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# 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.

## ASTR 1101-001 Spring 2008

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### 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.

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. 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!

16. 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!

17. 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!

18. 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!

19. 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.

20. 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.

21. 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.

22. 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.

23. 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°.

24. 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°.

25. 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°.

26. 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!

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