The Earth as a Planet and Earth Movements Practice Questions — JAMB UTME

Question 1Question →

During Earth's annual revolution around the Sun, the subsolar point (the position where the Sun is directly overhead at noon) continuously changes its latitude between the Tropics. Starting from the March equinox when the Sun is directly overhead at the Equator moving northward, arrange the following solar positions in chronological sequence throughout the year.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence starting from the March equinox is: Sun directly overhead at the Equator moving northward �� Sun directly overhead at the Tropic of Cancer → Sun directly overhead at the Equator moving southward → Sun directly overhead at the Tropic of Capricorn.

Because Earth's axis is inclined at

23.5^\circ

to its orbital plane, the overhead position of the noon Sun shifts continuously throughout the year. Starting at the March Equinox (around March 21), the overhead Sun is at the Equator moving northward. It reaches its northernmost boundary at the Tropic of Cancer (

23.5^\circ\text{N}

) during the June Solstice (around June 21). It then returns southward, crossing the Equator at the September Equinox (around September 23), and finally reaches its southernmost boundary at the Tropic of Capricorn (

23.5^\circ\text{S}

) during the December Solstice (around December 22).

Step-by-Step Solution

1

Identify the initial position of the overhead Sun at the start of the specified cycle.

The cycle begins at the March Equinox (approx. March 21), when the Sun is overhead at the Equator (

0^\circ

) heading into the Northern Hemisphere.

The question prompt specifies starting at the March equinox with the subsolar point moving northward.

2

Determine the northernmost limit of the overhead Sun.

Three months later (approx. June 21), the Sun reaches its maximum northern latitude at the Tropic of Cancer (

23.5^\circ\text{N}

).

Earth's axial tilt of

23.5^\circ

defines the maximum latitude where the Sun can be directly overhead at noon.

3

Trace the overhead Sun's retreat back across the Equator.

Three months after the June solstice (approx. September 23), the Sun crosses the Equator again, moving southward.

This event is the Autumnal Equinox.

4

Identify the southernmost limit of the overhead Sun.

Three months after the September equinox (approx. December 22), the Sun reaches its maximum southern latitude at the Tropic of Capricorn (

23.5^\circ\text{S}

).

This event marks the December (Winter) Solstice before the Sun begins moving back north toward the Equator.

Key Concept

Seasonal migration of the subsolar point due to Earth's axial tilt and revolution

Question 2Question →

A live transmission of an event originates from London, located on the Greenwich Meridian ( $0^\circ$ ), at 3:00 p.m. GMT. If a listener in Town X tunes in to the live broadcast at 8:00 p.m. local time on the same day, what is the longitude of Town X?

75^\circ\text{E}

75^\circ\text{W}

45^\circ\text{E}

45^\circ\text{W}

Show answer & explanation

Answer: $75^\circ\text{E}$

Answer

75^\circ\text{E}

The time difference between Town X (

8:00\text{ p.m.}

) and Greenwich (

3:00\text{ p.m.}

) is 5 hours. Because the Earth rotates

15^\circ

every hour, a 5-hour difference equals

5 \times 15^\circ = 75^\circ

of longitude. Since Town X is ahead of Greenwich time, it lies in the Eastern Hemisphere, making the longitude

75^\circ\text{E}

.

Step-by-Step Solution

1

Calculate the time difference between Town X and London (

0^\circ

).

8:00\text{ p.m.} - 3:00\text{ p.m.} = 5\text{ hours}

Determining the total solar time separation between the two locations.

2

Convert the time difference into longitudinal degrees using the Earth's rotation rate (

15^\circ

per hour).

5\text{ hours} \times 15^\circ/\text{hour} = 75^\circ

The Earth completes a

360^\circ

rotation in 24 hours, which equals

15^\circ

per hour.

3

Determine the direction (East or West) based on whether local time is ahead or behind GMT.

Town X time (

8:00\text{ p.m.}

) is ahead of GMT (

3:00\text{ p.m.}

), so Town X is located to the East.

Places to the east of the Prime Meridian experience sunrise and local noon earlier than places to the west ('East gain, West lose').

Key Concept

Calculating Longitude from Local Time Difference and GMT

Question 3Question →

A radio station in Town A ( $10^\circ\text{E}$ ) broadcasts a live news program at 1:45 p.m. local standard time. A researcher at Station B hears the broadcast live when the local standard time at Station B is 8:05 a.m. on the same day. What is the longitudinal position of Station B?

75^\circ\text{W}

95^\circ\text{E}

85^\circ\text{W}

65^\circ\text{W}

Show answer & explanation

Answer: $75^\circ\text{W}$

Answer

Station B is located at

75^\circ\text{W}

.

Station B experiences a local time that is 5 hours and 40 minutes behind Town A. Because Earth rotates

15^\circ

per hour (

1^\circ

every 4 minutes), this time lag equals an angular distance of

85^\circ

to the west. Moving

85^\circ

west from

10^\circ\text{E}

crosses the Prime Meridian (

0^\circ

) after

10^\circ

, placing Station B at

75^\circ\text{W}

.

Step-by-Step Solution

1

Calculate the time difference between Town A and Station B.

1:45 p.m. (13:45) minus 8:05 a.m. (08:05) = 5 hours and 40 minutes.

Determining the absolute local time difference between the two locations establishes the angular separation.

2

Convert the total time difference into degrees of longitude.

5 hours =

5 \times 15^\circ = 75^\circ

; 40 minutes =

40 / 4 = 10^\circ

. Total angular distance =

75^\circ + 10^\circ = 85^\circ

.

Earth rotates

15^\circ

per hour, which equals

1^\circ

every 4 minutes.

3

Determine direction and final meridian.

Since Station B's time is earlier (8:05 a.m. vs 1:45 p.m.), it lies west of Town A (

10^\circ\text{E}

).

85^\circ\text{ West of } 10^\circ\text{E} = 85^\circ - 10^\circ = 75^\circ\text{W}

.

Traveling west moves backward in local time relative to eastern longitudes.

Key Concept

Longitude and Time Difference Calculations

Question 4Question →

Town X is located on longitude $15^\circ\text{E}$ . If the local time at the Greenwich Meridian ( $0^\circ$ ) is 12:00 noon, what is the local time at Town X?

1:00 pm

11:00 am

12:15 pm

2:00 pm

Show answer & explanation

Answer: 1:00 pm

Answer

The local time at Town X is 1:00 pm.

Earth rotates

360^\circ

in 24 hours, which means

15^\circ

equals 1 hour. Because Town X is located at

15^\circ\text{E}

, it is ahead of Greenwich (

0^\circ

) by 1 hour. Adding 1 hour to 12:00 noon gives 1:00 pm.

Step-by-Step Solution

1

Calculate the longitudinal difference between Greenwich Meridian (

0^\circ

) and Town X (

15^\circ\text{E}

).

15^\circ - 0^\circ = 15^\circ

.

The rate of Earth's rotation is

15^\circ

per hour (

360^\circ / 24\text{ hours} = 15^\circ/\text{hour}

).

2

Convert the longitudinal difference into time difference.

15^\circ \div 15^\circ/\text{hour} = 1\text{ hour}

.

Every

15^\circ

of longitude corresponds to a time difference of

1

hour.

3

Determine whether to add or subtract time based on direction relative to GMT.

12:00 noon + 1 hour = 1:00 pm.

Places located East of the Greenwich Meridian are ahead in time ('East gain, West lose').

Key Concept

Local time calculation using longitude differences and Earth rotation rate

Question 5Question →

City P is located at longitude $38^\circ\text{E}$ where the local time is 4:20 p.m. At the exact same moment, the local time at City Q is 9:40 a.m. on the same day. What is the longitude of City Q in degrees West of the Greenwich Meridian?

Show answer & explanation

Answer: 62

Answer

City Q is located at longitude

62^\circ\text{W}

.

The calculation demonstrates that a time difference of 6 hours and 40 minutes equals

100^\circ

of longitude (

400 \text{ minutes} \div 4 \text{ minutes/degree}

). Because City Q is earlier in time (9:40 a.m.) than City P (4:20 p.m.), City Q is located

100^\circ

to the west of

38^\circ\text{E}

. Subtracting

38^\circ

to reach

0^\circ

leaves

62^\circ

in the Western Hemisphere, giving a final position of

62^\circ\text{W}

.

Step-by-Step Solution

1

Find the local time difference between City P and City Q

Local time difference is 6 hours and 40 minutes (400 minutes)

Converting both times to 24-hour format (16:20 and 09:40) allows direct subtraction:

16:20 - 09:40 = 6\text{ h } 40\text{ min}

.

2

Convert time difference into angular longitude difference

Longitude difference is

100^\circ

Earth rotates

1^\circ

every 4 minutes. Dividing 400 minutes by 4 yields

100^\circ

of total longitude separation.

3

Determine the relative direction of City Q from City P

City Q is located

100^\circ

west of City P

Places with earlier local times lie to the west because the Earth rotates from west to east.

4

Calculate the final longitude position west of Greenwich (

0^\circ

)

Longitude of City Q is

62^\circ\text{W}

Moving

100^\circ

west from

38^\circ\text{E}

covers

38^\circ

to reach

0^\circ

, and the remaining

62^\circ

extends into the Western Hemisphere (

100^\circ - 38^\circ = 62^\circ

).

Key Concept

Calculation of longitude position using local time differences across the Prime Meridian

Question 6Question →

City A is located at longitude $25^\circ\text{W}$ and City B is located at longitude $65^\circ\text{E}$ . Calculate the difference in local solar time between the two cities in hours.

Show answer & explanation

Answer: 6

Answer

The time difference between City A and City B is 6 hours.

Because City A (

25^\circ\text{W}

) and City B (

65^\circ\text{E}

) lie in opposite hemispheres relative to the Greenwich Meridian (

0^\circ

), the total longitudinal difference between them is the sum of their absolute longitudes (

25^\circ + 65^\circ = 90^\circ

). Given that Earth rotates

15^\circ

of longitude every hour (

360^\circ / 24\text{ hours}

), dividing

90^\circ

by

15^\circ\text{/hour}

gives a total time difference of 6 hours.

Step-by-Step Solution

1

Calculate the total angular distance between the two longitudes.

25^\circ\text{W} + 65^\circ\text{E} = 90^\circ

Because the two locations are in different hemispheres (West and East), their longitudinal values must be added to find the total separation across the Prime Meridian.

2

Convert longitudinal degrees into time difference.

90^\circ / 15^\circ\text{ per hour} = 6\text{ hours}

Earth completes one full rotation of

360^\circ

in 24 hours, which corresponds to an angular velocity of

15^\circ

per hour.

Key Concept

Calculating time difference from longitudinal distance across hemispheres.

Question 7Question →

An aircraft departs from Town X, located at longitude $120^\circ\text{W}$ , at 4:00 a.m. local time on Monday. The non-stop flight to Town Y, located at longitude $75^\circ\text{E}$ , takes exactly 14 hours. What is the local time and day at Town Y when the aircraft lands?

5:00 a.m. on Monday

9:00 p.m. on Monday

7:00 a.m. on Tuesday

5:00 p.m. on Tuesday

Show answer & explanation

Answer: 7:00 a.m. on Tuesday

Answer

7:00 a.m. on Tuesday

The correct answer is 7:00 a.m. on Tuesday. Town X (

120^\circ\text{W}

) and Town Y (

75^\circ\text{E}

) are separated by

195^\circ

of longitude (

120^\circ + 75^\circ

). Dividing by

15^\circ

per hour yields a 13-hour time difference. Because Town Y is located east of Town X, local time at Town Y is 13 hours ahead of Town X. Thus, when the plane takes off at 4:00 a.m. Monday in Town X, the local time in Town Y is 5:00 p.m. Monday. Adding the 14-hour flight duration to 5:00 p.m. Monday results in 7:00 a.m. on Tuesday.

Step-by-Step Solution

1

Calculate total longitudinal difference between Town X and Town Y

Longitudinal difference =

120^\circ + 75^\circ = 195^\circ

Since Town X is in the Western Hemisphere (

120^\circ\text{W}

) and Town Y is in the Eastern Hemisphere (

75^\circ\text{E}

), their angular distances from the Prime Meridian (

0^\circ

) must be added together.

2

Convert the longitudinal difference into time difference

Time difference =

\frac{195^\circ}{15^\circ/\text{hour}} = 13\text{ hours}

The Earth rotates

360^\circ

in 24 hours, which corresponds to

15^\circ

per hour.

3

Determine the local time at Town Y at the exact moment of departure from Town X

Departure time at Town Y = 4:00 a.m. Monday + 13 hours = 5:00 p.m. (17:00) Monday

Places to the east gain time relative to places to the west ('East gain, West lose'). Town Y is east of Town X.

4

Add the flight duration to find the arrival time at Town Y

Arrival time at Town Y = 5:00 p.m. Monday + 14 hours = 7:00 a.m. Tuesday

Adding 14 hours to 5:00 p.m. (17:00) yields 31:00 hours. Subtracting 24 hours for a full day rollover gives 7:00 a.m. on the following day (Tuesday).

Key Concept

Calculating local time differences across Eastern and Western hemispheres combined with flight elapsed time and calendar day rollover.

Estimated Time:3m 0s

Question 8Question →

Due to the Earth's rotation from west to east, local solar time varies with longitude across the globe. Arrange the following longitudes in order of their local solar time at any given moment, from the EARLIEST time of day to the LATEST time of day.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from earliest to latest local time is

90^\circ\text{W}

,

30^\circ\text{W}

,

45^\circ\text{E}

, and

120^\circ\text{E}

.

Because the Earth rotates on its axis from West to East at a rate of

15^\circ

per hour, local solar time increases as you move eastwards. Therefore, the longitude located furthest west (

90^\circ\text{W}

) has the earliest time of day, followed by

30^\circ\text{W}

, then

45^\circ\text{E}

, and finally the longitude furthest east (

120^\circ\text{E}

) which has the latest time.

Step-by-Step Solution

1

Identify the relationship between Earth's rotation direction and local time.

Earth rotates from West to East, meaning eastern longitudes experience sunrise earlier and are ahead in time, while western longitudes are behind in time.

The sun appears to rise in the east and set in the west due to Earth's axial rotation.

2

Compare longitudes relative to the Greenwich Meridian (

0^\circ

).

Longitudes in the Western Hemisphere (

\text{W}

) are behind

0^\circ

, while longitudes in the Eastern Hemisphere (

\text{E}

) are ahead of

0^\circ

.

Moving west reduces local solar time (

1^\circ = 4\text{ minutes}

behind), whereas moving east increases local solar time (

1^\circ = 4\text{ minutes}

ahead).

3

Order the given longitudes from furthest west to furthest east.

90^\circ\text{W} \rightarrow 30^\circ\text{W} \rightarrow 45^\circ\text{E} \rightarrow 120^\circ\text{E}

.

This spatial progression from west to east directly corresponds to chronological progression from earliest hour of the day to latest hour.

Key Concept

Earth's West-to-East rotation causes places located further East to have a local time ahead of places located further West.

Question 9Question →

Arrange the following geographical locations in order of INCREASING daylight duration (from shortest day length to longest day length) during the June Solstice (June 21st).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of locations from shortest to longest daylight duration on June 21st is: Antarctic Circle (

66\frac{1}{2}^\circ\text{S}

), Tropic of Capricorn (

23\frac{1}{2}^\circ\text{S}

), Equator (

0^\circ

), Tropic of Cancer (

23\frac{1}{2}^\circ\text{N}

), and Arctic Circle (

66\frac{1}{2}^\circ\text{N}

).

On June 21st (the June Solstice), the North Pole is inclined toward the Sun. As a result, day length increases continuously from south to north across the globe. Locations south of the Antarctic Circle (

66\frac{1}{2}^\circ\text{S}

) experience 0 hours of sunlight (24 hours of darkness). Mid-latitude southern regions such as the Tropic of Capricorn (

23\frac{1}{2}^\circ\text{S}

) experience short winter days (~10.5 hours). The Equator (

0^\circ

) always maintains an equal 12-hour day and night. Mid-latitude northern regions such as the Tropic of Cancer (

23\frac{1}{2}^\circ\text{N}

) experience long summer days (~13.5 hours), and the region within the Arctic Circle (

66\frac{1}{2}^\circ\text{N}

) receives continuous 24-hour daylight. Arranging these from shortest to longest daylight duration gives: Antarctic Circle, Tropic of Capricorn, Equator, Tropic of Cancer, and Arctic Circle.

Step-by-Step Solution

1

Identify Earth's orientation relative to the Sun on June 21st (June Solstice).

The Northern Hemisphere is tilted towards the Sun at an angle of

23.5^\circ

, making the Tropic of Cancer (

23\frac{1}{2}^\circ\text{N}

) the subsolar point.

Earth's axial tilt causes daylight duration to increase progressively from the South Pole toward the North Pole during the June solstice.

2

Determine daylight hours at the polar circles.

The Antarctic Circle (

66\frac{1}{2}^\circ\text{S}

) receives 0 hours of daylight, while the Arctic Circle (

66\frac{1}{2}^\circ\text{N}

) receives 24 hours of daylight.

The entire area south of the Antarctic Circle is in Earth's shadow, whereas the area north of the Arctic Circle remains continuously illuminated.

3

Determine daylight hours at intermediate latitudes.

The Tropic of Capricorn (

23\frac{1}{2}^\circ\text{S}

) has ~10.5 hours of daylight, the Equator (

0^\circ

) has exactly 12 hours, and the Tropic of Cancer (

23\frac{1}{2}^\circ\text{N}

) has ~13.5 hours.

Day length increases smoothly along the latitudinal gradient from south to north.

4

Order the locations from shortest to longest daylight duration.

Antarctic Circle (

0\text{ hrs}

) < Tropic of Capricorn (

\sim 10.5\text{ hrs}

) < Equator (

12\text{ hrs}

) < Tropic of Cancer (

\sim 13.5\text{ hrs}

) < Arctic Circle (

24\text{ hrs}

).

This sequence correctly reflects increasing daylight hours.

Key Concept

Latitudinal Variation of Daylight Hours during Solstices

Question 10Question →

A town is located at longitude $45^\circ\text{E}$ . If the Greenwich Mean Time (GMT) is 8:00 a.m., what is the local time in the town?

5:00 a.m.

11:00 a.m.

11:00 p.m.

2:00 p.m.

Show answer & explanation

Answer: 11:00 a.m.

Answer

The local time in the town is 11:00 a.m.

Earth rotates from west to east, meaning places to the east experience solar noon earlier than places to the west. For every

15^\circ

of longitude moving eastward, time increases by 1 hour. Since the town is at

45^\circ\text{E}

, it is

45^\circ / 15^\circ = 3

hours ahead of Greenwich Mean Time (8:00 a.m.), making the local time 11:00 a.m.

Step-by-Step Solution

1

Calculate the longitudinal difference between Greenwich Meridian (

0^\circ

) and the town (

45^\circ\text{E}

).

Longitudinal difference =

45^\circ - 0^\circ = 45^\circ

.

The longitudinal difference determines the total angular distance between the two locations.

2

Convert the longitudinal difference into time using the rate of

15^\circ = 1\text{ hour}

(or

1^\circ = 4\text{ minutes}

).

Time difference =

45^\circ \times 4\text{ minutes} = 180\text{ minutes} = 3\text{ hours}

.

Earth rotates

360^\circ

in 24 hours, which corresponds to

15^\circ

per hour.

3

Apply the rule 'East gain, West lose' relative to Greenwich Mean Time.

8:00 a.m. + 3 hours = 11:00 a.m.

Since the destination is east of the Greenwich Meridian, local time is ahead of GMT.

Key Concept

Longitude and Local Time Calculation

Question 11Question →

Arrange the following inner (terrestrial) planets of the solar system in order of INCREASING distance from the Sun (starting with the planet closest to the Sun).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Mercury → Venus → Earth → Mars

In order of increasing distance from the Sun, the terrestrial planets are Mercury (1st), Venus (2nd), Earth (3rd), and Mars (4th).

Step-by-Step Solution

1

Identify the terrestrial (inner) planets in the solar system

The four inner terrestrial planets are Mercury, Venus, Earth, and Mars.

These four planets lie within the asteroid belt and are composed primarily of rock and metal.

2

Recall their average distances from the Sun

Mercury lies at

0.39\text{ AU}

, Venus at

0.72\text{ AU}

, Earth at

1.00\text{ AU}

, and Mars at

1.52\text{ AU}

.

Distance from the Sun increases systematically outward along their orbits.

3

Arrange the planets from closest to farthest

The correct sequence is Mercury → Venus → Earth → Mars.

This places the planets in ascending order of their orbital radius from the Sun.

Key Concept

Planetary Sequence and Relative Distance from the Sun

Question 12Question →

A scientific research vessel positioned at longitude $48.5^\circ\text{W}$ records its local solar time as 3:34 p.m. At that exact instant, a terrestrial tracking station records its local solar time as 9:10 a.m. on the same day. What is the longitude of the tracking station in degrees West?

Show answer & explanation

Answer: 144.5

Answer

The longitude of the tracking station is

144.5^\circ\text{W}

.

The difference in local solar time between 3:34 p.m. (15:34) and 9:10 a.m. (09:10) is 6 hours and 24 minutes, or 384 minutes. Dividing 384 minutes by 4 minutes per degree gives a longitude difference of

96^\circ

. Since the tracking station is behind in time relative to the vessel, it must be located west of

48.5^\circ\text{W}

. Adding

96^\circ

westward yields a final longitude of

144.5^\circ\text{W}

.

Step-by-Step Solution

1

Calculate the total time difference between the vessel and the tracking station.

Time difference = 15:34 - 09:10 = 6 hours and 24 minutes = 384 minutes.

Determining the net difference in local time is required to find the angular displacement.

2

Convert time difference in minutes to degrees of longitude.

Longitudinal distance = 384 minutes / 4 minutes per degree = 96 degrees.

Earth rotates

360^\circ

in 24 hours, which corresponds to

1^\circ

of longitude every 4 minutes.

3

Apply the longitudinal directional rule ('West lose, East gain') to find the station's location.

Station longitude = 48.5 degrees W + 96 degrees = 144.5 degrees W.

Since the tracking station's local time is behind the vessel's local time, the station must be positioned further west of the vessel.

Key Concept

Longitude calculation from local solar time differences and Earth's axial rotation rate

Question 13Question →

The Earth completes a full rotation of $360^\circ$ on its axis every 24 hours. How many degrees of longitude does the Earth rotate through in 3 hours?

Show answer & explanation

Answer: 45

Answer

45 degrees

Because the Earth completes one full

360^\circ

rotation in 24 hours, its speed of rotation is

15^\circ

per hour (

360^\circ \div 24

). In a time span of 3 hours, it turns through

15^\circ \times 3 = 45^\circ

of longitude.

Step-by-Step Solution

1

Calculate the rotation rate of the Earth per hour.

15 degrees per hour

The Earth turns 360 degrees in 24 hours, so 360 divided by 24 equals 15 degrees per hour.

2

Calculate the total longitudinal rotation in 3 hours.

45 degrees

Multiplying the rate of 15 degrees per hour by 3 hours gives 45 degrees.

Key Concept

Earth's Rotational Angular Speed and Longitude

Estimated Time:45s

Question 14Question →

An ocean liner positioned at longitude $15^\circ\text{W}$ records its local solar time as 1:00 p.m. At the exact same moment, a cargo ship located at a different meridian records its local solar time as 5:00 p.m. What is the longitudinal position of the cargo ship?

45^\circ\text{E}

75^\circ\text{W}

60^\circ\text{E}

45^\circ\text{W}

Show answer & explanation

Answer: $45^\circ\text{E}$

Answer

The longitude of the cargo ship is

45^\circ\text{E}

.

Because the cargo ship's local time is 4 hours ahead of the ocean liner's time (5:00 p.m. vs 1:00 p.m.), it must be located east of the liner. Since Earth rotates

15^\circ

per hour, a 4-hour difference corresponds to an angular distance of

60^\circ

. Measuring

60^\circ

east from

15^\circ\text{W}

involves traveling

15^\circ

east to the Prime Meridian (

0^\circ

) and then another

45^\circ

east into the Eastern Hemisphere, placing the cargo ship at

45^\circ\text{E}

.

Step-by-Step Solution

1

Calculate the time difference between the two vessels.

Time difference =

\text{5:00 p.m.} - \text{1:00 p.m.} = 4\text{ hours}

.

Determining the time interval is necessary to find the total angular distance.

2

Convert the time difference into degrees of longitude.

Angular distance =

4\text{ hours} \times 15^\circ/\text{hour} = 60^\circ

.

Earth rotates

360^\circ

in 24 hours, which equals

15^\circ

per hour.

3

Determine the direction of movement and calculate the target longitude.

Since 5:00 p.m. is later than 1:00 p.m., the cargo ship is to the East. Starting at

15^\circ\text{W}

, moving

15^\circ

East reaches

0^\circ

(Greenwich Meridian), and moving the remaining

45^\circ

(

60^\circ - 15^\circ

) East reaches

45^\circ\text{E}

.

Locations with later local times are located further East.

Key Concept

Longitude and Local Time Calculations across Meridians

Estimated Time:1m 30s

Question 15Question →

A weather monitoring station located at longitude $35^\circ\text{E}$ records local solar noon ( $12:00\text{ p.m.}$ ) at the exact instant a UTC master clock displays $09:40\text{ a.m.}$ at the Greenwich Meridian ( $0^\circ$ ). At that identical moment, a remote research station records its local solar time as $06:20\text{ a.m.}$ on the same day. What is the longitudinal position of the remote research station?

50^\circ\text{W}

50^\circ\text{E}

85^\circ\text{W}

15^\circ\text{E}

Show answer & explanation

Answer: $50^\circ\text{W}$

Answer

50^\circ\text{W}

The remote research station's local solar time (

06:20\text{ a.m.}

) is 3 hours and 20 minutes (

200\text{ minutes}

) behind Greenwich Mean Time (

09:40\text{ a.m.}

). Since

1^\circ

of longitude equals 4 minutes of time, dividing 200 by 4 yields an angular difference of

50^\circ

. Because local time is behind Greenwich time, the station lies in the Western Hemisphere at

50^\circ\text{W}

.

Step-by-Step Solution

1

Determine the time difference between Greenwich (

0^\circ

) and the remote research station.

Greenwich Time =

09:40\text{ a.m.}

, Remote Station Time =

06:20\text{ a.m.}

. Time difference =

09:40 - 06:20 = 3\text{ hours } 20\text{ minutes} = 200\text{ minutes}

.

Longitude calculations must be referenced against Greenwich Mean Time (

0^\circ

Meridian) to determine absolute longitudinal position.

2

Convert the time difference into longitudinal degrees using the Earth's rate of rotation (

1^\circ = 4\text{ minutes}

).

\frac{200\text{ minutes}}{4\text{ minutes per degree}} = 50^\circ

.

The Earth rotates

360^\circ

in 24 hours, which corresponds to

15^\circ

per hour or

1^\circ

every 4 minutes.

3

Determine the longitudinal hemisphere (East or West).

Since

06:20\text{ a.m.}

is behind

09:40\text{ a.m.}

, the remote station is West of Greenwich. Therefore, the position is

50^\circ\text{W}

.

Places to the west of a given meridian experience local time earlier in the clock cycle (behind GMT), following the principle 'East gain, West lose'.

Key Concept

Longitude and Local Time Calculation across Meridians

Estimated Time:2m 0s

Question 16Question →

A live international cultural event is broadcast from a venue at longitude $15^\circ\text{E}$ starting at 4:00 p.m. local time. What is the local solar time for a viewer watching the live broadcast at longitude $45^\circ\text{W}$ ?

8:00 p.m.

12:00 p.m. (noon)

2:00 p.m.

6:00 p.m.

Show answer & explanation

Answer: 12:00 p.m. (noon)

Answer

12:00 p.m. (noon)

To find local solar time at

45^\circ\text{W}

relative to

15^\circ\text{E}

, calculate the total angular separation (

15^\circ + 45^\circ = 60^\circ

). Dividing by

15^\circ

per hour yields a 4-hour difference. Because

45^\circ\text{W}

lies to the west of

15^\circ\text{E}

, subtract 4 hours from 4:00 p.m., resulting in 12:00 p.m. (noon).

Step-by-Step Solution

1

Calculate total longitudinal difference between the two locations.

Since

15^\circ\text{E}

and

45^\circ\text{W}

are in opposite hemispheres relative to the Greenwich Meridian, add the values:

15^\circ + 45^\circ = 60^\circ

.

Longitudes in opposite eastern and western hemispheres must be summed to find total angular distance.

2

Convert longitudinal difference into time difference using the rate of Earth rotation (

15^\circ = 1\text{ hour}

).

\frac{60^\circ}{15^\circ/\text{hr}} = 4\text{ hours}

.

The Earth rotates

360^\circ

in 24 hours, which corresponds to

15^\circ

per hour.

3

Determine local time by applying the directional rule (East gain, West lose).

4:00\text{ p.m.} - 4\text{ hours} = 12:00\text{ p.m. (noon)}

.

The target location (

45^\circ\text{W}

) is west of the broadcasting venue (

15^\circ\text{E}

), so the time difference must be subtracted.

Key Concept

Longitude, Earth Rotation, and Local Time Calculation

Question 17Question →

A meteorological vessel measures its local solar time to be $3:20\text{ p.m.}$ when Greenwich Mean Time ( $0^\circ$ ) is $11:00\text{ a.m.}$ on the same day. What is the longitude of the vessel in degrees East?

Show answer & explanation

Answer: 65

Answer

The longitude of the vessel is

65^\circ\text{E}

.

The time difference between

3:20\text{ p.m.}

and

11:00\text{ a.m.}

is 4 hours and 20 minutes (

260\text{ minutes}

). Since the Earth rotates

1^\circ

every 4 minutes,

260\text{ minutes} \div 4 = 65^\circ

. Because the vessel's local solar time is ahead of Greenwich Mean Time, the vessel is located east of the Prime Meridian, yielding

65^\circ\text{E}

.

Step-by-Step Solution

1

Calculate time difference

4 hours 20 minutes (260 minutes)

Subtract GMT (11:00 a.m.) from local solar time (3:20 p.m.).

2

Convert time difference to longitude degrees

65 degrees

Divide total time difference in minutes (260) by 4 minutes per degree of Earth's rotation.

3

Determine longitudinal hemisphere

East

Local time is ahead of GMT, which indicates a position to the east of the Prime Meridian.

Key Concept

Calculation of longitude from local time and Greenwich Mean Time (GMT) using Earth's rotation rate

Question 18Question →

Earth's revolution around the Sun and its inclined axis cause the apparent movement of the subsolar point throughout the year. Arrange the following seasonal positions of the subsolar point (where the Sun is directly overhead at solar noon) in chronological sequence over a solar year, starting with the March Equinox.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence starting from the March Equinox is: (1) Sun overhead at the Equator moving north, (2) Sun overhead at the Tropic of Cancer, (3) Sun overhead at the Equator moving south, and (4) Sun overhead at the Tropic of Capricorn.

The sequence follows the natural annual progression of the overhead Sun (subsolar point). Starting at the March equinox (

0^\circ

heading north), Earth's revolution brings the Sun overhead at the Tropic of Cancer (

23.5^\circ\text{N}

) in June, back over the Equator (

0^\circ

heading south) in September, and finally overhead at the Tropic of Capricorn (

23.5^\circ\text{S}

) in December.

Step-by-Step Solution

1

Identify the initial benchmark event.

Around March 21 (Vernal Equinox), the Earth's axial tilt places the subsolar point directly over the Equator (

0^\circ

) as it migrates northward.

This is specified as the starting point for the annual cycle sequence.

2

Determine the next solstice event following three months of revolution.

Around June 21 (June Solstice), the subsolar point reaches its northernmost limit at the Tropic of Cancer (

23.5^\circ\text{N}

).

Earth's orbit advances 90 degrees, maximizing Northern Hemisphere solar inclination.

3

Trace the subsolar point as Earth continues its revolution toward the next equinox.

Around September 23 (Autumnal Equinox), the subsolar point moves back to cross the Equator (

0^\circ

) heading southward.

After the June solstice, the apparent position of the overhead sun recedes southward.

4

Identify the final solstice position completing the cycle.

Around December 21 (December Solstice), the subsolar point reaches its southernmost limit at the Tropic of Capricorn (

23.5^\circ\text{S}

).

Earth reaches the opposite point in its elliptical orbit where the Southern Hemisphere tilts most toward the Sun.

Key Concept

Apparent annual movement of the subsolar point due to Earth's axial tilt (

23.5^\circ

) and revolution around the Sun.

Question 19Question →

A radio station located at longitude $10^\circ\text{W}$ begins a live news broadcast at $2:15\text{ p.m.}$ local solar time. What is the local solar time at a receiving station located at longitude $35^\circ\text{E}$ when the broadcast is heard?

11:15\text{ a.m.}

3:55\text{ p.m.}

5:15\text{ p.m.}

12:35\text{ p.m.}

Show answer & explanation

Answer: $5:15\text{ p.m.}$

Answer

5:15\text{ p.m.}

The correct response of

5:15\text{ p.m.}

is derived by calculating the total longitudinal distance of

45^\circ

(

10^\circ\text{W} + 35^\circ\text{E}

), converting it to a 3-hour difference (

45^\circ / 15^\circ = 3\text{ hours}

), and adding 3 hours to the transmitting local time (

2:15\text{ p.m.}

) because the destination lies to the east.

Step-by-Step Solution

1

Calculate the total angular distance between the transmitting and receiving longitudes.

10^\circ\text{W} + 35^\circ\text{E} = 45^\circ

Since the two positions are in opposite hemispheres (West and East), their longitudinal values must be added together.

2

Convert the total angular distance into a time difference.

\frac{45^\circ}{15^\circ\text{ per hour}} = 3\text{ hours}

The Earth rotates

360^\circ

in 24 hours, which equals

15^\circ

of longitude per hour.

3

Adjust the initial local solar time according to relative direction.

2:15\text{ p.m.} + 3\text{ hours} = 5:15\text{ p.m.}

Because Earth rotates from west to east, locations situated further east experience solar time ahead of locations to the west.

Key Concept

Longitude and Local Solar Time Calculation

Estimated Time:1m 30s

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