
Activity #6
This is a series of activities which illustrate the principles behind hidden mass, rotation curves and the evidence for hidden mass.
Activity #6a: Modeling Mass in the Solar System and a Galaxy
In this activity, students will discover how mass is distributed in the solar system and a galaxy.
Materials:
1 10lb (4.54 kg) bag of kitty litter
Material such as sand or small pebbles
large paper
A balance capable of measuring a few grams
A balance capable of measuring a few kilograms.
The Student Worksheet gives the instructions for the activity.
Worksheet Answers
Part A
 To get the "kitty litter" equivalent masses of the planets, use ratios. For example, for Jupiter, use x/4536 = (1.90 x 10^{27})/(2 x 10^{30}). Here are the masses of the planets, and those masses as fractions of the Sun's mass:
Object 
Mass (kg) 
Kitty Litter Equiv. (g) 
Distance (km) 
Mass Inside Orbit (g) 
Sun 
2.00 x 10^{30} 
4536 
0 
4536 
Mercury 
3.30 x 10^{23} 
7.5 x 10^{4} 
57.9 x 10^{6} 
4536 
Venus 
4.87 x 10^{24} 
1.1 x 10^{2} 
108.2 x 10^{6} 
4536.01 
Earth 
5.97 x 10^{24} 
1.4 x 10^{2} 
149.6 x 10^{6} 
4540.02 
Mars 
7.35 x 10^{22} 
1.7 x 10^{4} 
227.9 x 10^{6} 
4540.02 
Jupiter 
1.90 x 10^{27} 
4.31 
778.4 x 10^{6} 
4540.34 
Saturn 
5.69 x 10^{26} 
1.29 
1427 x 10^{6} 
4541.63 
Uranus 
8.68 x 10^{25} 
0.20 
2871 x 10^{6} 
4541.83 
Neptune 
1.02 x 10^{26} 
0.23 
4498 x 10^{6} 
4542.06 
Pluto 
1.3 x 10^{22} 
2.9 x 10^{5} 
5906 x 10^{6} 
4542.06 
So if the Sun is represented by 4.55 kg of kitty litter, then Jupiter is 4.3 gms, Saturn is 1.3 gms, Uranus and Neptune are each about 0.2 gms, and all the terrestrial planets combined are about 0.025 gms. The "kitty litter" equivalents of the terrestrial planets individually are impossible to measure out; even their combined amount is most likely difficult. Computing these "kitty litter" equivalents provides the students a sense of how much more massive the Sun is.
 Note that in this exercise, the students are asked to imagine passing the orbits of the planets. The masses they are adding up is the mass within that distance to the sun. So it doesn't matter if the planet is actually on the opposite side of the Sun, just as long as its closer to the Sun than the distance being considered. Alternatively, you may ask students to imagine that the planets are lined up.
The mass within the distance between you and the Sun is the Sun's mass plus whatever planet orbits haven't been passed. For the most part, it's 4.536 kg. Details are in the chart above. Students should not record values less than one hundredth of a gram.
Part B
 You expect the matter in the galaxy to be where the light is. So a good deal of the mass should be in the central bulge, but a fair amount should be spread around in the spiral arms.
 Using the fact that the area of a pie segment is 0.5 R^{2} θ (where θ is in radians), advanced students might derive the appropriate radii for the portions.
 Now the students should note that the mass within their distance to the center will significantly decrease as they move toward the center.
 Distribute the new material uniformly throughout the model galaxy.
 Each third of the slice will be more massive. What the students should note is that the fraction by which the mass increases is larger for the outer thirds of the pie segment. This is because we distributed the "dark matter" uniformly, but the original visible matter is not. So the mass in the outer portion has more of an effect.
 Now the students will note that the mass does not change as much. If they could "see" the dark matter, the distribution of mass in the galaxy would look more uniform.
