Project in Mathematical Physics:

1. 0 Project in Mathematical Physics: Celestial Mechanics Professor Surace Margaret Senese N J G S S 2 0 0 5 A. Davis, M. Germino, E. Groveman, S. Kim, D. Morgan, K. Safin, J. Snell, A. Stephan, D. Voytenko, R. Yan, D. Yoo, E. Zhuang

2. 0 Celestial Mechanics??? Objective: to investigate the motion of celestial bodies • Project included: • derivation of Kepler’s Laws of Planetary Motion • a study of elliptical geometry • an investigation of spherical trigonometry • construction of a sundial

3. m (x,y) Fx F Fy r θ Deriving Kepler’s First Law 0 M

4. Deriving Kepler’s First Law (II) 0

5. 0 Deriving Kepler’s First Law (III)

6. 0 Deriving Kepler’s First Law (III)

7. 0 Kepler’s Second Law: Definition Earth's Elliptical Orbit Earth The line joining the planet to the sun sweeps out equal areas in equal intervals of time. Planet close to sun orbits faster… Planet far from sun orbits slower

8. 0 Kepler’s Second Law: Derived From We derive (constant) = (the rate the swept out area changes with respect to time)

9. 0 Kepler’s Third Law Relationship between T (period) and a (semimajor axis) T2 ~ a3 Through many substitutions and algebraic manipulation:

10. 0 Studying Elliptical Geometry Step 1:Move Sun to Origin Original Ellipse: Shifted Ellipse: Sun at Focus

11. 0 Studying Elliptical Geometry Step 2:Transfer to Polar Coordinates Standard Polar Definition: P r Final Result: θ

12. 0 Simplifying Irregularities in Elliptical Motion When a planet moves around the sun… speed differs depending on position in orbit (when furthest from sun, moves slowest; when closest, moves fastest) To account for difference: inscribe an ellipse into a circle (“Auxiliary Circle”) with radius = a

13. 0 Simplifying Irregularities in Elliptical Motion For every point P on ellipse: corresponding point P’ on the auxiliary circle While motion of point P on the ellipse is non-uniform, the motion of point P’ will be constant

14. p’ Auxiliary Circle p a r E c θ Focus 0 Simplifying Irregularities in Elliptical Motion It can be determined that: Therefore…P and P’ can be linked by defining two angles (E and θ), which will change in proportion to the motion to E and θ Recall that: a relationship linking r and θ

15. 0 Simplifying Irregularities in Elliptical Motion Substituting for r and solving for cosE, we get: Now, we can finally link Kepler’s Laws to a practical application.

16. 0 Applying Kepler's Laws to Ellipses (from ellipses) (integrating top equation)

17. 0 Applying Kepler's Laws to Ellipses WHAT’S THE POINT?!?!?! With this equation, we have related M to all constants and one variable, t plug in different values of t (time) and apply specific T values (different constants for each planet) to find M

18. 0 Finding an Angle in Orbit Given date, can find angle of planet in relation to perihelion Date: August 4 • Using: • M = E – e sin E • dM/dt = 2π/T • Substituting in values: • T = 365.25 • t = 212

19. 0 Finding an Angle in Orbit II cos θ ≈ -0.8826 θ ≈ 2.6521 rad ≈ 208.0430 degrees

20. 0 Spherical Trigonometry Spherical Triangle: defined by three intersecting arcs of distinct great circles on a sphere Angle of spherical triangle is given by angle between the lines tangent to the arc at a vertex Sides given by angles rather than lengths

21. 0 WHAT DO WE HAVE TO DO??!! We need to determinelaws for spherical trianglesanalogous to the Law of Cosines and the Law of Sines for Euclidian triangles

22. O a b B c A c R a b π/2 - b C π/2 - a A’ B’ 0 Translation to Euclidian Geometry Project the spherical triangle to a plane tangent to the sphere at point C

23. 0 Finding Side Lengths for the Planar Triangle Using the familiar Laws of Cosines and Sines, we determine the side lengths of the projected triangle…

24. 0 The Law of Cosines for Spherical Triangles C Equate the two expressions for side length A’B’ and solve for cos(c) to get a Law of Cosines for spherical triangles. B’ A’ Our final Law of Cosines: cos(c) = cos(a)∙cos(b) + sin(a)∙sin(b)∙cos(C)

25. 0 The Law of Sines for Spherical Triangles A Law of Sines for spherical triangles can be obtained by manipulating the law of cosines that we just determined. We end up with:

26. NCP 0 Locating Objects in the Celestial Sphere 0h α = 6h δ = ε= 23.5° α = 12h δ = 0° λ (α, δ) . θ Celestial Equator α = 0h δ = 0° KEY α= right ascension (latitude) δ = declination (longitude) Ecliptic SCP

27. North Celestial Pole Zenith π/2 - δ π/2 - φ π/2 δ H π/2 - δ S φ E Horizon Celestial Equator 0 A Complex Diagram...

28. 0 Becomes a Simple Triangle! π/2 - φ H π/2 - δ π/2 The final equation for the number of hours of light in one day…

29. The Sundial: A Diagram 0 Light from the sun will be emitted from point P and cross the tip of L, and hit the ground (Z’= 0), forming a shadow at an angle theta with respect to Y’.

30. 0 Solving for Tan θ

31. 0 Solving for Tan θ 1 pm and 11 am 15 ° 2 pm and 10 am 30 ° 4 pm and 8 am 60 ° 3 pm and 9 am 45 °

32. 0 Building the Sundial

33. In Conclusion... 0 • We have successfully unified our studies of Kepler’s Laws, the geometry of ellipses, and spherical trigonometry to pinpoint the sun’s position on the celestial sphere at any given time. • Our findings could be further applied to the movement of other celestial bodies.

34. 0 Thank You... The State of New Jersey Dr. S. Surace Margaret S. Dr. Miyamoto "Big" Paul and our audience!