Astronomy Assignment #1 Solutions
Text Problems:
Unit Number / Questions for Review / Problems1 / 4, 5, 7, 8, 9 / 11, 12, 13, 14
2 / 4, 5, 6, 7 / 10, 11, 14
3 / 9, 11, 12, 14
Problem Solutions: Unit 1
11. What would be the circumference and diameter of a ball that would represent the Moon if the Earth were a volleyball? What kind of object matches this size? (Note: See Problem #1 for the circumference of a volleyball.)
Use a proportion to solve questions like this where different objects appear at the same scale.
Required information
Circumference of a volleyball = 68 cm (Problem #1)
Circumference of the Earth = 2πR = 2π6,378 km = 40,074 km (See Appendix Table 5)
Circumference of the Moon = 2πR = 2π1,738 km = 10,920 km (See Appendix Table 6)
Solve the proportion
The scaled down Moon would be 18.5 cm in circumference equivalent to 5.90 cm in diameter. This diameter is about 2.3 inches, and this is a bit bigger than a golf ball, but smaller than a tennis ball. A regulation racquetball has a diameter of 2.25 inches and would be a good model for the Moon if the Earth were a volleyball.
12. If the Earth were a volley ball what would be the diameter of the Sun? What object matches
this size?
Use a proportion to solve questions like this where different objects appear at the same scale.
Required information
Circumference of a volleyball = 68 cm (Problem #1)
Circumference of the Earth = 2πR = 2π6,378 km = 40,074 km (See Appendix Table 5)
Circumference of the Sun = 2πR = 2π696,000 km = 4.37106 km (See App. Table 6)
Solve the proportion
The scaled down Sun would be 74.15 m in circumference equivalent to 23.6 m in diameter. This diameter is about 78 feet. A modest home has a length of about 80 feet. So if the Earth were a volleyball, the Sun would be the length of a modest sized home.
13. If the Earth were a volley ball how large would an astronomical unit be?
Use a proportion to solve questions like this where different objects appear at the same scale.
Required information
Circumference of a volleyball = 68 cm (Problem #1)
Circumference of the Earth = 2πR = 2π6,378 km = 40,074 km (See Appendix Table 5)
Length of and astronomical unit = 1 AU = 1.5 x 108km (See Appendix Table 1)
Solve the proportion
The scaled down Astronomical Unit would be 2.54 kilometers in length. This length is about 1½ miles. If the Earth were a volleyball then an AU would be about the distance from OCC to the Jreck Sub shop at Valley Drive and Seneca Turnpike in the Valley. Just for reference, on this scale you would be only 25 nanometers tall. A human hair is about 100,000 nanometers. (The moral of this problem is that an AU is really big, and it is the smallest of the special length units astronomers use.
14. During the 1960s and 1970s, The Apollo spacecraft took humans to the Moon in three days. Traveling to Mars requires a trip of around 2 AU in total. How long would this trip take, traveling at the same speed as to the Moon?
Use to solve problems like this.
Required information
Distance of the trip to Mars and back = 2 AU = 3.0 x 108km (See Appendix Table 1)
Speed of trip = Distance from Earth to Moon divided by 3 days
(See Appendix Table 6)
Solve the problem
The trip to Mars and back, at Apollo spacecraft speeds, would require over 6 years to complete. The moral of this problem is, again, that space is real big. A six year trip is required to reach one of the nearest planets and return. To reach the nearest stars and return would require a time 136,000 times longer, at this speed, or a total time of 816,000 years! Space is real big.
Problem Solutions: Unit 2
Scales of the UniverseTable 2.1
Object / Approximate Radius
Earth / R= 6,400 km
Moon’s Orbit / 380,000 km
Sun / R= 700,000 km
= 110 R
Earth’s Orbit / 1 AU = 150 Million km
= 210 R
= 23,100 R
Solar System to Pluto / 40 AU = 8500 R
Nearest Star / 4.2 ly = 270,000 AU
Milky Way Galaxy / 50,000 ly
Local Group / 1.5 Million ly
Local Super Cluster / 40 Million ly
Visible Universe / 13.2 Billion ly
2. If the Milky Way were the size of a nickel (about 2 centimeters).
- How big would the Local Group be?
- How big would the Local Supercluster be?
- How big would the visible Universe be?
Use a proportion to solve questions like this where different objects appear at the same scale.
Required information
Radius of a nickel = 2 cm
Radius of the Milky Way = 50,000 lyrs (See Table 2.1)
Radius of the Local Group = 1.5 million lyrs (See Table 2.1)
Solve the following proportionfor the diameter of the Local Group
The scaled down Local Group of Galaxies would be 60 cm in radius. Solving similar proportions for the Local Super Cluster and Visible Universe yields the following. The Local Super cluster would have a radius of 20 meters and the visible universe would have a radius of about 5.5 km. This last answer is worth discussing briefly. If the Milky Way galaxy were a nickel, then the distance to the edge of the visible universe would be from downtown Syracuse to Liverpool, approximately. One might well ask, what is beyond this visible limit? We don’t know, because the light from that great a distance has not yet reached us. We believe that the universe has a finite age of about 13.7 billion years. Thus, any light emitting object beyond that distance is un-“seeable “ because the light it emits has not yet traveled the distance between it and Earth. The light train, so to speak has not yet reached the Syracuse station. As the universe ages, its visible limit will grow.
2. If we detected radio signals of intelligent origin from the Andromeda Galaxy (M31) and immediately responded with a radio message of our own, how long would we have to wait for a reply?
This type of problem is particularly simple if we know the distance to the M31 in light years. Since light travels at the speed of light, the time to travel on e light year is just one year. Proportionally, the time for light to travel 100,000 light years is just 100,000 years.
Required information
Distance from the Milky Way to M31 = 0.8 Mpc (Mega parsecs) (See Appendix Table 11)
Convert 0.8 Mpc to light years
The one-way travel time for radios wave (radio waves are a form of light) from the Milky Way to the Andromeda Galaxy (M31) is about 2.6 million years. Thus, we would have to wait for 2.6 million years for our radio signal to arrive at M31 and then another 2.6 million years for there reply to reach us. Total communication travel time 5.2 million years – older that the fossil record of the human species. Guess what the moral of the problem is…you got it. Space is real big.
5. The Milky Way is moving toward the larger galaxy M31 at about 100 kilometers per second. M31 is about 2 million light years away. How long will it take before the Milky Way collides with M31 if it continues at this speed?
Use to solve problems like this.
Required information
Distance from the Milky Way to M31 = 2 million lys
Speed of approach = 100 km/sec = 1x105 m/s
Convert 2 million lyrs to meters
Solve the problem
We will collide with the Andromeda Galaxy in about 6 billion years.