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15.14. The equation for orbital velocity was provided in exercise 15.3. latitude = (zenith − altitude) − 23.5°
Enter the values for orbital radius (distance) and time, and Rearrange to solve for altitude.
solve for velocity.
−(latitude + 23.5°) = −zenith + altitude
7
r P = 5.8 × 10 km _ zenith − (latitude + 23.5°) = altitude
2πr P
V =
t P = 0.24 yr t P ∴ altitude = zenith − (latitude + 23.5°)
V = ? = 90° − (39.95° + 23.5°)
Convert time to h.
= 90° − 63.45°
yr)
3 h _ = 26.55°
t P = 0.24 yr ( 8.76 × 10

3
t P = 2.1 × 10 h 16.4. The method for determining latitude in the Northern
_ Hemisphere on the summer solstice applies:
2πr P
V =
t P zenith = 90°
7
2π(5.8 × 10 km)
__
= latitude = 21.0° N

3
2.1 × 10 h
7 altitude = ?
__ _
2π(5.8 × 10 ) km
= latitude = (zenith − altitude) + 23.5°

2.1 × 10 3 h
8
_ _ Rearrange to solve for altitude.
3.6 × 10 km

=

2.1 × 10 3 h
−(latitude − 23.5°) = −zenith + altitude
_
5 km
= 1.7 × 10

h zenith − (latitude − 23.5°) = altitude
∴ altitude = zenith − (latitude − 23.5°)
9
15.15. r P = 3.00 × 10 km _ = 90° − (21.0° − 23.5°)
2πr P
V =
t P = 84.01 yr t P = 90° − (−2.0°)
V = ? Convert time to h. = 92.0°
yr)

3 h _ 16.5. Take the difference between the length of daylight on the last
t P = 84.01 yr ( 8.76 × 10 date and the first date and divide by the number of days. Th e

5
t P = 7.36 × 10 h length of daylight is the difference between the sunset time and

_ the sunrise time. Converting to 24-hour time by adding
2πr P
V =

t P 12 hours to the p.m. time simplifies the calculation.
9
2π(3.00 × 10 km)
__

= sunrise (Sep 1) = 06:35
5
7.36 × 10 h
9 sunset (Sep 30) = 19:55
__ _
2π(3.00 × 10 ) km

= sunrise (Sep 30) = 07:15

7.36 × 10 5 h
sunset (Sep 30) = 18:55
10
1.89 × 10 _
_ km

= days (Sep 30 − Sep 1) = 30 days

7.36 × 10 5 h
average daylight change = ?
_
4 km
= 2.57 × 10

h Determine the length of each day.
CHAPTER 16 daylight Sep1 = 19:55 − 06:35 = 13 h 20 min
daylight Sep30 = 18 :55 − 07:15 = 11 h 40 min

16.1. The altitude is measured relative to the southern horizon, so the
Convert the hours of daylight to minutes.
observer is in the Northern Hemisphere. Determine the
_
(
min
)

difference between the zenith and the altitude and add 23.5°. daylight Sep1 = 13 h 60 + 20 m

h
The zenith is 90° above the horizon:

_
( )
min

= 13(60) h + 20 min

zenith = 90° latitude = (zenith − altitude) + 23.5° h
altitude = 110.0° = (90° − 110.0°) + 23.5° = 780 min + 20 min
latitude = ? = (−10.0°) + 23.5° = 800 min
(
min
)
= 13.5°N _

daylight Sep30 = 11 h 60 + 40 m

h

16.2. The observer is in the Southern Hemisphere because the _
min
( )

altitude is being measured relative to the northern horizon. = 11(60) h + 40 min
h
zenith = 90° latitude = (zenith − altitude) − 23.5° = 660 min + 40 min
altitude = 43.5° = (90° − 43.5°) − 23.5° = 700 min
latitude = ? = (46.5°) − 23.5° __
daylight Sep30 − daylight Sep1

average daylight change =
= 23.0° S days
__
700 min − 800 min

16.3. The method for determining latitude in the Northern = 30 days

Hemisphere on the winter solstice applies: _ _
(700 − 800) min

= 30

zenith = 90° day
_ _
(−100) min

latitude = 39.95°N =

30 day
altitude = ?
min
= −3.3 _

day
674 APPENDIX E Solutions for Group A Parallel Exercises E-32