Question 78879: Max is building a scale model of the solar system. In Max's model, the distance from Earth to the sun is 1.2 feet, and the distance from Venus to the sun is 0.86 feet.
If the approximate distance from Earth to the sun is 93 million miles, what is the approximate distance from Venus to the sun?
Answer by mathdoc314(58)   (Show Source):
You can put this solution on YOUR website! In a scale model, distances in the real world are scaled down.
For example in HO scale model trains the ratio is 3.5 millimeters in the model to 1 foot in the real world, or about 1:87 (although most HO scale model trains I have seen are not very accurate in their detail.)
Max has used a scale where 1.2 feet represents 93 million miles in the real world. The scale distance from Venus to the sun is 0.86 feet.
The equation you can use to solve this is:
Let R be the scale ratio of the model.
All scalings done in making the model should have the same ratio:
R mi:1 ft = 9.3 * 10^7 mi : 1.2 ft = V mi : 0.86 feet
It is helpful to solve for R first
R = 9.3 * 10^7 / 1.2 mi/ft
R = 7.75 * 10^7 mi/ft
This way R can be used to find V
R = V/0.86 ft
V = R * 0.86 ft
V = 7.75 * 10^7 * 0.86 mi
V = 6.665 * 10^7 mi
Venus is 66.65 million miles from the Sun
You can also do this without finding R
V/0.86 ft = 93 million mi/1.2 ft
V = 93 million mi * 0.86 / 1.2
V = 66.65 million mi
Venus is 66.65 million miles from the Sun
This is a little bit of a silly problem because it assumes Max has forgotten what V was, or gone away and left his model behind for you to investigate, and that this is easier than looking up the value of V directly, etc. But the idea can be valuable. For example in using a map if you are going to drive from Cleveland to Cincinnati and it is 4 inches on the map and the scale is 60 miles to an inch then you can find out it is 240 miles easily.
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