0, δ 0) ove wic te sun s tveled is lbeled λ As te sun jouneys ound te plnet, its igt scension nd declintion will constntly be cnging Ou objective ws to find te position of te sun (α, δ) t ny given time A ninety-degee ngle is fomed wen te line wee α 0 nd δ 0 intesect wit te celestil equto If we pply te Lw of Sines fo speicl tigonomety to tingle wit one vetex t (0, 0), note on te celestil equto, nd te lst vetex indicting te position of te sun, ten we get sin λ sinδ sin 90 sin ε Tis sttement cn be simplified to yield te eqution sin δ sin λ sin ε Finlly, we see tt δ Acsin (sin λ sin ε), sowing tt te declintion of te sun cn be found using given vlues fo λ nd ε Te igt scension of te sun cn lso be obtined by pplying te speicl Lw of Cosines, cos c cos cos b + sin sin b cos C, to te sme tingle used to find δ By substituting in vibles, we find cos λ cosα cosδ + sinα sinδ cos π, cos λ cosα cosδ cos λ cos α (1 sin δ) Using te elie eqution sin δ sinλ sinε, we deive We finlly detemine tt cos λ cos α (1 sin λ sin ε), cos cos λ α 1 sin λ sin ε α ccos 1 sin cos λ λ sin 1 π, ε nd we cn conclude tt te igt scension of te sun cn lso be obtined if λ nd ε e given [7-14]

Fig 1: Te celestil spee wit cicul plnes fo te oizon, equto, nd sun Speicl tingles cn be used to detemine ow long te sun is bove te oizon t given ltitude Te oizon depends on te ltitude of one s position on Et As te et s dius is negligible wen comped wit te celestil spee, we view te et s point Te oizon is defined s plne pssing toug te point wit its slope detemined s elting ngle φ, te ltitude of te et Tis is epesented on Fig 1 s te ngle between te line pependicul to te oizon nd te equto Define plne H s te plne pssing oizontlly toug te celestil spee Cossing tis get cicle e two pllel cicles on te celestil spee Te equto is pojected onto te celestil spee nd nmed plne E, wile te ote cicle depicts te pt of te sun on te celestil spee t given ltitude, nmed plne S Te c between tese two pllel cicles is defined s δ Te ngle nd c lengt between te equto cicle nd te oizon is (π/-φ) Te point Z is defined s te intesection of te line pependicul to te oizon emeging fom te cente of tt cicle nd te celestil spee Te not celestil pole (NCP) is similly defined s te intesection of te line pependicul to te equto cicle emeging fom te cente of tt cicle nd te celestil spee Fom Fig 34, we cn find te mount of time te sun spends bove te oizon in one dy Tis vlue cn be clculted fom te tio of te cicumfeence of get cicle S bove plne H to te totl cicumfeence To ccomplis tis, fist conside point A on plne H, wee te sun s pt cosses te oizon Becuse te get cicle contining Z is pependicul to H t point A, c AZ s te lengt π/ Joining point N wit A poduces side of lengt (π/-δ) tt connects c δ wit te point pependicul to oizon As te ngle between te oizon nd te NCP is φ, te ngle between point N nd Z is (π/-φ) Finlly, we must detemine te mesue of ngle H in Fig 34, fomed by te sides AN nd NZ Fist, we cn connect point N nd plne E wit two cs of lengt π/ Te fist will pss toug point A, wile te second will pss toug point Z We cn define point B s te intesection of te c contining point Z nd plne S Reclling te oiginl question, conside te side lengts AB nd DF, s well s ngle H Becuse DF lies on get cicle, nd ngle H is on te line pependicul to te cente of plne E, te side lengt of DF coesponds to ngle H Since bot te side lengts nd ngle mesues e in degees, we ecognize tt te ngle mesue of DF will equl tt of AB Since AB epesents only one-lf of te sun s pt ove te oizon duing one dy, (AB/180)4 will [7-15]

equl te totl numbe of ous te sun spends bove gound in one full dy To find tis vlue, we use te speicl tigonometic Lw of Cosines (7), wit ngle H being equl to lengt AB We get Cos( π/) Cos( π / δ ) Cos( π / ϕ) + Sin( π / δ ) Sin( π / ϕ) Cos( AB) Using bsic tigonometic identities, we detemine tt 0 Sin( δ ) Sin( ϕ) + Cos( δ ) Cos( ϕ) Cos( AB) Tis povides te finl nswe, expessed in dins We get AB Ac cos( Tn( δ ) Tn( ϕ)) Te numbe of ous spent by te sun bove te oizon cn teefoe be defined s: PART IV: THE SUNDIAL Hous ( /15) Ac cos( Tn( δ ) Tn( ϕ)) Fig 13: Te initil efeence fme Fig 14: Vecto L is pplied to te efeence fme Since te focus of te poject ws to undestnd te complexities of celestil motion, smll goup of us deived te necessy ngles, nd set out to mke sundil Sundils use n object ligned t specific ngle to cst sdow on te gound Te position of te sdow eltive to te object csting te sdow is used to epesent time Fig 13 is te efeence fme used initilly Tis efeence fme ws cosen becuse te points used cn be epesented in one o two dimensions Te vecto L descibes te distnce fom te gound to te top of te object, nd te point P epesents te point wee te sun is (Fig 14) Cuently, we e in dyligt svings time, wic diffes fom sun time (o ctul time) by one ou Teefoe, wen te sun is t te 11:00 position in te sky, clocks in dyligt svings time ed 1:00 Te point t wic te sun is in te 11:00 position will be epesented by te point P [7-16]

Fig 15: Te new coodinte system Assuming tt te et ottes unifomly, te sun sould move unifomly coss te sky eltive to te et Since tee e 4 ous in dy nd 360 degees in cicle, ec ou, te sun sould move 15 degees long its pt in te sky We ten poceeded to detemine te loction of te points cosen eltive to te efeence fme selected Using simple tigonometic functions, we cn now esily detemine te positions of te points P nd L L <0, 0, L> (Fig 14) P <-*sin(15), *cos(15), 0> (Fig 14) Now tt we ve tese coodintes, otte te fme of efeence ound te X xis by Φ degees, te ngle to otte te Y nd Z xes, suc tt te X nd Y xes fom coodinte plne on te sufce of te et (Fig 15) Tus, we cn find new coodintes <X,Y,Z > fo L nd P in eltion to te et s sufce, wit X X Consideing X X, now solve fo te points P nd L in two dimensions Unde te new coodinte system, P <-sin(15), cos(15)sin(φ), cos(15)cos(φ)>, nd L <0, -Lcos(Φ), v v Lsin(Φ)> Using te stndd fom of line fo tee dimensions, + t wit L nd v ( P L), we get v < o, L cosφ, L sinφ > + < sin15,( cos15sinφ + L cosφ),( cos15 cosφ L sinφ) > t Wit tt sid, we now cn solve fo te oiginl gol Ligt fom te sun will be emitted fom point P nd coss te tip of L, nd it te gound (Z 0), foming sdow t n ngle θ wit espect to Y Since t 1:00, θ 0, t ote time points, θ is equl to te ngle tt te sdow mkes wit te 1:00 line Using tt dt, it is possible to constuct te clock We solved fo θ Using te eqution of te line fomed between te points P nd L, we cn esily solve fo te point t wic te line its te gound (Z 0) Once we know tt point, we would be ble to solve fo θ using te tngent function Te genel eqution fo te diffeent components of line in tee dimensions is s follows: Fig 16: Ligt its te gound t point L x x y y z z 0 0 0 + v 1 + v + v 3 t t t [7-17]

We know tt since Z 0, z 0, so we cn solve fo t wen z 0 like so: Now tt we ve t wen z 0 Lsinφ + ( cos15 cosφ Lsinφ) t Lsinφ t cos15cosφ Lsinφ 0, we cn get te ote components of tt point o X Since tn θ, tn θ Wit tt infomtion, we ten solve fo tn θ s follows: Y sin15( Lsinφ) cos15cosφ Lsinφ tnθ Lsinφ L cosφ + ( cos15sinφ + L cosφ) cos15cosφ Lsinφ L sin15sinφ tnθ L cosφ( cos15 Lsinφ) + L cos15sin L sin15sinφ tnθ L cos15(sin φ + cos φ) tnθ tn15sinφ φ + L cosφ sinφ (5) Using tt eqution (8), we now obtin vlues of θ fo positions of te sun ote tn 15 degees To figue out te ote positions of te sun, we just substituted in 30 fo 10:00/:00, 45 fo 9:00/3:00, nd 60 fo 8:00/4:00 As stted elie, φ te degee of ltitude tt te sundil is t Wit ll tt infomtion, we wee edy to go out to te field nd put ou pedictions to pcticlity Building te Sundil We found efeence line (te 1:00 PM line fo te clock) te following dy using long pole tt we mde pependicul to te idel et s sufce ( pefect spee) Te efeence line is te line ceted by te sdow fomed by pole pependicul to te et t 1:00 sun time (1:00 PM) Using sting nd nils, we mked tis line We clculted te necessy ngles using eqution (8) Using te tn function nd simple distnce mesuements long te efeence line, we mpped out te ngles we clculted nd used nils nd sting to mk tem We used bod to old te pole csting te sdow t 41 degees, te ppoximte vlue fo ou degee of ltitude Te eson te pole is t slnt is to ccount fo ou position on te et in eltion to te equto, suc tt te sun s ys it te pole pependicully We ten cecked ou sundil ginst te sun nd found tt we d te coect time, nd spy pinted te finl poduct onto te gound, emoving te stings of te oiginl dil [7-18]

CONCLUSION Studying te lws of celestil mecnics equies te integtion of pysics, geomety, tigonomety, clculus, nd stonomy Using ou collective knowledge of tese elted fields we deived te oiginl equtions of Jonnes Keple in ode to bette undestnd plnety motion Te infomtion equied divided ou tem into tee pimy es; te fist goup tckled Keple s pimy woks nd succeeded in deiving is tee lws, te second goup exploed te pttens nd equtions ssocited wit te geomety of ellipses nd simplified tem fo moe pcticl pplictions, nd te tid goup tnsfeed tei knowledge of tingles to speicl tigonomety wic moe ccutely depicts te nigt sky s viewed fom Et Ten coming togete, we povided moe coeent view of celestil mecnics Te study of ellipses ceted equtions tt, wit Keple s Lws, llowed te goup to ccteize ny plnet s motion s constnt Tis could be pplied to find plnet s position wit espect to te sun Te speicl tigonomety goup could ten tke tis infomtion nd plce ec plnet in te nigt sky fo ou obsevtion Ou tem ten deived te equtions nd mesuements ssocited wit sundil by studying ow te sun moves coss te sky nd ten constucting woking model Ou next step would ve been to detemine te loctions of te plnets, bot long tei obit nd in te sky We could ve ten tested tese esults toug te Dew Univesity obsevtoy Tody, te discoveies tt we mde in tem poject e used by stonomes eveywee to pedict te motions of te evenly bodies In ecent mission, NASA scientists wee ble to ccutely pedict te motion of distnt comet nd it it wit n impct pobe lunced fom Et, no doubt using some fom of Keple s Lws nd te elted equtions tt we discoveed in te clssoom Tus, te pcticl impotnce of ou wok in tis tem poject is unquestionble nd will suely continue to seve society s we to look to te sky wit wonde Aside fom te wide ffects of tis wok, te fct emins tt te poject s wole went quite smootly Te fct tt it ws ely necessy to sk fo elp in te deivtion of some of te most impotnt fomuls in bsic stonomy is cedit to te tlent nd d wok of tis tem Regdless of te ge of tese fomuls, tis poject on te wole ws n ovewelming success [7-19]