Solutions to End of the Unit Quantitative Problems

Solutions to End of the Unit Quantitative Problems

Part I: The Cosmic Landscape

Solutions Manual for Pathways to Astronomy, 6e Stephen Schneider (All Chapters) Unit 1

• Using the value of the

circumference of Earth given in the chapter, Circumference (Volleyball) / Circumference (Real Earth) = 68 cm / 40,000 km = 0.0017 cm/km

• km would be 0.0017 cm or 0.017 mm. This is about the diameter of extremely fine hair.

11.There are several ways to solve this problem. One easy way is to compare Earth’s radius or diameter to the Moon’s radius or diameter. The chapter notes the Moon’s diameter is about ¼ Earth’s. Since C = π × D, then the circumference is also ¼ Earth’s, so in the model where Earth is a volleyball, the Moon must have a circumference 0.25 × 68 cm = 17 cm. The Moon’s diameter is then C/π = 17 cm / π = 5.4 cm.A slightly more accurate calculation using the values of Earth and the Moon’s size from the Appendix (the Moon’s radius is 1738 km; Earth is 6378) gives the Moon’s radius is 0.273 times Earth’s; therefore the diameter and circumference are also 0.273 times Earth’s; this gives a circumference for the model of the Moon of 18.6 cm and a diameter of 5.9 cm.Either way, the model of the Moon would be a little smaller than a tennis ball or about the size of a small orange.

12.The volleyball is about 22 cm across (D = C/π = 68 cm /π = 21.6 cm). The diameter of the Sun is about 109 times that of Earth (R Sun /R Earth

= 696,000 km/6378 km = 109) so the model of the Sun would be 22 cm × 109 = 2398 cm = 23.98 m, or about 24 meters. This would be roughly the size of a house.(Also okay to use R Sun /R Earth

~ 100 as given in chapter, for 2200 cm).

13.An AU is approximately 1.5 × 10 8

km, while the Moon is d = 384,400 km away from Earth (Appendix Value). Performing a standard unit conversion gives a distance of d = (384,400 km) × (1 AU/1.5 × 10 8

km) = 0.0026 AU.

14.The Moon is 384,400 km away from Earth (Appendix value). A three-day (72-hour) flight gives an average speed of 384,400 km/3 day = 128,000 km/day (= 5339 km/hr). A trip to Mars 2

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on a trajectory 2 AU long would cover a distance 2 × 1.5×10 11 m = 3.0 × 10 11 m = 3.0 × 10 8 km.At the speed of an Apollo spacecraft, a trip to Mars would therefore take time = distance/speed = 3.0 × 10 8 km / (1.28 × 10 5 km/day) = 2.34 × 10 3 days This is about 6.4 years to reach Mars. (Note: if using the rounded value of 400,000 km distance to the Moon in the chapter, the speed is about 130,000 km/day, which yields about 6 yr.)

• Based on the speed calculated in the last problem, a 2 AU flight took 6.4 years. Pluto is 20

times farther away, so the flight would take 20 times longer: 20 × 6.4 yr = 128 yr.

Unit 2

• Model scale: Diameter(nickel) / Diameter(Milky Way) = 2 cm / 100,000 ly = 2 × 10

-5

cm/ly.

Using the sizes in table 2.1:

a) The Local Group is about 3 million ly in diameter, which in the model is:

3 × 10

6 ly × 2 × 10 -5 cm/ly = 60 cm.60 cm is about the size of a beach ball.

b) The Local Supercluster is 50 million ly × 2 = 100 million ly in diameter, scaled to the

model this is:

100 × 10

6 ly × 2 × 10 -5 cm/ly = 2000 cm = 20 m.20 m is about the size of a house.

c) The visible universe would be about 13.7 billion ly × 2 = 27.4 billion ly in diameter,

which in the model is:

27.4 × 10

9 ly × 2 × 10 -5 cm/ly = 5.5 × 10 5 cm = 5500 m = 5.5 km 5.5 km is about the size of a town or the height of Mt. McKinley.

• Radio waves, like all forms of electromagnetic radiation, travel at the speed of light. We’d

have to wait for the message to get to Andromeda, and for the response to make it back to us— twice the distance (2.5 million light years), divided by the speed of light. Because light travels

1 light year in a year, so:

time = distance / speed = (2 × 2.5 million ly) / (1 ly/yr) = 5 million years.There is an introduction to solving distance-velocity-time problems at the beginning of the appendix.

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• An AU is approximately 1.5 × 10

8 km, and the speed of light is about 3.0× 10 5 km/sec.Therefore, the time it takes light to travel from the Sun to Earth is time = distance / speed = 1.5 × 10 8 km / 3.0× 10 5 km/sec = 500 sec, which is about 8.3 minutes.

13. This is another distance-velocity-time problem: d = 1 ly = 9.46 × 10

15 m = 9.46 × 10 12 km; V = 1000 km/sec. Therefore, we find the time for the distance to increase by 1 light-year is t = d/V = 9.46 × 10 12 km / (1000 km/sec) = 9.46 × 10 9 sec.

Converting this to years:

9.46 × 10

9 sec × 1hr/(3600 sec) × 1 day / (24 hr) × 1 yr/(365.25 day) = 300 yr.

• This problem is like the last one, but with d = 2.5×10

6

ly and V = 100 km/sec:

t = (2.5×10 6 ly × 9.46 × 10 12 km/ly) / (100 km/sec) = 2.37 × 10 17 sec.Converting this to years gives

1.89 × 10

17 sec × 1hr/(3600 sec) × 1 day / (24 hr) × 1 yr/(365.25 day) = 7.5 × 10 9 yr.So the Milky Way and Andromeda will collide in 7.5 billion years, or sooner with acceleration.

• At a scale of Figure 2.4, 1 light year = 1 meter, so 12 billion light years = 12 billion meters:

12 × 10

9 m = 12 × 10 6 km = 12 million km. So on the scale where the Milky Way is about the size of a large metropolitan area, the most distant galaxies we can see are about 30 times the Moon’s distance [12×10 6 km/384,000 km = 31], or about a tenth of an AU distant [12×10 9 m

/(1.5×10

11 m/AU) = 0.8 AU].

Unit 3

• The ratio of the Sun’s radius to Earth’s radius is 7 × 10

5 km / (6.4 × 10 3 km) = 7/6.4 × 10 5−3

≈ 1 × 10 2 or about 100 times larger.

• From Table 3.3 a solar radius is 6.97 × 10

8 m, or in kilometers 6.97 × 10 5 km. The radius of a hypergiant star would be about 1500 × 6.97 × 10 5 km = 1.05 × 10 9 km.An astronomical unit is 1.50 × 10 8 km making the radius of the hypergiant star (1.05 × 10 9 km) × (1 AU/1.50 × 10 8 km) = 6.97 AU ≈ 7.0 AU, which is larger than Jupiter’s orbit!

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• The light travel time to Earth from the Sun can be calculated from the distance-speed-time

formula t = d/V, where V = c = 3.0 × 10 5 km sec -1 , and d = 1 AU = 1.496 × 10 8

km:

t = 1.496 × 10 8 km/ (3.0 × 10 5 km sec -1

) = (1.5/3.0) × 10

8-5 sec = 0.50 × 10 3

sec = 5.0 × 10 2 sec

Next, we convert this to minutes, also using scientific notation:

t = 5.0 × 10 1 sec × 1min / (6 × 10 1 sec) = 50/6 min = 8.3 min (about 8 1/2 min).

• Sun’s luminosity is 3.86 × 10

26

W = 3.86 × 10

26 J/sec. Divide the by the energy output of a

1 kiloton bomb to get the equivalent number of these explosions each second:

(3.86 × 10

26 J/sec)/(4.18 × 10 12 J/kt) = 9.23 × 10 13 kt per second.

• The Andromeda galaxy is 2.5 million light years away. The Sun is 25,000 light years (2.5 ×

10 4

= 0.025 × 10

6 = 0.025 million light years) from the center of the Milky Way galaxy. Since

2.5 × 10

6

– 0.025 × 10

6

= 2.475 × 10

6

≈ 2.5 × 10

6 , it would not affect the distance estimate to within the 2-digit precision of the distance of the Andromeda galaxy.

• The time for the galaxies to move this distance is given by the distance-speed-time equation:

t = d/V = (3 × 10 8 ly)/(6000 km/sec) = (3 × 10 8 ly × 9.46 × 10 12 km/ly)/(6000 km/sec)

= 4.73 × 10

17 sec = 4.73 × 10 17 sec × 1y/(3.16 × 10 7 sec) = 1.50 × 10 10 years.This time of 15 billion years assumes the galaxies have moved at a constant speed.

• To get the units to work out properly in the MKS system, the 3.10-gram mass of the penny

must be expressed in kilograms:

E = m c 2

= (3.10 × 10

‒3 kg) × (3.00 × 10 8 m/sec) 2

= 2.79 × 10

14 kg m 2 /sec 2

= 2.79 × 10

14 J Note: this means the conversion of a penny to pure energy is equivalent to a 66.7 kiloton bomb.

16. Using scientific notation:

(1.4 × 10

9 ) 3

/ (9.3 × 10

8 ) 2

= (1.4

3 × 10

9 × 3

) / (9.3

2 × 10

8 × 2

)

= 2.74 × 10

27

/ 86.5 × 10

16

= 2.74/86.5 × 10

27 -16

= 0.0317 × 10

11

= 3.2 × 10

9

• Paris: 5 ft(P) 2 in(P) = 5 × 12 + 2 = 62 in(P). Since the Parisian units are 6.63% larger,

1 Parisian inch = 1.0663 English inches. To convert this:

62 in(P) × 1.0663 in / 1 in(P) = 66 in = (5 × 12 + 6) in

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