On the grounds of the space center in Jackson (Michigan) is a scale model of the solar system. You can begin at a large sphere that represents the sun, mounted on a post stuck in the ground.
Nearby is a dot that is the planet Mercury. Then you walk, eventually coming to Earth. Then you walk and walk
some more. There is a lot of walking and very small objects. You get some sense of the space between the sun and the planets.
Suppose that you wish to build such a model in which the earth is represented by a sphere 4mm in diameter, the size of the ball E above.
Step 1. Establish a scale Make a ratio of the actual diameter of the Earth to the diameter of the model of the Earth (Be sure units are the same in the numerator and denominator before canceling them) SCALE = [1.2x10^{7}m] / [4 x 0.001m] = [3x10^{9}] : [1] |
The SCALE of this model is:
[3x10^{9}] : [1] The Scale factor is 3x10^{9} |
Notice that the scale factor is a number without any units.
It is a ratio between two lengths both in meters.
Now apply your result: Question (A): How large will the sun be in your model? Diameter of the sun in your model is the actual diameter of the sun, divided by the scale factor. Solution: Diameter of the sun in your model is: 1.5x10^{9}m / 3x10^{9} = 0.5m. |
In your JOB 1 model, the sun will be a balloon a half meter in diameter, a 20 inch balloon. |
Question (B): How far will the earth be from the sun in your model? Express the answer in meters, and in lengths of a football field (A football field is 100 yards long, about 100m). (Hint: Use the same scale factor as long as you are with Model #1) Distance from the sun to the Earth in your model is the actual distance, divided by the scale factor. Solution: Distance from the sun to the Earth in your model is: 1.5x10^{11}m / 3x10^{9} = 50m |
In your JOB 1 model, the sun-to-earth distance is 50 meters or, about one half the length of a football field. |
Question (C): What is the diameter of the solar system in your model? Half of this is the distance from the sun to the farthest planet, Neptune. (Pluto weaves in and out of Neptune's orbit) Express in meters and in miles. Diameter of the solar system in your model is the actual diameter of the solar system, divided by the scale factor. Solution: Diameter of the solar system in your model is 9.6x10^{12}m / 3x10^{9} = 3,200 meters |
In your JOB 1 model, the diameter of the solar system is 3,200m or 2 miles. (The distance from the sun to Neptune in your model is a one mile walk.) |
Question (D): What is the diameter of that farthest planet, Neptune, in your model? (The actual diameter of the planet Neptune is 4.5x10^{7} meters) Express the diameter of your model of Neptune in meters. Which ball (A-H) is closest to the size of your model of Neptune? Diameter of Neptune in your model is the actual diameter of Neptune, divided by the scale factor. Solution: Diameter of Neptune in your model is 4.5x10^{7}m / 3x10^{9} = 0.015m or 1.5cm |
In your JOB 1 model, the diameter of the planet Neptune is 1.5cm. (The size of Neptune in your model is about the size of ball G.) |
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Go to 'JOB 3' The Sun a grain of fine sand.
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Go to 'JOB 5' The Universe an exploding 4^{th} of July firecracker.
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