An astronomical observatory located at longitude 64^\circ\text{E} records a cosmic ray pul… — JAMB UTME

An astronomical observatory located at longitude $64^\circ\text{E}$ records a cosmic ray pulse at $04:12\text{ PM}$ local solar time. Simultaneously, a ground monitoring station records the exact same event at $11:48\text{ AM}$ local solar time. What is the longitude of the ground monitoring station in degrees West ( $^\circ\text{W}$ )?

Cevap: 2 ��W

Cevap

The longitude of the ground monitoring station is

2^\circ\text{W}

The time difference between

04:12\text{ PM}

(

16:12

) and

11:48\text{ AM}

(

11:48

) is

4\text{ hours and } 24\text{ minutes}

, which corresponds to an angular separation of

66^\circ

(

4 \times 15^\circ + 24/4^\circ

). Because the time at the ground station is earlier, it lies to the west of

64^\circ\text{E}

. Subtracting

66^\circ

from

64^\circ\text{E}

crosses the

0^\circ

Prime Meridian to reach

2^\circ\text{W}

Adım Adım Çözüm

Calculate the difference in local solar time between the two stations.

Time difference =

16:12 - 11:48 = 4\text{ hours and } 24\text{ minutes}

Determining the time interval is the first step to finding angular separation.

Convert the total time difference into degrees of longitude.

(4 \times 15^\circ) + (24 \div 4^\circ) = 60^\circ + 6^\circ = 66^\circ

The Earth rotates

15^\circ

per hour (

1^\circ

every 4 minutes).

Determine the relative directional position of the ground station.

The ground station is

66^\circ

west of the observatory.

Earlier local solar time indicates a position further to the west.

Calculate the absolute longitude across the Prime Meridian.

66^\circ - 64^\circ\text{E} = 2^\circ\text{W}

Moving west from

64^\circ\text{E}

uses

64^\circ

to reach

0^\circ

, leaving

2^\circ

in the Western Hemisphere.

Anahtar Kavram

Calculating longitude position across hemispheres using local solar time differences.

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