Solution Manual For
Investigating Oceanography 2026 Release by Keith Sverdrup, Raphael Kudela
Chapters 1-16
Chapter 1
Answers to Study Problems
1) Earth’s mean radius can be defined as the radius of a sphere having the same
volume as Earth. This is called the volumetric radius (R v ) and it is easily calculated
from the equatorial radius (Re) and the polar radius (R p ) using:
R v = cube root (Re2 ×R p )
Using the values given in figure 1.8 for the lengths of Earth’s equatorial and polar
radii, calculate Earth’s volumetric radius, or mean radius, in kilometers and miles.
Solution:
Re = 6378.1 km and R p = 6356.8 km
R v = cube root [(6378.1)2 ×6356.8]= 6371 km (3959 mi)
2) If it is 2:30 P.M. at your location when it is noon along the Prime Meridian, what is
your longitude?
Solution:
First calculate the difference in time = (12:00 – 14:30) = −(2 hours and 30 minutes)
Difference in longitude for 2 hours = 2 hr × 15o /hr = 30o
Difference in longitude for 30 minutes = 30 min × 15/min = 450 = 7 ×30 = 7.5
o o
Total difference in longitude is = 37.5°
Since the original time difference was negative, longitude is West.
Answer: 37.5° West
3) If it is 8:40 A.M. at your location when it is noon along the Prime Meridian, what is
, your longitude?
Solution:
First calculate the difference in time = (12:00 – 08:40) = +(3 hours and 20 minutes)
Difference in longitude for 3 hours = 3 hr × 15o /hr = 45o
Difference in longitude for 20 minutes = 20 min × 15/min = 300 = 5
o
Total difference in longitude is = 50°
Since the original time difference was positive, longitude is East.
Answer: 50° East
4) Use Table 1.3 to determine the volume of water held in the atmosphere. Use Figure
1.16 to determine how much water is removed from the atmosphere by
precipitation over the oceans and the land each year. Using these two estimates,
calculate how many times the water in the atmosphere is replaced in a year.
3 3
Solution: (380,000 km /yr) / (13,000 km )= 29 times / yr
5) The mean depth of the ocean can be calculated by knowing the average depth of
each individual ocean and the percent of the five ocean basins, and the percent of
the total world ocean area they cover. Use estimates of these values given in table
1.4 to verify the average depth of the world ocean in meters and feet.
Solution:
Pacific(3972 m × 50.1%) + Atlantic(3646 m × 23.3%) + Indian(3741 m × 19.8%) +
Southern(3270 m × 5.4%) + Arctic(1205 × 1.4%) = 3774 m or 12,381 ft
6) Determine the distance between two locations: 110°W, 38.5°N and 110°W, 45°N.
Express this distance in nautical miles and kilometers.
Solution:
Because the longitudes of the two points are the same, the distance between the
points is based on the change in latitude only. Students need to recall that 1° of
latitude equals 60 nautical miles and that 1° of latitude is divided into 60 minutes.
45.0° N - 38.5° N = 6.5° of latitude,
6.5o × 60 nm/o latitude = 390 nm = 722.3 km
1 km = .540 nm; therefore, 1 nm = 1.852 km
7) Use the volume of the oceans and Earth’s surface area to calculate the depth of a
hypothetical ocean covering the entire globe. Express this depth in meters and feet.
, Solution:
9 3 8 2
Volume of the oceans is ~ 1.335×10 km ; surface area of Earth is ~ 5.098×10 km .
Single ocean depth = volume/surface area
(~ 1.335×109 km3 )/(~ 5.098×108 km2 )= 2.619 km = 2619 m = 8593 ft
Chapter 2: Earth Structure and Plate
Tectonics
Answers to Study Problems
1) Estimate the thickness of oceanic lithosphere that is a) 5 million, b) 10 million, c) 20
million, and d) 50 million years old.
Solution:
Thickness (km) = 10 × sqrt(age in millions of years)
a) Thickness = 10 × sqrt(5) = 22.4 km
b) Thickness = 10 × sqrt(10) = 31.6 km
c) Thickness = 10 × sqrt(20) = 44.7 km
d) Thickness = 10 × sqrt(50) = 70.7 km
2) Given the results you obtained in the previous problem, how would you describe a
plot of oceanic lithosphere thickness as a function of age of the lithosphere?
Solution:
The plot is not linear. The thickness increases more rapidly at younger ages than at
older ages.
Near the Hawaiian Islands, the Pacific Plate is moving to the northwest at a speed of
about 7 cm (~2.76 in) per year. How far will the Island of Oahu move in 30 million
years? Give your answer in km and miles. Compare this with the distance between Los
Angeles, CA and New York, NY.
, Solution:
Distance km = 7 cm / yr × 30,000,000 yr × 1 km / 100,000 cm
= 2100 km
Distance mi = 2100 km × 1 mi / 1.609 km = 1305 mi
The distance between Los Angeles and New York is about 3944 km or 2451 mi. This is
roughly twice the distance that Oahu would move.
Chapter 3: The Seafloor and its
Sediments
Answers to Study Problems
1) Assume an accumulation rate of 0.8 cm per 1000 years for deep-ocean pelagic
sediment. How long would it take to accumulate 500 m of sediment?
Solution:
Time = amount accumulated / accumulation rate
= (500 m (100 cm/m)) / (0.8 cm/1000 yr) = 62,500,000 yr
2) Assume an accumulation rate of 30 cm per 1000 years for continental shelf
sediment. How long would it take to accumulate 500 m of sediment?
Solution: Time = amount accumulated / accumulation rate
= (500 m ×(100 cm/m)) / (30 cm/1000 yr) = 1,666,666 yr
3) If underwater cables are spaced 14 km apart on the sea floor, and if monitoring
equipment shows that they break in sequence from the shallowest to the deepest at
15-minute intervals, what can you determine about the event that caused the
breaks?
Solution:
Investigating Oceanography 2026 Release by Keith Sverdrup, Raphael Kudela
Chapters 1-16
Chapter 1
Answers to Study Problems
1) Earth’s mean radius can be defined as the radius of a sphere having the same
volume as Earth. This is called the volumetric radius (R v ) and it is easily calculated
from the equatorial radius (Re) and the polar radius (R p ) using:
R v = cube root (Re2 ×R p )
Using the values given in figure 1.8 for the lengths of Earth’s equatorial and polar
radii, calculate Earth’s volumetric radius, or mean radius, in kilometers and miles.
Solution:
Re = 6378.1 km and R p = 6356.8 km
R v = cube root [(6378.1)2 ×6356.8]= 6371 km (3959 mi)
2) If it is 2:30 P.M. at your location when it is noon along the Prime Meridian, what is
your longitude?
Solution:
First calculate the difference in time = (12:00 – 14:30) = −(2 hours and 30 minutes)
Difference in longitude for 2 hours = 2 hr × 15o /hr = 30o
Difference in longitude for 30 minutes = 30 min × 15/min = 450 = 7 ×30 = 7.5
o o
Total difference in longitude is = 37.5°
Since the original time difference was negative, longitude is West.
Answer: 37.5° West
3) If it is 8:40 A.M. at your location when it is noon along the Prime Meridian, what is
, your longitude?
Solution:
First calculate the difference in time = (12:00 – 08:40) = +(3 hours and 20 minutes)
Difference in longitude for 3 hours = 3 hr × 15o /hr = 45o
Difference in longitude for 20 minutes = 20 min × 15/min = 300 = 5
o
Total difference in longitude is = 50°
Since the original time difference was positive, longitude is East.
Answer: 50° East
4) Use Table 1.3 to determine the volume of water held in the atmosphere. Use Figure
1.16 to determine how much water is removed from the atmosphere by
precipitation over the oceans and the land each year. Using these two estimates,
calculate how many times the water in the atmosphere is replaced in a year.
3 3
Solution: (380,000 km /yr) / (13,000 km )= 29 times / yr
5) The mean depth of the ocean can be calculated by knowing the average depth of
each individual ocean and the percent of the five ocean basins, and the percent of
the total world ocean area they cover. Use estimates of these values given in table
1.4 to verify the average depth of the world ocean in meters and feet.
Solution:
Pacific(3972 m × 50.1%) + Atlantic(3646 m × 23.3%) + Indian(3741 m × 19.8%) +
Southern(3270 m × 5.4%) + Arctic(1205 × 1.4%) = 3774 m or 12,381 ft
6) Determine the distance between two locations: 110°W, 38.5°N and 110°W, 45°N.
Express this distance in nautical miles and kilometers.
Solution:
Because the longitudes of the two points are the same, the distance between the
points is based on the change in latitude only. Students need to recall that 1° of
latitude equals 60 nautical miles and that 1° of latitude is divided into 60 minutes.
45.0° N - 38.5° N = 6.5° of latitude,
6.5o × 60 nm/o latitude = 390 nm = 722.3 km
1 km = .540 nm; therefore, 1 nm = 1.852 km
7) Use the volume of the oceans and Earth’s surface area to calculate the depth of a
hypothetical ocean covering the entire globe. Express this depth in meters and feet.
, Solution:
9 3 8 2
Volume of the oceans is ~ 1.335×10 km ; surface area of Earth is ~ 5.098×10 km .
Single ocean depth = volume/surface area
(~ 1.335×109 km3 )/(~ 5.098×108 km2 )= 2.619 km = 2619 m = 8593 ft
Chapter 2: Earth Structure and Plate
Tectonics
Answers to Study Problems
1) Estimate the thickness of oceanic lithosphere that is a) 5 million, b) 10 million, c) 20
million, and d) 50 million years old.
Solution:
Thickness (km) = 10 × sqrt(age in millions of years)
a) Thickness = 10 × sqrt(5) = 22.4 km
b) Thickness = 10 × sqrt(10) = 31.6 km
c) Thickness = 10 × sqrt(20) = 44.7 km
d) Thickness = 10 × sqrt(50) = 70.7 km
2) Given the results you obtained in the previous problem, how would you describe a
plot of oceanic lithosphere thickness as a function of age of the lithosphere?
Solution:
The plot is not linear. The thickness increases more rapidly at younger ages than at
older ages.
Near the Hawaiian Islands, the Pacific Plate is moving to the northwest at a speed of
about 7 cm (~2.76 in) per year. How far will the Island of Oahu move in 30 million
years? Give your answer in km and miles. Compare this with the distance between Los
Angeles, CA and New York, NY.
, Solution:
Distance km = 7 cm / yr × 30,000,000 yr × 1 km / 100,000 cm
= 2100 km
Distance mi = 2100 km × 1 mi / 1.609 km = 1305 mi
The distance between Los Angeles and New York is about 3944 km or 2451 mi. This is
roughly twice the distance that Oahu would move.
Chapter 3: The Seafloor and its
Sediments
Answers to Study Problems
1) Assume an accumulation rate of 0.8 cm per 1000 years for deep-ocean pelagic
sediment. How long would it take to accumulate 500 m of sediment?
Solution:
Time = amount accumulated / accumulation rate
= (500 m (100 cm/m)) / (0.8 cm/1000 yr) = 62,500,000 yr
2) Assume an accumulation rate of 30 cm per 1000 years for continental shelf
sediment. How long would it take to accumulate 500 m of sediment?
Solution: Time = amount accumulated / accumulation rate
= (500 m ×(100 cm/m)) / (30 cm/1000 yr) = 1,666,666 yr
3) If underwater cables are spaced 14 km apart on the sea floor, and if monitoring
equipment shows that they break in sequence from the shallowest to the deepest at
15-minute intervals, what can you determine about the event that caused the
breaks?
Solution:
