SCI-103 Introductory Astronomy Summer 2013
Homework #1a Ratio and Proportion Problems
1.
How many Earths could fit side by side across the Sun? Look up the radii of the Earth and Sun
Radius of the Earth = 6.378  10 m
6
Radius of the Sun = 6.96  10 m
8
The Ratio
384.4  10 6 m
 109
6.378  10 6 m
109 Earths could fit side by side across the Sun.
2.
How many Earth’s could fit between the Earth and the Moon? Look up the distance to the Moon
Distance to the Moon = 384.4  10 m
6
Diameter of the Earth = 1.276  10 m
7
The Ratio
384.4  106 m
 30.1
1.276  10 7 m
About 30 Earths could fit side by side Between the Earth and the Moon.
3.
How many Earth’s could fit between the Sun and the Earth? Look up the AU
1 AU = 1.5  10 m
11
Diameter of the Earth = 1.276  10 m
7
The Ratio
1.5  1011 m
 11,755
1.276  107 m
Almost 12,000 Earths could fit side by side between the Earth and the Sun.
4.
How many Suns could fit side by side across the solar system? Look up Pluto’s Distance from the Sun (There is an
error in this solution. Mathew Thomas found it. Can you? The correct answer is twice the answer below.)
Pluto’s distance from the Sun = 39 AU = 5.85  10
12
m
Diameter of the Sun = 1.39  10 m
9
The Ratio
5.85  1012 m
 4,209
1.39  109 m
About 4,200 Suns could fit side by side across our solar system.
5.
How many solar systems could fit side by side between the Sun and the next nearest star Proxima Centauri? Look up
the distance to Proxima Centauri
Distance to Proxima Centauri = 4.3 ly = 4.07  10
16
m
Diameter of the Solar System = 78 AU = 1.17  10
13
m
4.07  1016 m
The Ratio
 3,479
1.17  1013 m
About 3,400 of our solar systems could fit side by side between the Sun and the nearest star Proxima Centauri.
6.
How many solar systems could fit side by side across the Milky Way galaxy? Look up the diameter of the Milky Way
galaxy.
Diameter of the Solar System = 78 AU = 1.17  10
m
Diameter of the Milky Way = 100,000 ly = 9.46  10
20
13
m
9.46  10 20 m
 8.09  107  81 Million
The Ratio
13
1.17  10 m
About 81 Million of our solar systems could fit side by side across the Milky Way Galaxy.
Proportion Problems
1.
If the Earth were 1 foot in diameter, how far away would the Sun be?
1 AU

x
2  6,378 km  1 ft
x  2  6,378 km   1 AU  1 ft
x
1 AU  1 ft
2  6,378 km 
1.5  108 km  1 ft
2  6,378 km 
x  11,759 ft
x
x  2.23 miles
If the Earth were 1 foot in diameter, the Sun be 2.23 miles away.
2.
If an AU were shrunk to 1 foot, how far away would the next nearest star be?
4.23 ly
x

1 AU 1 ft
x  1 AU  4.23 ly  1 ft
4.23 ly  1 ft
1 AU
4.23  63,240 AU  1 ft
x
1 AU
x  267,505 ft
x
x  50.7 miles
If the AU were 1 foot in diameter, the nearest star be 50.7 miles away.
3.
If an AU were shrunk to 1 foot, what would be the diameter of the Milky Way galaxy?
100,000 ly
x

1 AU
1 ft
x  1 AU  100,000 ly  1 ft
100,000 ly  1 ft
1 AU
100,000  63,240 AU  1 ft
x
1 AU
x  6.324  109 ft
x
x  1.20  10 6 miles
If the AU were 1 foot in diameter, the Milky Way Galaxy would be 1.20 million miles across.
4.
If an AU were shrunk to 1 foot, what would be the diameter of the Earth?
2  6,378 km  
x
1 AU
1 ft
x  1 AU  2  6,378 km   1 ft
2  6,378 km  1 ft
x
1 AU

2  6,378 km   1 ft
x
1.5  108 km
x  8.5  10 5 ft
x  .001 inches
If the AU were 1 foot in diameter, the Earth would be 0.001 inch across.
5.
If a light year were shrunk to 1 foot, what would be the distance to the nearest large galaxy, M31?
2.5  10 6 ly
x

1 ly
1 ft
x  1 ly  2.5  10 6 ly  1 ft
x
2.5  10 6 ly  1 ft
1 ly
x  2.5  10 6 ft
x  473 miles
If the light year were 1 foot in diameter, the nearest large galaxy, M31, would be 473 miles away.
6.
If the Milky Way galaxy were shrunk to 1 foot in diameter, what would be the diameter of the observable Universe?
27.4  109 ly
x

100,000 ly
1 ft
x  100,000 ly  27.4  109 ly  1 ft
x
27.4  109 ly  1 ft
100,000 ly
x  27.4  10 4 ft
x  51.9 miles
If the Milky Way Galaxy were 1 foot in diameter, the diameter of the visible universe would be 51.9 miles.