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A hora civil de Greenwich 10:30 se encuentran en una zona de calma, parados y sin arrancada, siendo su meridiano de situación= 119º 50,0´W, y tras calcular el horario en Greenwich del Sol (hGO) de ese instante= 337º 24,1´, calculan el intervalo de tiempo que tardará el Sol en alcanzar dicho meridiano fijo de encalmada:At Greenwich civil time 10:30, becalmed without way at longitude 119º 50.0' W, given the Sun's Greenwich hour angle (hGO) = 337º 24.1', calculate the time interval until the Sun reaches that fixed meridian:
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Correct answer: a) 09:29:43 h.
Why it is correct
The Sun's local hour angle is hL = 337º 24.1′ − 119º 50.0′ = 217º 34.1′. To reach the upper meridian (360º) it still has 360º − 217º 34.1′ = 142º 25.9′ (P East) to go, which at 15º per hour takes 09h 29m 43.6s.
Why the others are not
- b) Converts only the longitude (119º 50′ / 15 ≈ 7h 59m 20s), ignoring the Sun's hour angle.
- c) Converts hGO directly to time without subtracting longitude or finding the remaining angle.
- d) Misconverts the arcminutes (25.9′) into minutes and seconds of time.
How to work it out in the exam
- Time to meridian transit is the East meridian angle: P(E) = 360º − hL, divided by 15º/h.
Where this question comes from
- Región de Murcia · 2024 · 2ª convocatoria · CY Tipo 1 · question 14 Official paper · Official answer key
Pending instructor review. Machine translation, not reviewed.