Astronomy 115 Summary Questions
Lab #1 – Scale of the Universe
1. What is meant by “scale factor?”
A comparison of two values so that a larger value can be modeled on a smaller scale. Example: on a map, 1 inch =
20 miles.
2. If the distance between the Sun and Earth is 150,000,000 km and the Sun has a diameter of
about 1,400,000 km, determine a scale factor in terms of km/solar diameter, and then determine
the Earth-Sun distance in terms of solar diameters. Show all work and logic.
Scale Factor: 1.4 x 106 km/solar diameter
1.5 x 108 km/1.4 x 106 km/solar diameter = 107 solar diameters
3. Using your scale factor from #2 above, how many Suns will fit between the position of the
Sun and Jupiter? Show all work and logic.
Jupiter is 5 times the distance from the Sun to the Earth.
So, using the answer from number 2 above
5 x 107 solar diameters = 535 solar diameters
4. Why do astronomers measure distances in light-years?
The Universe is very large and a kilometer is very small in comparison to the distances between objects in space. So
as to not get entangled in extremely large numbers, astronomers use units of distance that represent a large number
of smaller units of distance. A light-year is almost 1000 trillion kilometers. Also, the light-year is based on a very
important concept in science, the speed of light.
Lab #1 – Scale of the Universe
1. What is meant by “scale factor?”
A comparison of two values so that a larger value can be modeled on a smaller scale. Example: on a map, 1 inch =
20 miles.
2. If the distance between the Sun and Earth is 150,000,000 km and the Sun has a diameter of
about 1,400,000 km, determine a scale factor in terms of km/solar diameter, and then determine
the Earth-Sun distance in terms of solar diameters. Show all work and logic.
Scale Factor: 1.4 x 106 km/solar diameter
1.5 x 108 km/1.4 x 106 km/solar diameter = 107 solar diameters
3. Using your scale factor from #2 above, how many Suns will fit between the position of the
Sun and Jupiter? Show all work and logic.
Jupiter is 5 times the distance from the Sun to the Earth.
So, using the answer from number 2 above
5 x 107 solar diameters = 535 solar diameters
4. Why do astronomers measure distances in light-years?
The Universe is very large and a kilometer is very small in comparison to the distances between objects in space. So
as to not get entangled in extremely large numbers, astronomers use units of distance that represent a large number
of smaller units of distance. A light-year is almost 1000 trillion kilometers. Also, the light-year is based on a very
important concept in science, the speed of light.
