Review of the Scale Problems
• Scale problems are problems of the style
that ask how big or small would something
be if all objects were scaled up or down in
size.
• We’ll use a simple proportion to solve these
types of scale problems
The working equation for all scale problems is
Model Size of Object 1 Model Size of Object 2

Actual Size of Object 1 Actual Size of Object 2
where one object is given in its scaled size and you need to find the scaled size of the
other object.
Practice Problem 1: If every object in the Solar System were scaled down so that the
Earth were shrunk to a diameter of 1 foot, how far from the Earth would the Sun be?
Solution: Let the diameter
of the Earth object 1 and
the distance from the Earth
to Sun be object 2.
Model Size of Object 1 Model Size of Object 2

Actual Size of Object 1 Actual Size of Object 2
1 foot
x

2  6,378 km 150 106 km
1 foot
x
 150 106 km  11,759 ft  2.23 miles
2  6,378 km


Answer in sentence: The Sun would be 11,759 feet or 2.23 miles from the Earth if
the Solar System were scaled down so that the Earth had a 1 foot diameter.
The working equation for all
scale problems is
Model Size of Object 1 Model Size of Object 2

Actual Size of Object 1 Actual Size of Object 2
where one object is given in its scaled size and you need to find the scaled size of the
other object.
Practice Problem 2: If every object in the Universe were scaled down so that the
Astronomical Unit were shrunk to a diameter of 1 inch, how far from the Earth
would the nearest star be? Note the distance to Proxima Centauri (the nearest star
beyond the Sun) is 4.3 ly.
Solution: Let the AU
be object 1 and the
distance to Proxima
Cenaturi be object 2.
Model Size of Object 1 Model Size of Object 2

Actual Size of Object 1 Actual Size of Object 2
1 inch
x

150 106 km 4.3 ly

1 inch
9.46 1012 km 
  271,186 in  4.28 miles
x
  4.3 ly 
6
150 10 km 
1 ly

Answer in sentence: The nearest star beyond the Sun, Proxima Centauri, would be
4.28 miles from the Sun if the Universe were scaled down so that the 1 AU was equal
to 1 inch.
The working equation for all
scale problems is
Model Size of Object 1 Model Size of Object 2

Actual Size of Object 1 Actual Size of Object 2
where one object is given in its scaled size and you need to find the scaled size of the
other object.
Practice Problem 3: If every object in the Universe were scaled down so that the
Solar System (80 AU actual diameter) were shrunk to a diameter of 1 meter, how far
from the Sun would the center of the Milky Way Galaxy be? Note the distance to
the center of the galaxy is about 28,000 ly.
Solution: Let the 80
AU be object 1 and
the distance to the
galactic center be
object 2.
Model Size of Object 1 Model Size of Object 2

Actual Size of Object 1 Actual Size of Object 2
1m
x

180 AU 28,000 ly
x

1m
63,240 AU 
  9.84 106 m  9,840 km
  28,000 ly 
180 AU 
1 ly

Answer in sentence: The center of the Milky Way would be 9,840 kilometers from
the Sun if the Universe were scaled down so that the Solar System was 1 meter in
diameter.
The working equation for all scale problems is
Model Size of Object 1 Model Size of Object 2

Actual Size of Object 1 Actual Size of Object 2
where one object is given in its scaled size and you need to find the scaled size of the
other object.
Practice Problem 1: If every object in the Solar System were scaled down so that the
Solar System (80 AU actual diameter) had the diameter of a dinner plate (10.5 inch
diameter), what would be the diameter of the Milky Way Galaxy 100,000 ly actual
diameter?
Model Size of Object 1 Model Size of Object 2
Actual Size of Object 1
Solution: Let the diameter
10.5 inches
x
of the Solar System be

1105 ly
object 1 and the diameter of 80 AU
the Milky Way Galaxy be
object 2.
x

Actual Size of Object 2
10.5 inches 
63,240 AU 
  8.30 108 in  13,100 miles
 1105 ly 
80 AU
1 ly


Answer in sentence: The Milky Way Galaxy would have a diameter of 13,100 miles if
the Solar System were scaled down so that it was the size of a dinner plate. This is
equivalent to 50% greater than the diameter of the Earth.