# the solar system planetary orbits. johannes kepler december 27, 1571 – november 15, 1630 german...

Post on 27-Dec-2015

218 views

Category:

## Documents

Embed Size (px)

TRANSCRIPT

• Slide 1
• The Solar System Planetary Orbits
• Slide 2
• Johannes Kepler December 27, 1571 November 15, 1630 German mathematician, astronomer and astrologer Keplers 3 Laws of Planetary Motion
• Slide 3
• Law #1 #1 -The orbit of every planet is an ellipse with the Sun at one of the two foci Explanation: Explanation: Planets revolve in an ellipse around the sun. Planets revolve in an ellipse around the sun. An ellipse has that are on either side of the center of the axis An ellipse has two fixed points called foci that are on either side of the center of the axis The sun lies at and is not the center of Earths orbit The sun lies at one focus and is not the center of Earths orbit Planet
• Slide 4
• Law #2 #2 - A line joining a planet and the Sun sweeps out equal areas during equal intervals of time Planet moves faster Planet moves slower PerihelionAphelion Covers equal areas in Equal amount of time Closer to Sun = faster in orbit because of gravitational pull of Sun on planet
• Slide 5
• Law #3 Law #3 - Mathematical formula to prove Law 1 + 2 Eccentricity = distance between the foci length of the major axis length of the major axis e = d/L In ESRT Length of Major Axis Foci All measurements are in cm
• Slide 6
• Slide 7
• E= 2.2 cm In ESRT 4.1 cm 2.2 cm Eccentricity = distance between the foci length of the major axis length of the major axis 4.1 cm E= 0.53658 E= 0.537 Eccentricity is always 3 numbers past the decimal pointto the nearest 1000ths place. NEVER more than 1 NEVER less than 0 Eccentricity has no label!
• Slide 8
• If the two foci are located near the ends of the axis, an ellipse is long and narrow highly elliptical Many comets have this type of path If the foci move closer together, the shape of the ellipse becomes more circular A circle has NO eccentricity e = 0 A straight line is the maximum eccentricity e = 1 e=0 e=0.150 e=0.305
• Slide 9
• E= 0 E= 0.015 E= 0.356 E= 0.569E= 0.897
• Slide 10
• Eccentricity of an Orbit Lab This lab is on the Regents Exam Video
• Slide 11
• December 2012 December 2012
• Slide 12

Recommended