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Exploring Mars

Learn about Tycho Brahe and Johannes Kepler, their contributions to astronomy and space exploration, Kepler's laws of planetary motion, and the significance of Mars. Explore the role of conic sections in planetary and spacecraft orbits. Follow-up activities included.

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Exploring Mars

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  1. Exploring Mars . Part 4 Pathfinder’s Path I • Mr. K. NASA/GRC/LTP • Edited: Ruth Petersen

  2. Preliminary Activities • (Use the URL’s provided on Slide 21 to complete the preliminary activities.) • Who were Tycho Brahe and Johannes Kepler? What did they contribute to modern astronomy and space exploration? • 2. Write down Kepler’s three laws of planetary motion. Why are these laws significant today? • 3. What role did Mars play in the discovery of Kepler’s law of planetary orbits?

  3. 4. Why is Mars significant today? 5. In your algebra class, discuss the conic sections. Write the equation for an ellipse with its center at the origin. 6. What role do the conic sections play in planetary and spacecraft orbits?

  4. Tycho Brahe (1546 - 1607)

  5. Uraniborg - Tycho’s Famous Observatory

  6. Three Laws of Planetary Motion Johannes Kepler (1571 - 1630) Every planet travels in an ellipse with the sun at one focus. The radius vector from the sun to the planet sweeps equal areas in equal times. The square of the planet’s period is proportional to the cube of its mean distance from the sun.

  7. Kepler’s study of Mars’ orbit lead him to the discovery that planetary orbits were ellipses. Actually, we know now that orbits can be any conic section, depending on the total energy involved.

  8. The Conic Sections Circle Ellipse Parabola Hyperbola

  9. P = any point on the ellipse (0,b) p s2 s1 (a,0) (-a,0) f1 f2 (0,-b) f1 & f2 = specific points on the x-axis The Ellipse y x S1 + S2 = Constant

  10. v Velocity Vector Radius Vector r Kepler’s First law: Elliptical Orbits Planet Sun The sun is located at one of the two foci of the ellipse.

  11. “Vis Viva” Conservation of Energy: ½mv2 - GMm/r = K v m r M v = {2(K + GMm/r)/m }1/2

  12. “Vis Viva” (Continued) v m r How are v and r related? Faster Slower M As r increases, v decreases & vice versa.

  13. Pathfinder’s Path: Hohmann Transfer Ellipse Departure: December 1996 Earth's Orbit Mars' Orbit

  14. Pathfinder’s Path: Arrival: July 1997 Departure: December 1996

  15. The Conic Sections - Revisited Circle Ellipse Parabola Hyperbola Open orbits: Some comets Parabolic velocity = escape velocity Closed orbits: Planets, moons, asteroids, spacecraft.

  16. Follow-Up Activities 1. Earth orbits the sun at a mean distance of 1.5 X 108 km. It completes one orbit every year. Compute its orbital velocity in km. sec. 2. The Pathfinder required a greater velocity than Earth orbital velocity to achieve its transfer orbit. Why? Since additional velocity costs NASA money for fuel, can you explain why we launched the spacecraft eastward? (Hint: When viewed from celestial north, the Earth and planets orbit the sun counter-clockwise.)

  17. 3. The equation for an ellipse with its center at the origin is (x/a)2 + (x/b)2 = 1 Under what mathematical condition does the ellipse become a circle? (Check with your algebra teacher if necessary.) 4. Plot the ellipse choosing different values of a and b. (a < b; a = b; a > b). What do you observe? 5. In the Vis-Viva equation for velocity, how does the velocity vary around a CIRCULAR orbit?

  18. 6.Extra Credit:The ellipse is defined as a locus of points p such that for two points, f1 and f2 (the foci), the sum of the distances from f1 and f2 to p is a constant. Use this definition and your knowledge of algebra to show that the equation of an ellipse follows: i.e., that (x/a)2 + (y/b)2 = 1 where a and b are the x and y intercepts respectively.

  19. Solution to #6: The Setup y P(x,y) (0,b) s2 s1 (a,0) x (f,0) (f,0) (-a,0) (0,-b)

  20. Solution to #6: The Algebra • Given: s1 + s2 = k • (f - x)2 + y2 = s12 … (eq. i) • (f + x)2 + y2 = s22 … (eq. ii) • 1.) Let (x,y) = (a,o). This gives k = 2a, and • s1 = 2a - s2 • 2.) Let (x,y) = (0,b). This gives s1 = s2 = (f2+b2)1/2, and f2 = a2 - b2 • 3.) Result 2.)  eq. ii gives • s2 = a + (x/a)(a2 - b2)1/2 • 4.) Result 2.) and 3.)  eq. ii gives • (x/a)2 + (y/b)2 = 1 From geometry: Be careful: The algebra gets messy!

  21. Johannes Kepler: csep10.phys.utk.edu/astr161/lect/history/kepler.html www.vma.bme.hu/mathhist/Mathematicians/Kepler.html Tycho Brahe: http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Brahe.html Hohmannn Transfer Orbits: http://www.jpl.nasa.gov/basics/bsf-toc.htm

  22. joseph.c.kolecki@grc.nasa.gov

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