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4
16.6. Take the difference between the length of daylight on the last d = 1.928 × 10 km

date and the first date, and divide by the number of days. Th e t = 23 h 56 min 4 s

length of daylight is the difference between the sunset time and v = ?

the sunrise time. Converting to 24-hour time by adding d _
v =

12 hours to the p.m. time simplifies the calculation. t
Convert time to minutes.
sunrise (Nov 21) = 07:25 _ _
1 min
60 min

sunset (Nov 21) = 16:25 t = 23 h ( 1 h ) + 56 min + 4 s ( 60 s )
sunrise (Dec 21) = 07:55 _ 4 _ _
( )
min
min

sunset (Dec 21) = 16:20 = (23)(60) h + 56 min + s ( )
s
60
h
days (Dec 21 − Nov 21) = 30 days = 1,380 min + 56 min + 0.067 min
average daylight change = ? = 1.436067 × 10 min
3
4
__
1.928 × 10 km
Determine the length of each day. v =

3
1.436067 × 10 min
daylight Nov21 = 16:25 − 07:25 = 9 h 00 min = 1.343 × 10 _
1 km

daylight Dec21 = 16 :20 − 07:55 = 8 h 35 min min
Convert the hours of daylight to minutes. 16.9. Orbital velocity can be determined by using equation 2.1 in
_
(
)
min

daylight Nov21 = 9 h 60 + 00 m chapter 2. Add the orbital altitude of the satellite to Earth’s polar

radius from Figure 16.4 to calculate orbital circumference,
h
_
min
( )

= 9(60) h + 00 min which is the distance the satellite travels in each orbit.
h
First, calculate the orbital circumference:

= 540 min + 00 min 3
r Earth = 6.357 × 10 km
= 540 min
satellite altitude = 850 km
(
)
min
_

daylight Dec21 = 8 h 60 + 35 m orbital circumference = ?

h
min
_
( )

= 8(60) h + 35 min orbital circumference = 2π(r Earth + satellite altitude)
3
= 2π(6.357 × 10 km + 850 km)
h
3
= 480 min + 35 min = 2π(6.357 × 10 + 850) km
4
= 515 min = 4.53 × 10 km
4
d = 4.53 × 10 km d _
___ v =
daylight Dec21 − daylight Nov21

average daylight change = t = 0.99 h t
days 4
4.53 × 10 km
__ v = ? = __

515 min − 540 min

= 0.99 h

31 days
_
4 km
_ _ = 4.6 × 10

(515 − 540) min

= h

31 day
(−25) min
_ _

= 16.10. Because latitude lines are parallel, distance along Earth from

31 day
the equator can be calculated by multiplying Earth’s polar
_
min
= −0.81 diameter from Figure 16.4 by the fraction of the circumference
day
represented by the latitude.
16.7. Rotational velocity can be determined by using equation 2.1 in
)
_
latitude = 39.95°N latitude
(

chapter 2. Use the circumference for distance and the length of a d = πD polar 360°
D polar = 12,714 km
sidereal day converted to the appropriate units for time. 39.95°
d = ? = π12,714 km ( _ )

4
d = 3.072 × 10 km 360°
t = 23 h 56 min 4 s = 12,714π(0.1110) km
v = ? = 4,434 km
d _
v = 16.11. Distance along Earth at a given latitude can be calculated by
t
multiplying the circumference at the latitude by the fraction of
Convert time to hours.
the circumference represented by the longitude angle.
1 min
_ _ _ )
1 h
1 h

t = 23 h + 56 min ( ) + 4 s ( ) ( Circumference is determined from the radius.

60 min 60 min 60 s
longitude
56 _ h _ 4 _ h _ _ longitude = 1° _ )
min

= 23 h + min ( ) + s ( ) ( ) d = 2πradius 10° (

60 min 60 2 min s radius 10° = 6.28 × 10 km 360°
3
1°
= 23 h + 0.9333 h + 0.0011 h d = ? = 2π6.28 × 10 km _
( )
3

= 23.9344 h 360°
−3
3
4
__ = 2π6.28 × 10 (2.7778 × 10 ) km
3.072 × 10 km
v =

23.9344 h = 1.10 × 10 km
2
3 km
_
= 1.284 × 10 = 110 km

h
16.8. Rotational velocity can be determined by using equation 2.1 in
chapter 2. Use the circumference for distance and the length of
a sidereal day converted to the appropriate units for time.
E-33 APPENDIX E Solutions for Group A Parallel Exercises 675