Fall 2004
Dr. Mike Fanelli
Solutions to Assigned Problems
Chapter 1
PROBLEM 1-1:
In one second light leaving Los Angeles reaches about how far ? Five
possibilities are listed.
ANSWER: Light travels at 300,000 kilometers per second, a large but not infinite
speed. Of the five choices, the correct one is (c) the Moon, since the moon’s
average distance from Earth is given as 384,000 km. Light (or radio waves)
takes about 1.3 seconds to reach the Moon.
PROBLEM 1-6:
Through how many degrees, arcminutes, and arcseconds does the Moon move
in (a) one hour, (b) one minute, (c) one second ? How long does it take the
Moon to move a distance equal to its own diameter on the sky ?
ANSWER: In this problem you are asked to think about angular motion — the
motion of an object across the sky with respect to the background of stars. The
Moon takes 27.3 days to complete 1 orbit of the Earth with respect to the stars, a
period known as a sidereal month. Through a telescope, the moon’s motion is
readily apparent, but to the eye, it requires careful observation over several hours
to notice the Moon’s changing position against the background of stars. This
problem exercises your understanding of (1) angular units such as degrees,
arcminutes, and arcseconds, (2) the conversion between one set of units into
another, and, (3) the use of ratios.
Note that the symbol for arcminutes is  and the symbol for arcseconds is .
(a) How far does the Moon move in one hour ?
 The moon completes one orbit in about a month, or more quantitatively,
moves through 360 degrees in 27.3 days. 360 divided by 27.3 = 13.19
degrees per day. Each day the moon will appear about 13 to the east of its
position at the same time on the previous day.
 Now convert to motion per hour. 13.19 degrees/day divided by 24 hrs/day =
0.55 degrees/hr. The moon passes through a little more than ½ of a degree in
one hour of time.
 Since 1 = 60 arcminutes, then 0.55 = 33 arcminutes (this is a ratio, 1:60 as
0.55:33). Therefore the moon moves through 33 arcminutes per hour of time.
 Since 1 arcminute = 60 arcseconds, then the moon moves through 33 times
60 arcseconds in a hour = 1978 per hour of time.
(b) How far does the Moon move across the sky in 1 minute ?
 From (a) we found that the Moon moves 0.55 degrees per hour. To find out
how far the Moon moves in a minutes, divide by 60 minutes per hour which =
0.0092 degrees per minute. The Moon moves through a small fraction of a
degree in one minute of time.
 As 1 = 60 arcminutes, then 0.0092 = 0.55 arcminutes. The moon moves
through 0.55 arcminutes per minute of time.
 Since 1 arcminute = 60 arcseconds, the moon moves through 33 arcseconds
per minute of time. This motion is easily visible through a small telescope.
(c) How far does the Moon move across the sky in 1 second ?
 From (b) we found that the Moon moves through 0.55 degrees per hour. To
find how far the Moon would move in 1 second, divide by 3600 seconds per
hour, which = 0.000153 degrees in a second of time.
 If the Moon moves 0.55 arcminute per minute of time (from part b) then the
moon will move 0.55 / 60 = 0.0092 arcminutes in one second of time.
 0.0092 arcminutes X 60 arcseconds per arcminute = 0.55 per second of
time.
(d) How long does it take the Moon to move a distance equal to its own
diameter ?
 The diameter of the Moon is about 30 arcminutes, which = 0.5 degree (given
in the notes).
 From part (b), we found that the Moon moves through 0.55 arcminutes per
minute of time
 The time in takes to move a specific distance = that distance divided by the
rate of motion. A familiar analogy is estimating travel time on a car trip. The
time it takes to travel 100 miles at 50 mph = distance (100 miles) divided by
the rate (50 mph), which = 2 hours. By analogy, the time it take the Moon to
move one lunar diameter = the distance traveled (30 arcminutes) divided by
the rate of motion (0.55 arcminutes per minute of time), which = 54.5 minutes.
The Moon moves its own diameter in just under an hour.
PROBLEM 1-13
The distance to the Sun is much greater than the distance to the Moon. Given
that the Sun & Moon have about the same angular diameter on the sky, we can
infer that the Sun must be physically larger than the Moon. Otherwise, the Sun
would appear smaller on the sky. Given the Earth-Moon distance and the EarthSun distance, you are asked to determine how much larger is the Sun compared
to the Moon.
ANSWER: There are two methods to solve this problem, the conceptual
approach and the mathematical approach. Thinking about the problem, you
might realize that since the angular diameters of the Sun and Moon are the
same, the ratio of their physical sizes will just be the ratios of their distances.
Why is this ? Simply because the angular size of an object decreases in direct
proportion to distance, as implied by the small angular formula. If the Sun and
Moon were at the same distance, lets say both located at the Moon’s distance
from Earth (yikes!), the Sun would be XXX times larger in the sky. The Sun’s
much greater distance causes it to appear with the same angular size as the
Moon on the sky. The ratio of distances = 150,000,000 km / 384,000 km = 391.
The mathematical approach is to use the small angle formula:
Sun’s diameter (km) = distance (km)  (0.5  57.3) [from MP 1-4]
= 150  106 km 
8.7  10-3 = 1.3  106 km
Sun’s diameter  Moon’s diameter = 1.31  106 km  3400 km [from the book]
= 385
These numbers are slightly different because of small inaccuracies in the quoted
values.