Astronomy 110: Stars, Galaxies, & Cosmology - Spring 2004

Solutions to Chapter 1 Problems (Universe, 6th edition, Freedman & Kaufmann)

7. 3600

8. Draw a line from your eye to one side of the Moon. Draw another line from your eye to the other side. The angle between those lines measures one-half of a degree.

11. a) 108 b) 4 x 104 c) 3 x 10-3 d) 2.2 x 1010 e) "May you live to be 1.2 x 102." -- Jewish proverb

12. An AU is the average distance between the Sun and the Earth, approximately 1.496 x 108 km. It is useful for measuring distances within the solar system.

14. A parsec is a unit of distance, equal to 3.26 light-years. It is defined as the distance at which 1 AU subtends an angle of 1" (that is, if you were 1 parsec away from an object with a linear size of 1 AU, it would appear to have an angular size of 1"). A kiloparsec is 103 parsecs, and a megaparsec is 106 parsecs.

15. a) kilometer b) centimeter c) second d) kilometers-per-second (a unit of speed) e) miles-per-hour (a unit of speed) f) meter g) meters-per-second (a unit of speed) h) hour i) year j) gram k) kilogram

16. Han Solo appears to believe that a parsec is a unit of time, rather than distance. His statement is equivalent to saying, "I drove from New York to Los Angeles in four miles!"

24. An object traveling at speed v for a time t will cover a distance d given by

d = v t
Solving this for the time t, we have
t = d / v
In this problem, the distance d is (1.496 x 108 km)(103 m/km) = 1.496 x 1011 m. The speed v is given as 3.00 x 108 m/s. Thus,
t = (1.496 x 1011 m) / (3.00 x 108 m/s)
t = 4.99 x 102 s

25. Radio signals travel at the speed of light, so v = (3.00 x 108 m/s)(10-3 km/m) = 3.00 x 105 km/s. The time of travel t is given as (4 h)(3.6 x 103 s/h) = 1.44 x 104 s. Thus,

d = v t
d = (3.00 x 105 km/s) (1.44 x 104 s)
d = 4.32 x 109 km

26. a) (3.22 pc)(3.09 x 1013 km/pc) = 9.95 x 1013 km
b) (3.22 pc)(3.26 ly/pc) = 10.5 ly. Therefore it takes 10.5 years for light from Epsilon Eridani to reach the Earth.

29. Solve the small-angle formula for the distance d:

d = 206,265 D / a
The linear size of the nickel D is given as (2.0 cm)(10-2 m/cm) = 0.02 m. Remember that the angular size a must be in arcseconds.
a)
d = (206,265) (0.02 m) / (3,600)
d = 1.1 m
b)
d = (206,265) (0.02 m) / (60)
d = 69 m
c)
d = (206,265) (0.02 m) / (1)
d = 4.1 x 103 m

31. The distance to the Moon d is given as 384,000 km, and the angular size a of the smallest features your telescope can make out is 2". Using the small-angle formula:

D = a d / 206,265
D = (2) (384,000 km) / (206,265)
D = 3.7 km

Last edited 25 Jan 04 M. A. Weinstein.