1 / 46

ASTR 1101-001 Spring 2008

ASTR 1101-001 Spring 2008. Joel E. Tohline, Alumni Professor 247 Nicholson Hall [Slides from Lecture03]. Assignment: “ Construct” Scale Model of the Solar System. Sun is a basketball. Place basketball in front of Mike the Tiger’s habitat.

diem
Download Presentation

ASTR 1101-001 Spring 2008

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. ASTR 1101-001Spring 2008 Joel E. Tohline, Alumni Professor 247 Nicholson Hall [Slides from Lecture03]

  2. Assignment: “Construct” Scale Model of the Solar System • Sun is a basketball. • Place basketball in front of Mike the Tiger’s habitat. • Walk to Earth’s distance, turn around and take a picture of the basketball (sun). • Walk to Jupiter’s distance, take picture of sun. • Walk to Neptune’s distance, take picture of sun. • Assemble all images, along with explanations, into a PDF document. • How far away is our nearest neighbor basketball? Due via e-mail (tohline@lsu.edu): By 11:30 am, 25 January (Friday) You may work in a group containing no more than 5 individuals from this class.

  3. Assignment:

  4. Worksheet Item #1 • A basketball has a circumference C = 30”, so its radius is … • For all circles, the relationship between circumference (C) and radius (R) is: C = 2pR • Hence, R = C/(2p) = 4.78” • But there are 2.54 centimeters (cm) per inch, so the radius of the basketball is: R = (4.78 inches)x(2.54cm/inch) = 12.1 cm = 0.121 meters.

  5. Worksheet Item #1 • A basketball has a circumference C = 30”, so its radius is … • For all circles, the relationship between circumference (C) and radius (R) is: C = 2pR • Hence, R = C/(2p) = 4.78” • But there are 2.54 centimeters (cm) per inch, so the radius of the basketball is: R = (4.78 inches)x(2.54cm/inch) = 12.1 cm = 0.121 meters.

  6. Worksheet Item #1 • A basketball has a circumference C = 30”, so its radius is … • For all circles, the relationship between circumference (C) and radius (R) is: C = 2pR • Hence, R = C/(2p) = 4.78” • But there are 2.54 centimeters (cm) per inch, so the radius of the basketball is: R = (4.78 inches)x(2.54cm/inch) = 12.1 cm = 0.121 meters.

  7. Worksheet Item #1 • A basketball has a circumference C = 30”, so its radius is … • For all circles, the relationship between circumference (C) and radius (R) is: C = 2pR • Hence, R = C/(2p) = 4.78” • But there are 2.54 centimeters (cm) per inch, so the radius of the basketball is: R = (4.78 inches)x(2.54cm/inch) = 12.1 cm = 0.121 meters.

  8. Worksheet Items #3 & #4 • The sun-to-basketball scaling ratio is … • f = Rsun/Rbasketball = (7 x 108 m)/(0.121 m) = 5.8 x 109 • What is the Earth-Sun distance on this scale? • dES = 1 AU/f = (1.5 x 1011 m)/5.8 x 109 = 26 m Note: Textbook §1-6 reviews “powers-of-ten” (i.e., scientific) notation. Textbook §1-7 explains that 1 astronomical unit (AU) is, by definition, the distance between the Earth and the Sun. 1 AU = 1.496 x 108 km = 1.496 x 1011 m.

  9. Worksheet Items #3 & #4 • The sun-to-basketball scaling ratio is … • f = Rsun/Rbasketball = (7 x 108 m)/(0.121 m) = 5.8 x 109 • What is the Earth-Sun distance on this scale? • dES = 1 AU/f = (1.5 x 1011 m)/5.8 x 109 = 26 m Note: Textbook §1-6 reviews “powers-of-ten” (i.e., scientific) notation. Textbook §1-7 explains that 1 astronomical unit (AU) is, by definition, the distance between the Earth and the Sun. 1 AU = 1.496 x 108 km = 1.496 x 1011 m.

  10. Worksheet Items #3 & #4 • The sun-to-basketball scaling ratio is … • f = Rsun/Rbasketball = (7 x 108 m)/(0.121 m) = 5.8 x 109 • What is the Earth-Sun distance on this scale? • dES = 1 AU/f = (1.5 x 1011 m)/5.8 x 109 = 26 m Note: Textbook §1-6 reviews “powers-of-ten” (i.e., scientific) notation. Textbook §1-7 explains that 1 astronomical unit (AU) is, by definition, the distance between the Earth and the Sun. 1 AU = 1.496 x 108 km = 1.496 x 1011 m.

  11. What about the Dime?

  12. What about the Dime? NOTE: A dime held 1 meter from your eye subtends an angle of 1°.

  13. Calendar See §2-8 for a discussion of the development of the modern calendar.

  14. Calendar • Suppose you lived on the planet Mars or Jupiter and were responsible for constructing a Martian or Jovian calendar.

  15. Information on Planets[Drawn principally from Appendices 1, 2 & 3]

  16. Information on Planets[Drawn principally from Appendices 1, 2 & 3]

  17. Information on Planets[Drawn principally from Appendices 1, 2 & 3]

  18. Information on Planets[Drawn principally from Appendices 1, 2 & 3]

  19. Information on Planets[Drawn principally from Appendices 1, 2 & 3]

  20. Information on Planets[Drawn principally from Appendices 1, 2 & 3]

  21. Information on Planets[Drawn principally from Appendices 1, 2 & 3]

  22. Information on Planets[Drawn principally from Appendices 1, 2 & 3]

  23. Information on Planets[Drawn principally from Appendices 1, 2 & 3]

  24. Earth’s rotation • Responsible for our familiar calendar “day”. • Period (of rotation) = 24 hours = (24 hours)x(60 min/hr)x(60s/min) =86,400 s • Astronomers refer to this 24 hour period as a meansolarday (§2-7), implying that this time period is measured with respect to the Sun’s position on the sky. • A sidereal day (period of rotation measured with respect to the stars – see Box 2-2) is slightly shorter; it is shorter by approximately 4 minutes. • The number of sidereal days in a year is precisely one more than the number of mean solar days in a year!

  25. Earth’s rotation • Responsible for our familiar calendar “day”. • Period (of rotation) = 24 hours = (24 hours)x(60 min/hr)x(60s/min) =86,400 s • Astronomers refer to this 24 hour period as a meansolarday (§2-7), implying that this time period is measured with respect to the Sun’s position on the sky. • A sidereal day (period of rotation measured with respect to the stars – see Box 2-2) is slightly shorter; it is shorter by approximately 4 minutes. • The number of sidereal days in a year is precisely one more than the number of mean solar days in a year!

  26. Earth’s rotation • Responsible for our familiar calendar “day”. • Period (of rotation) = 24 hours = (24 hours)x(60 min/hr)x(60s/min) =86,400 s • Astronomers refer to this 24 hour period as a meansolarday (§2-7), implying that this time period is measured with respect to the Sun’s position on the sky. • A sidereal day (period of rotation measured with respect to the stars – see Box 2-2) is slightly shorter; it is shorter by approximately 4 minutes. • The number of sidereal days in a year is precisely one more than the number of mean solar days in a year!

  27. Earth’s rotation • Responsible for our familiar calendar “day”. • Period (of rotation) = 24 hours = (24 hours)x(60 min/hr)x(60s/min) =86,400 s • Astronomers refer to this 24 hour period as a meansolarday (§2-7), implying that this time period is measured with respect to the Sun’s position on the sky. • A sidereal day (period of rotation measured with respect to the stars – see Box 2-2) is slightly shorter; it is shorter by approximately 4 minutes. • The number of sidereal days in a year is precisely one more than the number of mean solar days in a year!

  28. Earth’s orbit around the Sun • Responsible for our familiar calendar “year”. • Period (of orbit) = 3.155815 x 107 s = 365.2564 mean solar days (§2-8). • Orbit defines a geometric plane that is referred to as the ecliptic plane (§2-5). • Earth’s orbit is not exactly circular; geometrically, it is an ellipse whose eccentricity is e = 0.017 (Appendix 1). • Because its orbit is and ellipse rather than a perfect circle, the Earth is slightly farther from the Sun in July than it is in January (Fig. 2-22). But this relatively small distance variation is not responsible for Earth’s seasons.

  29. Earth’s orbit around the Sun • Responsible for our familiar calendar “year”. • Period (of orbit) = 3.155815 x 107 s = 365.2564 mean solar days (§2-8). • Orbit defines a geometric plane that is referred to as the ecliptic plane (§2-5). • Earth’s orbit is not exactly circular; geometrically, it is an ellipse whose eccentricity is e = 0.017 (Appendix 1). • Because its orbit is and ellipse rather than a perfect circle, the Earth is slightly farther from the Sun in July than it is in January (Fig. 2-22). But this relatively small distance variation is not responsible for Earth’s seasons.

  30. Earth’s orbit around the Sun • Responsible for our familiar calendar “year”. • Period (of orbit) = 3.155815 x 107 s = 365.2564 mean solar days (§2-8). • Orbit defines a geometric plane that is referred to as the ecliptic plane (§2-5). • Earth’s orbit is not exactly circular; geometrically, it is an ellipse whose eccentricity is e = 0.017 (Appendix 1). • Because its orbit is and ellipse rather than a perfect circle, the Earth is slightly farther from the Sun in July than it is in January (Fig. 2-22). But this relatively small distance variation is not responsible for Earth’s seasons.

  31. Earth’s orbit around the Sun • Responsible for our familiar calendar “year”. • Period (of orbit) = 3.155815 x 107 s = 365.2564 mean solar days (§2-8). • Orbit defines a geometric plane that is referred to as the ecliptic plane (§2-5). • Earth’s orbit is not exactly circular; geometrically, it is an ellipse whose eccentricity is e = 0.017 (Appendix 1). • Because its orbit is and ellipse rather than a perfect circle, the Earth is slightly farther from the Sun in July than it is in January (Fig. 2-22). But this relatively small distance variation is not responsible for Earth’s seasons.

  32. Tilt of Earth’s spin axis • Responsible for Earth’s seasons (§2-5) • Tilt of 23½° measured with respect to an axis that is exactly perpendicular to the ecliptic plane. • Spin axis points to a fixed location on the “celestial sphere” (§2-4); this also corresponds very closely to the position of the north star (Polaris) on the sky. • This “fixed location” is not actually permanently fixed; over a period of 25,800 years, precession of the Earth’s spin axis (§2-5) causes the “true north” location to slowly trace out a circle in the sky whose angular radius is 23½°.

  33. Tilt of Earth’s spin axis • Responsible for Earth’s seasons (§2-5) • Tilt of 23½° measured with respect to an axis that is exactly perpendicular to the ecliptic plane. • Spin axis points to a fixed location on the “celestial sphere” (§2-4); this also corresponds very closely to the position of the north star (Polaris) on the sky. • This “fixed location” is not actually permanently fixed; over a period of 25,800 years, precession of the Earth’s spin axis (§2-5) causes the “true north” location to slowly trace out a circle in the sky whose angular radius is 23½°.

  34. Tilt of Earth’s spin axis • Responsible for Earth’s seasons (§2-5) • Tilt of 23½° measured with respect to an axis that is exactly perpendicular to the ecliptic plane. • Spin axis points to a fixed location on the “celestial sphere” (§2-4); this also corresponds very closely to the position of the north star (Polaris) on the sky. • This “fixed location” is not actually permanently fixed; over a period of 25,800 years, precession of the Earth’s spin axis (§2-5) causes the “true north” location to slowly trace out a circle in the sky whose angular radius is 23½°.

  35. Moon’s orbit around the Earth • Responsible for our familiar calendar month. • Period (of orbit) = 2.36 x 106 s = 27.32 days (Appendix 3). • Moon’s orbital plane does not coincide with the ecliptic plane; it is inclined by approximately8° to the ecliptic (§2-6). • Much more about the Moon’s orbit in Chapter 3!

More Related