Download Astronomy Lecture Day 02 Scale, Ratios and Proportions Intro

Astronomy Lecture Day 02 Scale, Ratios and Proportions
1. Intro
a. Collect Extra-credit HW
b. Problems or questions
c. Today’s goals
i. Introduce the structures on the Universe
ii. Review ratios and proportions to better understand the scale of the Universe
iii. See the First exam study guide
2. Structures of the Universe
a. List the principle structures in increasing size
i. Moons and Planets
ii. Stars: The Sun and star fields (http://apod.nasa.gov/apod/ap070508.html ,
http://apod.nasa.gov/apod/ap120106.html )
iii. Planetary Systems
iv. Star clusters: an open cluster and a globular cluster
Galaxies: The Whirlpool galaxy
v. Galaxy Clusters: rich cluster and poor clusters
b. Scale Video: Astronomy Picture of the day on 20 Jan 2010:
http://apod.nasa.gov/apod/ap100120.htm (6 min)
c. Bookmark or favorite this web site, check it daily
d. Another scale exercise: APOD 12 Mar 2012 http://apod.nasa.gov/apod/ap120312.html
3. Working with ratios to better understand the scale
a. Distance to the Sun compared to the diameter of the Earth
i. Distance to Sun is 1 AU = 1.51011 m = 1.5108 km
ii. How to convert from m to km: divide by 1,000 or reduce the exponent by 3
iii. How to convert from km to m: multiple by 1,000 or increase the exponent by 3
iv.
Distance to Sun
1 AU
1.5 108 km


 11,759  12,000
Diameter of Earth 2  6,378 km 2  6,378 km
v. The distance between the Sun and Earth (center to center) is equivalent to 12,000
Earth diameters. (Imagine if the Earth was $1!)
b. Diameter of the Milky Way compared to the diameter of the Solar System
i. Diameter of Milky Way is 100,000 ly = 100,000(9.461015 m) = 9.461020 m
=9.461017 km
ii. Diameter of the Solar System is about 80 AU = 80(1.51011 m) = 1201011 m =
120108 km
iii.
Diameter of Milky Way
100,000 ly 9.46  1017 km


 7.88  107  79 million
8
Diameter of Solar System
80 AU
120  10 km
iv. The diameter of the Milky Way galaxy is equivalent to about 79 million Solar
System diameters. (Imagine if the Solar System diameter was $1!)
c. Distance to Whirlpool Galaxy (M51 or NGC 5194) compared to the diameter of the Milky Way
i. Distance to Whirlpool galaxy is about 23 million ly = 23,000,000 ly = 23106 ly
=(23106)(9.461015 m) = 2.181023 m =2.181020 km
ii. Diameter of the Milky Way is about 100,000 ly =9.461017 km
iii.
Distance to Whirlpool Galaxy
23 million ly 23  106 ly


 23  101  230
Diameter of Milky Way Galaxy
100,000 ly
1 105 ly
d. The Whirlpool galaxy is about 230 Milky Way diameters away.
4. Working with proportions to better understand scale
a. If the Earth had a diameter of 1 foot, what would be the diameter of the Sun?
i. General rule for scale problems solved with proportions:
Actual #1 Model #1

Actual #2 Model #2
ii.
iii.
iv.
v.
Actual #1 = Diameter of the Earth = 26,378 km
Actual #2 = Diameter of the Sun = 2696,000 km
Model #1 = 1 foot
Model #2 = unknown model size of the Sun, x
2  6,378 km
1 ft
, cancel the 2’s, cross multiply and solve for x.

2  696,000 km
x
1 ft  696,000 km 
vii. x 
 109 ft  36 yards
6,378 km
vi.
viii. If the Earth had a diameter of 1 foot, the Sun’s diameter would be about 36 yards.
Image a football (1 foot length) on the 36 yard line of a football field.
b. If the Earth had a diameter of 1 inch, what would be the distance to the stars in Orion’s
Belt?
i. Actual #1 = Diameter of the Earth = 26,378 km
ii. Actual #2 = Distance of stars in Orion’s Belt = 1340 ly = 1340  (9.461015 m) =
1.271019 m = 1.271016 km
iii. Model #1 = 1 inch
iv. Model #2 = unknown model distance to stars in Orion’s Belt, x
v.
2  6,378 km 1 inch

, cross multiply and solve for x.
1.27  1016 km
x
vi.
x


1 inch  1.27 1016 km
 9.96 1011 inch  8.30 1010 ft  1.57 107 miles
2  6,378 km
vii. If the Earth had a diameter of 1 inch, the most distance star in Orion’s Belt would
be about 16 million miles away. (16 million miles is 17% of an AU!)
c. If the Sun had a diameter of 1 cm, what would be the diameter of the Milky Way?
i. Actual #1 = Diameter of the Sun = 2696,000 km
ii. Actual #2 = Diameter of Milky Way = 100,000 ly = 100,000  (9.461015 m) =
9.461019 m = 9.461016 km
iii. Model #1 = 1 cm (about ½ an inch)
iv. Model #2 = unknown model diameter of the Milky Way galaxy, x
v.
2  696,000 km 1 cm

, cross multiply and solve for x.
9.46  1016 km
x
vi.
x


1 cm  9.46 1016 km
 6.80 1010 cm  6.80 108 m  6.80 105 km  680,000 km
2  696,000 km
vii. If the Sun had a diameter of 1 cm, the Milky Way would have a diameter of
680,000 km. (If the Sun where 1 cm in diameter, the Milky Way would be
almost as big as the real Sun!!
5. Summary
a. Pale Blue Dot image from the Cassini Spacecraft orbiting Saturn: APOD 16 Oct 2006
http://apod.nasa.gov/apod/ap061016.html
b. Handout the HW, due on Tuesday
c. We begin Units 5, 6 & 7 next week. Please look these over.
d. Look at Dr. Fred’s Five Rules for the Apparent Motion of the Stars on Hot Tips Week 2